Strenght Of Materials(es-64)

(Mindanao State University-General Santos City, Philippines) BSEE-(G.R.F) students ‘04 Solved Problems in Mechanics of Materials ES-64

1. 12-7. Determine the equations of the elastic curve using the x1 and x2 coordinates. Specify the slope at A and the maximum deflection. EI is constant. 1. 12-10.The beam is made of two rods and is subjected to the concentrated load P. Determine the slope at C. The moments of inertia of the rods are IAB and IAC, and the modulus of elasticity is E. 2. 12-13. Determine the elastic curve for the cantilevered beam, which is subjected to the couple moment Mo. Also compute the maximum slope and maximum deflection of the beam. EI is constant. 3. 12-15. Determine the deflection at the center of the beam and the slope at B. EI is constant. 4. 12-16. Determine the elastic curve for the simply supported beam, which is subjected to the couple moments Mo. Also, compute the maximum slope and the maximum deflection of the beam. EI is constant. 5. 12-19. Determine the equations of the elastic curve using the coordinates x1 and x2, and specify the slope at A. EI is constant. 6. 12-22.The floor beam of the airplane is subjected to the loading shown. Assuming that the fuselage exerts only vertical reactions on the ends of the beam, determine the maximum deflection of the beam. EI is constant. 7. 12-27. Determine the elastic curve for the simply supported beam using the x coordinates 0 ≤ x ≤ L/2. Also, determine the slope at A and the maximum deflection of the beam. EI is constant. 8. 12-34.The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 9. 12-35.The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft, and at C by a thrust bearing that exerts horizontal and vertical reactions on the shaft. Determine the equation of the elastic curve. EI is constant.

10. 12-37.The shafts support the two pulley loads shown. Determine the equation of the elastic curve. The bearing at A and B exerts only vertical reactions on the shaft. EI is constant. 11. 12-38.The shafts support the two pulley loads shown. Determine the equation of the elastic curve. The bearing at A and B exerts only vertical reactions on the shaft. EI is constant. 12. 12-43.The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 13. 12-54.The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 14. 12-55. Determine the slope and deflection at B. EI is constant. 15. 12-61. Determine the maximum slope and the maximum deflection of the beam. EI is constant. 16. 12-40.The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 17. 12-45.The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 18. 12-50. Determine the equation of the elastic curve. Specify the slope at A. EI is constant. 19. 12-25.The beam is subjected to the linearly varying distributed load. Determine the maximum slope of the beam. EI is constant. 20. 12-42.The beam is subjected to the load shown. Determine the equation of the slope and elastic curve. EI is constant. 21. 12-45. The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 22. 12-46.The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. EW=1.6(103) ksi. 23. 12-63. Determine the deflection and slope at C. EI is constant. 24. 12-66. Determine the deflection at C and the slope of the beam at A, B, and C. EI is constant. 25. 12-73. Determine the slope at B and deflection at C. EI is constant. 26. 12-86.The beam is subjected to the load shown. Determine the slope at B and deflection at C. EI is constant.

1. 5-5.The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and the three torques are applied to it as shown., determine the shear stress developed at points A and

B. These points lie on the pipe’s outer surface. Sketch the shear stress on the volume elements located at A and B. 2. 5-9. The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller pipe has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is tightly secured to the wall at C, determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench. 3. 5-10.The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear stress distribution over the cross section. 4. 5-2.The solid shaft of radius r is subjected to a torque T. Determine the radius r of the inner cone of the shaft that resists one-half of the applied torque (T/2). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear stress distribution. 5. 5-3.The solid shaft of radius r is subjected to a torque T. Determine the radius r of the inner core of the shaft that resists one-quarter of the applied torque (T/4). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear stress distribution. 6. 5-13. A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev/min. Determine the inner diameter d of the tube to nearest 1/8 in. if the allowable shear stress is τ allow = 10 ksi. 7. 5-17. The steel shaft has a diameter of 1 in. and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft if the couple forces have a magnitude of F = 30 lb. 8. 5-21. The 20-mm-diameter steel shafts are connected using a brass coupling. If the yield point for the steel is (τY)st = 100 MPa, determine the applied torque T necessary to cause the steel to yield. If d = 40 mm, determine the maximum shear stress in the brass. The coupling has an inner diameter of 20 mm. 9. 5-22. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d. 10. 5-29. The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function N⋅m/m, where x is in meters. Determine the minimum torque T0 t = 25 x x

(

2

)

needed to overcome friction and cause it to twist. Also, determine the absolute maximum stress in the shaft. 11. 5-30. The solid shaft has a linear taper from rA at one end to rB at the other. Derive and equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis. 12. 5-32. The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow = 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev/min, determine the minimum required thickness of the shaft’s wall. 13. 5-33. The drive shaft AB of an automobile is to be designed as a thin walled tube. The engine delivers 150 hp when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The material has an allowable shear stress τallow = 7 ksi. 14. 5-34. The drive shaft of a tractor is to be designed as a thin-walled tube. The engine delivers 200 hp when the shaft is turning at 1200 rev/min. Determine the minimum thickness of the wall of the shaft if the shaft’s outer diameter is 3 in. The material has an allowable shear stress of τallow = 7 ksi. 15. 5-35. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. If it is rotating at 200 rad/s. determine its largest inner diameter to the nearest 1/8 in. If the allowable shear stress for the material is τallow = 25 ksi. 16. 5-36. The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 6 ksi. If the outer diameter is 3 in. and the engine delivers 175 hp to the shaft when it is turning at 1250 rev/min. determine the minimum required thickness of the shaft’s wall. 17. 5-37. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi. 18. 5-38. The 0.75 in. diameter shaft for the electric motor develops 0.5 hp and runs at 1740 rev/min. Determine the torque produced and compute the maximum shear stress in the shaft. The shaft is supported by ball bearings at A and B. 19. 5-42. The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi. 20. 5-46. The solid shaft of radius c is subjected to a torque T at its ends. Show that the maximum shear strain developed in the shaft is γmax = Tc/JG. What

is the shear strain on an element located at point A, c/2 from the center of the shaft? Sketch the strain distortion of this element. 21. 5-74. A rod is made from two segments: AB is steel and BC is brass. It is fixed a its ends and subjected to a torque of T = 680 N⋅m. If the steel portion has a diameter of 30mm. determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa. 22. 5-78. The composite shaft consists of a mid-section that includes the 1-in.diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 800 lb⋅ft. The material is A-36 steel.

The column is subjected to an axial force of 8 kN at its top. If the cross-sectional area has the dimensions shown in the figure, determine the average normal stress acting at section a-a. Show this distribution of stress acting over the area’s cross section.

A = 2[10mm(150mm ) + 10mm(140mm )] A = 4400 mm 2

(

)

A = 4.4 10 −3 m 2 σ = =

P A 8000 N 4.4 10 −3 m 2

(

)

σ = 1.82 MPa

The anchor shackle supports a cable force of 600 lb. If the pin has a diameter of 0.25 in., determine the average shear stress in the pin.

d = 0.25 in r = 0.125 in A = πr 2 = π (0.125)

2

= 0.04909 in 2 For double shear stress : τ =V

2A

=

0.6 lb 2 0.04909 in 2

(

= 6.11 ksi

)

The small block has a thickness of 5 mm. if the stress distribution at the support developed by the load varies as shown, determine the force F applied to the block, and the distance d to where it is applied.

( )

[

( )]

1 1 F =  (40) 10 6 (0.06)(0.005) +   (40 + 60) 10 6 (0.12)(0.005)  2 2 F = 36,000 N F = 36 kN Solving for d :  1   1  2   1 2 F ⋅ d =  (40)(60 )(5) (60) + (40)(120)(5) (120) + 60 +  (20)(120)(5 ) (120) + 60  2 3  2    2  3   F ⋅ d = 3,960,000 3960000 d= 36000 d = 110 m

A 175-lb woman stands on a vinyl floor wearing stiletto high-heel shoes. If the heel has the dimensions shown, determine the average normal stress she exerts on the floor and compare it with the average normal stress developed when a man having the same weight is wearing flat-heel shoes. Assume the load is applied slowly, so that dynamic effect can be ignored. Also assume the entire is supported only by the heel of one shoe.

( 2 )(1.2)

AM = (0.5in )(2.4in ) + π

2

AM = 3.462 in 2 σM =

175 lb FM = AM 3.462 in 2

σ M = 50.5 psi

( 2 )(0.3in)

AW = (0.1in )(0.6in ) + π AW = 0.201in 2 σW =

FW 175 lb = AW 0.201in 2

σ W = 869 psi

2

The 50-lb lamp is supported by three steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take θ=30°. The diameter of each rod is given in the figure.

ΣFX = 0 FAC cos 30° = FAD cos 45° FAC =

FAD cos 45° cos 30°

ΣFY = 0 FAC sin 30° + FAD sin 45° = 50 lb FAD =

50

(

cos 45° sin 30°

FAD = 44.83 lb FAC = 36.6 lb 44.83 π 0.3 2 2 = 634.21 psi 36.6 = π 0.25 2 2 = 745.61 psi

σ AD =

( )

σ AC

(

)

cos 30°

)+ sin 45°

∴Rod AC is subjected to greater average normal stress at 745.61 psi.

The 50-lb lamp is supported by three steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take θ=45°. The diameter of each rod is given in the figure.

ΣF X = 0 FAC cos 45° = FAD cos 45° FAC = FAD ΣFY = 0 FAC sin 45° + FAD sin 45° = 50 lb FAD = 35.36 lb FAC = 35.36 lb 35.36 π 0.3 2 2 = 500.24 psi 35.36 = π 0.25 2 2 = 720.3 psi

σ AD =

( )

σ AC

(

)

∴Rod AC is subjected to greater average normal stress at 745.61 psi.

The two steel members are joined together using a 60° scarf weld. Determine the average normal and average shear stress resisted in the plane of the weld.

ΣFY = 0 N sin 30° = V sin 60° V=

N sin 30° sin 60°

ΣFx = 0 − 8 + N cos 30° + V cos 60° = 0 − 8 + N cos 30° + N = 6.93 kN V = 4 kN

N sin 30° cos 60° = 0 sin 60°

 30mm  A = 25mm   sin 60°  A = 866.025 mm 2 6.93kN (1000) = 8MPa 866.025mm 2 3

σ =N

A

=

4kN (1000) = 4.62MPa 866.025mm 2 3

τ =V

A

=

The build-up shaft consist of a pipe AB and solid rod BC. The pipe has an inner diameter of 20mm and outer diameter of 28mm. The rod has a diameter of 12mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of this points.

 282 AD = π   4

  20 2   −     4 

AD = 301.5929 mm2 AE = π (36) AE = 113.0973 mm 2 4kN (1000) σD = 301.5929mm 2 σ D = 13.26 MPa (C ) 3

(

)

8kN 10003 113.0973mm 2 σ E = 70.75 MPa (T ) σE =

The plastic block is subjected to an axial compressive force of 600N. Assuming that the caps at the top and bottom distribute the load uniformly throughout the block. Determine the average normal and average shear stress acting along section a-a.

ΣFx = 0 N sin 30° = V cos 30° N sin 30° V= cos 30° Σ Fy = 0 N sin 30° sin 30° − N cos 30° = 0 cos 30° N = 519.62 N V = 300 N 600 −

519.62 N (1000) = = 90kPa A 50mm 100mm cos 30° 2

σ a−a

=N

τ a−a

=V

(

)

300 N (1000) = = 52kPa A 50mm 100mm cos 30° 2

(

)

The specimen failed tension test at an angle of 52° when the axial load was 19.80 kip. If the diameter of the specimen is 0.5 in., determine the average normal and average shear stress acting on the area of the inclined failure plane. Also, what is the average normal stress acting on the cross section when failure occurs?

Σ Fy = 0 N sin 38° = V sin 52° V=

N sin 38° sin 52°

τ ave = 0

ΣFx = 0 − 19.8kip + N cos 38° + V cos 52° = 0 N sin 38° − 19.8kip + N cos 38° + cos 52° = 0 sin 52° N = 15.6 kip V = 12.19 kip

( 2)

2

A = π 0.5

A = 0.196 in 2 A' = A

sin 52° A' = 0.2492 in 2 ∴σ ' = N

A'

=

15.6 kip 0.2492 in 2

σ ' = 62.6 ksi 12.19 kip A' 0.2492 in 2 τ ' ave = 48.9 ksi τ ' ave = V '

=

When failure occurs no shear stress exists, since the shear force at the section is 0.

σ =P

A

=

σ = 101 ksi

19.8 kip 0.196 in 2

A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine the maximum average shear stress in the specimen and indicate the orientation θ of a section on which it occurs.

τ =V

AS

; where AS = A

sin θ

V = P cosθ P cosθ A sin θ θ sin θ P cos =

τ =

A dτ P = (− sin 2θ + cos 2θ ) dθ A P 0 = (cos 2θ − sin 2θ ) A P cos 2θ = P sin 2θ cosθ = sin θ tan θ = 1 θ = 45° P τ =  (cosθ sin θ )  A P τ =   cos 45° sin 45°  A P τ = 2A

The joint is subjected to the axial member force of 5 kN. Determine the average normal stress acting on sections AB and BC. Assume the member is smooth and is 50 mm thick.

ΣFx = 0 − 5kN cos 45° + FAB cos 30° = 0 FAB = 4.08kN ΣF y = 0 5kN sin 45° − 4.08 sin 30° = FBC FBC = 1.49kN 4.08kN (1000) = 50mm(40mm ) = 2.04 MPa

3

σ AB σ AB

1.49kN (1000) 50mm(50mm ) = 596 kPa 3

σ BC = σ BC

The joint is subjected to the axial member force of 6kip. Determine the average normal stress acting on section AB and BC. Assume the member is smooth and is 1.5in. thick.

ΣFy = 0 6sin60º = FBCsin70º FBC = 5.53 kip ΣFx = 0 FAB = 6cos60º+FBCcos70º FAB = 4.89 kip For section AB: σAB=FAB/A = 4.89 kip 2(1.5 in2) σAB=1.63 ksi For section BC: σBC=FBC/A = 5.53 kip 4.5 in(1.5 in) σBC=0.819 ksi

Rod AB and BC have diameters of 4 mm and 6 mm, respectively. If the load of 8 kN is applied to the ring at B, determine the average normal stress in each rod if θ = 60°.

ΣFy = 0 -8 kN+FBCsin60º = 0

ΣFx = 0 - FBC = 9.24 kN FAB+FBCcos60º = 0 FAB = 4.62 kN σAB=4.62 kN(10003) = 367.65 MPa π(4mm2) σBC=9.24 kN(10003) = 326.8 MPa π(9mm2)

Rods AB and BC have diameters of 4mm and 6mm respectively. If the vertical load of 8kN is applied to the ring at B, determine the angle θ of rod BC so that the average normal stress in each rod is equivalent. What is this stress?

Given: dAB=4mm dBC=6mm Find: θ & σ ΣFy = 0 -8 kN+FBCsinθ = 0 FBC=8 kN/sinθ ΣFx = 0 -FAB+FBCcosθ = 0 FAB = FBCcosθ = 8 kNcosθ/sinθ σBC= 8 kN (sinθ)π(0.006m/2)2 σAB= 8 kNcosθ πsinθ(0.004m/2)2 σBC=σAB 8 kN = 8 kNcosθ πsinθ(0.006m/2)2 πsinθ(0.004m/2)2 θ=cos-1(0.0042/0.0062) θ=63.61º σ= 8 kN

πsin63.61º(0.006m/2)2 =316 MPa

The bars of the truss each have a cross-sectional area of 1.25 in2. Determine the average normal stress in each member due to the loading P=8kip. State whether the stress is tensile or compressive.

ΣFx=0 Cx-Bx=0 Cx=Bx ΣFy=0 Cy+Dy-0.75-P=0 Cy+Dy=1.75 P At joint A ΣFy=0 AB(4/5)=P AB=(5/3)P (T) ΣFx=0 AB(4/5)=AE AE=(4/3)P (C) At joint B ΣFy=0 -AB(3/5)-BE+BD(3/5)=0 BD=()5/3[.75P+(3/5)(5/3)P] BD=(35/12)P (C) ΣFx=0 -AB(4/5)-BD(4/5)+BC=0 BC=(11/3)P (T) ΣMD=0 -3Cx+4(.75P)+8P=0 Cx=(11/3)P Dx=(11/3)P

At joint E ΣFx=0 AE=ED ED=(4/3)P (C) ΣFy=0 BE=.75P (P)

σAB= (5/3)P 1.25 =(4/3)(8)=10.7 ksi (T) σBC=(11/3)P/1.25=(44/5)(8) = 23.5 ksi (T) σAE=(4/3)P/1.25=(16/15)(8) =8.53 ksi (C) σBE=.75P/1.25=(3/5)(8) =4.8 ksi (T) σED=(4/3)P/1.25=(16/15)(8) =8.53 ksi (C) σBD=(35/12)P/1.25=(7/3)(8) =18.7 ksi (C)

The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 12 mm2 and 8 mm2, respectively. If d = 1 m, determine the average normal stress in each rod.

ΣMC = 0 3ΤAB = 12 kN.m TAB=4 kN ΣFy = 0 TCD = 6 kN-4 kN = 2 kN σAB=

4000 N 12(1m2/10002) σAB=333 MPa σCD= 2000 N 8(1m2/10002) =250 MPa

The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 12 mm2 and 8 mm2, respectively. If d = 1 m, determine the average normal stress in each rod.

ΣMA = 0 6d = TCD(3) TCD = 2d ΣFy = 0 TAB = 6-2d σAB = σCD 6-2d = 2d 12 m2 8 m2 10002 10002 (6-2d)(8)(10002) = (12)(2d)(10002) 48 = 40d :ּd = 1.2m

The railcar dock light is supported by the 1/8 in diameter pin at A. If the lamp weighs 4lb, and the extension arm AB has a weight of 0.5 lb/ft, determine the average shear stress in the pin needed to support the lamp. Hint: The shear force is cost by the couple moment required for equilibrium at A.

0.5 lb.ft M

Ay

4 lb 3 ft

ΣFy = 0 Ay = 4+0.5(3) Ay = 5.5 lb ΣMA = 0 M = 4(3)+0.5(3)(1.5) M = 14.25 lb.ft[12 in/1ft] M = 171 lb.in T = V/A = M/1.25 πr2 = 11147.47 lb/in2 = 11.1 ksi

The two member frame is subjected to the distributed loading shown. Determine the intensity w of the largest uniform loading that can be applied to the frame w/o causing either the average normal stress or the average shear stress at section b-b to exceed σ = 15MPa and τ = 16MPa, respectively. Member CB has a square crosssection of 35mm on each side.

ΣM A = 0 Ay Ax

A

 4 (1.5) 3BC   = 3W BC 5 4 5 3 3m BC = 1.875W B BC x W σ b −b = W l 3 5 3 (1.875W ) = 15 352  5  5  3  

( )

W = 27,222.22 N W = 27.2 kN τ b −b =

BC y W l 3 5

4 (1.875W ) = 16 352 5 

( ) 53 

W = 21,777.78 N W = 21.8 kN ∴W = 21.8 kN

 

1-81. The 60mm × 60mm oak post is supported on the pin block. If the allowable bearing stresses for this material are σ oak =43MPa and σ pine =25MPa, determine the greatest load P that can be supported. If a rigid bearing plate is used between this material, determine its required area so that the maximum load P that can be supported. What is this load?

σOAK=43 MPa σPINE=25 MPa σOAK=P/A [43 N/mm2][602 mm2]=P 154,800 N=P P=154.8 kN σPINE=P/A [25 N/mm2][602 mm2] 90000 N=P P=90 kN :ּ P=90 kN since this is the greatest load the pine block can told. Since there is a bearing plate between the oak block and the pine block, the maximum load is the allowable load by the pine block that is P≈155 kN

AALLOW= PMAX/σOAK = 154.8[1000/1 kN] 25N/mm2[(1000mm)2/1 m2

=6.19x10-3 m

1-82. The join is fastened together using two bolts. Determine the required diameter of the bolts if the allowable shear stress for the bolts is τallow = 110 MPa. Assume each bolt supports an equal portion of the load.

Τallow = V/4A 110 MPa = 80 kN(1000) 4π(d2/4) d = √80(1000)/π(110) d = 15.2 mm

1-83. The lever is attached to the shaft A using a key that has a width d and length of 25mm. If the shaft is fixed and a vertical force of 200N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is τ allow = 35MPa. d

A 20 mm 500 mm

200 N

ΣFy=0 Ay=200 N ΣMA =0 M=200(500 mm)[1 m/1000 mm] M=100 N.m V=M/20 mm=100 N.m/.02 m =5000N A=V/TALLOW dL=V/TALLOW d=5000 N/(25x25) N d=5.71 mm

1-86. The tension member is fastened together using two bolts, one on each side of the member as shown. Each bolt has a diameter of 0.3 inch. Determine the maximum P that can be applied to the member if the allowable shear stress for the bolts is τ allow =12ksi and the allowable average normal stress is σ allow = 20ksi. 60° P

ΣFy = 0 Vsin60º = Nsin30º V = Nsin30º sin60º ΣFx = 0 -P+Ncos30º+Vcos60º = 0 N = 0.866 P V = 0.5 P σ=

0.866 P 2π(0.3 in/2)2 Τ = 0.5 P/2π(0.3 in/2)2 If σ=20 ksi P = 3.26 kip Τ = 11.55 ksi (safe) If Τ=12 ksi P = 3.39 kip σ=20.8 ksi (fail) :ּP = 3.26 kip

P

1-88. The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile stress of σallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5 kN.

C B

5

60°

3

4

A

P

ΣFx = 0 -FABsin60º+FAC(4/5) = 0 FAB = 4FBC/5sin60º ΣFy = 0 -5+FABcos60º+FAC(3/5) = 0 FAC = 4.71 kN FAB = 4.35 kN σAB=4.35(10004) π(dAB)2 dAB=√4.35(10004)/π(200) dAB = 5.26mm σAC=4.71(10004) π(dAC)2 dAC = √4.71(10004)/π(200) = 5.48 mm

1-89. The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile stress of σ allow = 180MPa, and wire AB has a diameter of 6mm and AC has a diameter of 4mm, determine the greatest force P that can be applied to the chain before one of the wires fails.

C B

5

60°

3

4

A

P

ΣFx = 0 -FABsin60º+FAC(4/5) = 0 FAB = 4FAC 5sin60º ΣFy = 0 -P+FABcos60º+FAC(3/5) = 0 FAC = 0.942 P FAB = 0.87 P σAB = 0.87 P (1000) π(9m2) σAC=0.942 P(1000) π(4m2) If σAB = 180 MPa

P = 5.85 kN

2-1. An air filled rubber ball that has a diameter of 6in. If the air pressure w/in it is increase until the ball’s diameter becomes 7in., determine the average normal strain in the robber. 6 in.

P

∈=

δ 7 − 6 1 in = = 6 in L 6

2-2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5in., determine the average normal strain in the strip. ∈=

δ 0.708 = = 0.0472 in in L 15

2-3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10mm downward, determine the normal strain developed in wires CE and BD. D

E

P

4m C

B

A

3m

2m

2m

δC δ B = 7 3 δ B = 30 mm 7 δC 10 mm = L 4 × 10 3 mm ∈CE = 2.5 × 10 −3 mm mm 30 mm 7 ∈BD = 4 × 10 3 mm ∈ = 1.07143 × 10 −3 mm

∈CE =

BD

mm

2-5. The wire AB is unstretched when θ = 45°. If the load is applied to the bar AC, w/c causes θ = 47°, determine the normal strain in the wire. B θ

L A

C L

( )

sin 47 2 L = x x = 1.034290324 L 1.034290324 L − L L ∈= 0.343 ∈=

2-7. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2mm, determine the normal strain developed in each wire. C 30

0m m

30° 30° 30

B

A

P

m 0m

150 (300 cos 30° ) + 2 φ = 29.81° tan φ =

300 + x =

(300 cos 30°) + 2 cos φ

x = 1.734 ∈=

1.734 30°

∈= 5.779 × 10 − 3 mm

mm

2-9. Part of a control linkage for an airplane consists of rigid member CBD and a flexible cable AB. If the force is applied to the end D of the member and causes a normal strain in the cable of 0.0035mm/mm, determine the displacement of D. Originally the cable is unstretched. θ D

P

300 mm

B

300 mm

C

A 400 mm

2-13. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5mm. Determine the average normal strain ∋x along the x axis. y

45°

x'

A

tan φ =

A' 5 mm

45° 800 mm

x

800 sin 45° (800 cos 45°) + 5 φ = 44.75°

800 mm

45°

(800 cos 45°) + 5 − 800

cos 44.75° 800 − 3 mm ∈= 4.43 × 10 ∈=

mm

2-14 The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5mm. Determine the average normal strain ∋x’ along the x’ axis. y

800 mm

45°

45°

x'

A

A' 5 mm

45° 800 mm

x δ 5 = L 800 cos 45° ∈x ' = 8.84 × 10 −3 mm ∈x ' =

mm

2-15. The corners of the square plate are given the displacements indicated. Determine the average normal strains ∋x and ∋ y along the x and y axes. y 2 in. A

10 in. D

A

3 in.

x 3 in

10 in.

C 10 in.

10 in.

10 − 10.3 δ x = L 10 in ∈x = −0.03 in ∈x =

10.2 − 10 δ y = 10 L ∈y = 0.02 in in ∈y =

2 in.

2-21. A thin wire, lying along the x axis, is strained such that each point on the wire is displaced x = kx2 along the x axis. If k is constant, what is the normal strain at any point P along the wire?

P x

1 kx 2 ∈= 2 dx 1 1 dx = 2 ∈ kx 2 1 1 1 x −2 = x dx 2 ∈ k ∫0 11 1 =− 2∈ kx 1 ∈= −kx 2 ∈= −2kx

x

2-22. The wire is subjected to a normal strain that is defined by ∋ - xe-x2 , where x is millimeters. If the wire has an initial length L, determine the increase in its length.

∈= xex

2

x

x L

∆L = ∫ xe − x dx 2

LET

u = e −x

2

( )

du = −2 x e − x dx 2 1 − du = e − x xdx 2 1 ∆L = ∫ − du 2 1 =− u 2 2 L 1 = − e −x 0 2 1 2 1 = − e −L + 2 2 2 1 ∆L = 1 − e − L 2 2

( )

(

)

2-25. The piece rubber is originally rectangular. Determine the average shear strain γxy if the corners B and D are subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD. y 3 mm

C D

400 mm

x

B

A

2 mm 300 mm

2-26. The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD. y 3 mm

C D

tan φ = 3

400 φ = 0.43°

400 mm

x

B

A

2 mm

400 − 400 ° cos 0 . 43 ∈ AD = 400 ∈ AD = 2.81 × 10 −3 mm mm

300 mm

tan θ = 2

300 θ = 0.38° 500 − 500 ∈ AB = cos 0.229° 500 ∈ AB = 8.00 × 10 − 6 mm mm

2

2

 400   300   400  300  w =  +  − 2   cos 89.188°  cos 0.43°   cos .382°   cos 0.43°  cos 0.382°  w = 496.6 2

496.6 − 500 500 ∈BD = −6.8 × 10 −3 mm ∈BD =

mm

2-28. The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF. y 10 mm C 25 mm

15 mm D

B

10 125 θ = 4.5739 tanθ =

100 mm 50 mm

x

F

A

x 2 = (125mm) + (10mm) 2

2 mm

x = 125.4mm

80 mm

(

)

CF = 80 2 + (125 .4 ) − 2(80 ) 15725 cos (90 − 405739 ) 2

2

CF = 143 .2655 ∈CF =

143 .2655 −

2

2-29. The non uniform loading causes a normal strain in the shaft that can be expressed as ∋ = kx2 , where k is a constant. Determine the displacement of the end B. Also, what is the average normal strain in the rod?

∫ kx

L A

B x

2

=∫

2-30. The non uniform loading causes a normal strain in the shaft that can be expresses as ∋ = k sin ((π/L)x), where k is a constant. Determine the displacement of the center C and the average normal strain in the entire rod. C A

B L/2

L/2 π  ∈x = k sin  x  L  L  π  (∆x )C 2 ∫0 k sin L x  = dx

(∆x )C = ∫0 2 k sin π x dx L  L

(∆x )C

L π  = k ∫ 2 sin  x dx 0 L 

(∆x )C

 π  2  L  = k  − cos x   L  0  π 

(∆x )C

  π  = k − cos  + cos(0) 2   kL = π

L

(∆x )C

NORMAL STRAIN : kL 1 ∈X = π 2 L 2k ∈x = π

2-31. The curved pipe has an original radius of 2ft. If it is heated non uniformly, so that the normal strain along its length is ∋ = 0.05cosθ, determine the increase in length of the pipe.

L = rθ

2 ft θ

0.05 cosθ =

∆x rdθ

∆x = ∫ 0.05r cosθ dθ 90

A

0

∆x = 0.1∫ cosθdθ 90

0

= 0.1(sin θ )0

90

∆x = 0.1 ft

4-27. Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P. d 2 d1 − 2 2

P d2 t

x 2

dx h

h

y

x d1 − 2 2

d1 P

x d1 d 2 d1 − − 2 2 = 2 2 y h x − d 1 d 2 − d1 = y h xh − d 1 h = d 2 y − d 1 y  d − d1  x= 2  y + d1  h  d − d1 dx = 2 dy h h dy = dx d 2 − d1

d1 2

Pdy AE  h  dx P  d 2 − d 1   = AE  h  dx P  d 2 − d 1   = xtE  h   P  d2 d 2 − d 1  dx  δ =∫ d1 tE x  h   P d d − 2 1  d 2 dx =  ∫d1 x tE P h  [ln x ]dd12 =  tE  d 2 − d 1  dδ =

P h   ln (d 2 − d 1 )  tE  d 2 − d 1  P  h  d2  ln δ =  tE  d 2 − d 1  d 1 δ =

4-50. The three suspender bars are made of the same material and have equal cross-sectional areas A. Determine the average normal stress in each bar if the rigid beam ACE is subjected to the force P.

B

D

F

P

L

A

E

C

d/2

d/2

d

4-53. The 10mm diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20mm, and its inner diameter is 10mm. If the bolt is subjected to a compressive force of P= 20kN. Determine the average normal stress in the steel and the bronze. Est = 200Gpa, Ebr = 100GPa. 20kN = Pst + Pb

P

δB =

Pb(1000)L(0.4) π 0.02 − 0.012 100 × 109 Pa

(

2

)(

)

δ B = 4.2441× 10−5 PB L

δ ST = 6.3662 × 10−5 L(20 − PB ) 10 mm 20 mm

δ ST = δ B

4.2441× 10 −5 LPB = 6.3662 × 10 −5 L(20 − PB ) PB = 12kN PST = 8kN δ ST E ST L = 6.3662 × 10 −5 (8kN ) 200 × 109

σ ST =

(

P

σ ST = 102MPa

(

σ B = 4.2441× 10 −5 (12) 100 × 109 σ B = 50.9MPa

)

)

4-110. A 0.25in diameter steel rivet having a temperature of 1500°F is secured between two plates such that at this temperature it is 2in long and exerts a clamping force of 250lb between the plates. Determine the approximate clamping force between the plates when the rivet cools to 70°F. For the calculation, assume that the heads of the rivet and the plates are rigid. Take αst = 8(10-6)/°F, Est = 29(103)ksi. Is the result a conservative estimate of the actual answer? Why or why not? 2 in.

An air filled rubber ball that has a diameter of 6in. If the air pressure w/in it is increase until the ball’s diameter becomes 7in., determine the average normal strain in the robber. 6 in.

P

∈=

δ 7 − 6 1 in = = 6 in L 6

A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5in., determine the average normal strain in the strip.

∈=

δ 0.708 = = 0.0472 in in L 15

The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10mm downward, determine the normal strain developed in wires CE and BD.

D

E

P

4m C

B

A

3m

2m

2m

δC δ B = 7 3 δ B = 30 mm 7 δC 10 mm = L 4 × 10 3 mm ∈CE = 2.5 × 10 −3 mm mm 30 mm 7 ∈BD = 4 × 10 3 mm ∈ = 1.07143 × 10 −3 mm

∈CE =

BD

mm

The wire AB is unstretched when θ = 45°. If the load is applied to the bar AC, w/c causes θ = 47°, determine the normal strain in the wire.

B θ

L A

C L

( )

sin 47 2 L = x x = 1.034290324 L 1.034290324 L − L L ∈= 0.343 ∈=

The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2mm, determine the normal strain developed in each wire.

C 30

0m m

30° 30° 30

A

P

m 0m

B

150 (300 cos 30° ) + 2 φ = 29.81° tan φ =

300 + x =

(300 cos 30°) + 2 cos φ

x = 1.734 ∈=

1.734 30°

∈= 5.779 × 10 − 3 mm

mm

Part of a control linkage for an airplane consists of rigid member CBD and a flexible cable AB. If the force is applied to the end D of the member and causes a normal strain in the cable of 0.0035mm/mm, determine the displacement of D. Originally the cable is unstretched.

θ D

P

300 mm

For the wire: ∆AB 500 ∆AB = 1.75 mm 0.0035 =

B

x 600 x = 4.383 mm

tan θ = 300 mm

C

A 400 mm

(501.75)2

= 4002 + 3002 − 2(400)(300) cos(90 + θ ) 1753.0625 = −240000cos(90 + θ ) θ = 0.4185°

The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5mm. Determine the average normal strain ∋x along the x axis.

y

800 mm

45°

800 sin 45° (800 cos 45°) + 5 φ = 44.75° tan φ =

45°

x'

A

5 mm

45° 800 mm

x

A'

(800 cos 45°) + 5 − 800

cos 44.75° 800 − 3 mm ∈= 4.43 × 10 ∈=

mm

The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5mm. Determine the average normal strain ∋x’ along the x’ axis.

y

800 mm

45°

45°

x'

A

A' 5 mm

45° 800 mm

x δ 5 = L 800 cos 45° ∈x ' = 8.84 × 10 −3 mm ∈x ' =

mm

The corners of the square plate are given the displacements indicated. Determine the average normal strains ∋ x and ∋ y along the x and y axes. y 2 in. A

10 in. D

A

3 in.

x 3 in

10 in.

C 10 in.

10 − 10.3 δ x = L 10 in ∈x = −0.03 in ∈x =

10.2 − 10 δ y = 10 L ∈y = 0.02 in in ∈y =

10 in.

2 in.

A thin wire, lying along the x axis, is strained such that each point on the wire is displaced x = kx2 along the x axis. If k is constant, what is the normal strain at any point P along the wire?

P x

1 kx 2 ∈= 2 dx 1 1 dx = 2 ∈ kx 2 1 1 1 x −2 = x dx 2 ∈ k ∫0 11 1 =− 2∈ kx 1 ∈= −kx 2 ∈= −2kx

x

The wire is subjected to a normal strain that is defined by ∋ - xe-x2 , where x is millimeters. If the wire has an initial length L, determine the increase in its length.

∈=xex

2

x

x L

∆L = ∫ xe − x dx 2

LET

u = e−x

2

( )

du = −2 x e − x dx 2

2 1 − du = e − x xdx 2 1 ∆L = ∫ − du 2 1 =− u 2 2 L 1 = − e−x 0 2 1 2 1 = − e −L + 2 2 2 1 ∆L = 1 − e − L 2

( )

(

)

The piece rubber is originally rectangular. Determine the average shear strain γxy if the corners B and D are subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD. y 3 mm

C D

400 mm

x

B

A

2 mm 300 mm

The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD. y 3 mm

C D

400 mm

tan φ = 3

400 φ = 0.43°

x

B

A

2 mm 300 mm

400 − 400 ° cos 0 . 43 ∈ AD = 400 ∈ AD = 2.81 × 10 −3 mm mm tan θ = 2

300 θ = 0.38° 500 − 500 ∈ AB = cos 0.229° 500 ∈ AB = 8.00 × 10 − 6 mm mm

2

2

 400   300   400  300  w2 =   +  − 2   cos 89.188°  cos 0.43°   cos .382°   cos 0.43°  cos 0.382°  w = 496.6 496.6 − 500 500 ∈BD = −6.8 × 10 −3 mm ∈BD =

mm

The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF. y 15 mm

10 mm C 25 mm

D

10 125 θ = 4.5739° tanθ =

B

x 2 = (125mm) + (10mm) 2

100 mm 50 mm F

A

x = 125.4mm

x 2 mm

80 mm

(

)

CF = 80 2 + (125.4 ) − 2(80) 15725 cos(90 − 405739) 2

2

CF = 143.2655 mm ∈CF =

143.2655 − 22025

22025 ∈CF = −0.03465 mm mm 15 125 φ = 6.843° tan φ =

y = 15850 = 125.9 mm

(

)

AD = 80 2 + 15850 − 2 15850 (80) cos(90 + φ ) 2

AD = 157.003 mm

∈AD =

157.003 − 15850

15850 ∈AD = 0.24708 mm

mm

2

The non uniform loading causes a normal strain in the shaft that can be expresses as ∋ = k sin ((π/L)x), where k is a constant. Determine the displacement of the center C and the average normal strain in the entire rod.

C A

B L/2

L/2

π  ∈x = k sin  x  L  L  π  (∆x )C 2 ∫0 k sin L x  = dx

(∆x )C = ∫0 2 k sin π x dx L  L

(∆x )C

L π = k ∫ 2 sin  0 L

(∆x )C

π  L  = k  − cos L  π 

(∆x )C

  π  = k − cos  + cos(0) 2   kL = π

(∆x )C

 x dx 

NORMAL STRAIN : kL 1 ∈X = π 2 L 2k ∈x = π

L

 2 x   0

The curved pipe has an original radius of 2ft. If it is heated non uniformly, so that the normal strain along its length is ∋ = 0.05cosθ, determine the increase in length of the pipe.

2 ft θ

A

L = rθ 0.05 cosθ =

∆x rdθ

∆x = ∫ 0.05r cosθ dθ 90

0

∆x = 0.1∫ cosθdθ 90

0

= 0.1(sin θ )0

90

∆x = 0.1 ft

Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P.

d 2 d1 − 2 2

P d2 t

x 2

dx h

h

y

x d1 − 2 2

d1 P

x d1 d 2 d 1 − − 2 2 = 2 2 y h x − d1 d 2 − d1 = y h xh − d 1 h = d 2 y − d 1 y  d − d1  x= 2  y + d1  h  d − d1 dx = 2 dy h h dy = dx d 2 − d1

d1 2

Pdy AE  h  dx P  d d − 1  =  2 AE  h  dx P  d 2 − d 1   = xtE  h   P  d2 d 2 − d 1  dx  δ =∫ d1 tE x  h   P d 2 − d 1  d 2 dx  = ∫d1 x tE P h  [ln x ]dd12 =  tE  d 2 − d 1  dδ =

δ =

P h   ln (d 2 − d 1 )  tE  d 2 − d 1 

δ =

P  h  d2  ln  tE  d 2 − d 1  d 1

The three suspender bars are made of the same material and have equal crosssectional areas A. Determine the average normal stress in each bar if the rigid beam ACE is subjected to the force P.

B

D

σ =

F

σ AB

P

L

σ CD A

E

C

σ EF d/2

d/2

d

ΣFy = 0solve FEF u sin g (4) FAB + FCD 7+ FEF = P p − P + FEF = − 12 2 ΣM E = F0EF: = p 12  3d  FAB (2d ) + FCD (d ) = P   2  solve FCD u sin g (1) 3P 2 FAB + F7CD = P =P P + F2Cd + 12 12 P CD = σ − σ EF σ AB − σFEF ` = 3CD d 2d FAB FEF FP FEF − σ = 7 CD − A AAB = 12AA A d 2d P − FEF FAB − FσEFCD = F3CD = A 2 P1 σ = EF FAB − FEF = 212 FCDA − 2FEF

FAB − 2 FCD + FEF = 0 FAB + FCD + FEF = P

eqn.1

FAB − 2 FCD + FEF = 0

eqn.2

2 FAB + FCD =

3P 2

eqn.3

e lim inate FCd u sin g (1 & 3) F AB + FCD + FEF = P 2 F AB + FCD − FAB + FEF =

= −p 2

3 P 2 eqn.4

e lim inate FCD u sin g (2 & 3) F AB − 2 FCD + FEF = 0 2 F AB + FCD 5 FAB + FEF = 3P

3 P 2 eqn.5

=

e lim inate FEF u sin g (4 & 5) −p 2 5 FAB + FEF = 3P 7 − 6 FAB = − P 2 7 FAB = P 12 − FAB + FEF =

P , A P = A P = A P = A

P=F , P = F AB , P = FCD , P = FEF

(Mindanao State University-General Santos City, Philippines) BSEE-(G.R.F) students The 10mm diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20mm, and its inner diameter is 10mm. If the bolt is subjected to a compressive force of P= 20kN. Determine the average normal stress in the steel and the bronze. Est = 200Gpa, Ebr = 100GPa.

P

20kN = Pst + Pb δB =

Pb(1000)L(0.4) π 0.02 − 0.012 100 × 109 Pa

(

)(

2

)

−5

δ B = 4.2441× 10 PB L 10 mm 20 mm

δ ST = 6.3662 × 10−5 L(20 − PB ) δ ST = δ B

4.2441× 10 −5 LPB = 6.3662 × 10 −5 L(20 − PB ) PB = 12kN PST = 8kN δ ST E ST L = 6.3662 × 10 −5 (8kN ) 200 × 109

σ ST =

P

(

σ ST = 102MPa

(

σ B = 4.2441× 10 −5 (12) 100 × 109 σ B = 50.9MPa

)

)

A 0.25in diameter steel rivet having a temperature of 1500°F is secured between two plates such that at this temperature it is 2in long and exerts a clamping force of 250lb between the plates. Determine the approximate clamping force between the plates when the rivet cools to 70°F. For the calculation, assume that the heads of the rivet and the plates are rigid. Take αst = 8(10-6)/°F, Est = 29(103)ksi. Is the result a conservative estimate of the actual answer? Why or why not?

2 in.

αL(∆T ) = −

PL AE

(8 × 10 )(2)(70 − 1500 ) = −6

PTH = 16.29 kips PT = 16.289 + 0.25 PT = 16.54 kips

PTH (2 )  π  2 3  4 (0.25 )  29 × 10   

(

)

The rubber block is subjected to an elongation of 0.03 in along the x axis, and its vertical faces are given a tilt so that θ = 89.3° .Determine the strains ∈x , ∈x , γ xy . Take vr = 0.5

y

3 in.

θ

x 4 in.

Finding ∈x ∈x =

0.03 in 4 in

∈x = 0.0075 in

in

Finding ∈ y ∈y = −υ ∈x

∈y = 0.5(0.0075)

∈y = −0.00375 in

Finding γ xy

in

(

γ xy = (90° − 89.3°) π

180

)

γ xy = 0.0122 rad

The non uniform loading causes a normal strain in the shaft that can be expressed as ∋ = kx2 , where k is a constant. Determine the displacement of the end B. Also, what is the average normal strain in the rod?

∫ kx

L A

B x

2

=∫

5- 2. The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-half of the applied torque (T/2).

r

T

r'

by ratio and proportion: τ max τ = r r' T 2T = 4 4 πr πr ' 4 2r ' = r 4 r '4 =

r4 2

r4 2 r ' = 0.841r

r' = 4

r ' = r cos 32.7651

5-3. The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-quarter of the applied torque (T/4).

r

T

T 2( ) T τ = 23 = 3 πr ' πr ' 2T τ max = 3 πr b.) T 2πr '3 τ max τ = r r' 2T T = 4 4 πr πr ' 4 2r ' = r 4

τ =

r'4 =

r4 4

r4 4 r ' = 0.707r r ' = r sin 45 r' = 4

r'

5-5. The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and the three torques are applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe’s outer surface. Sketch the shear stress on the volume elements located at A and B.

B

A

C

450 lb-ft

450 lb⋅ft = 5400 lb⋅in 350 lb⋅ft = 4200 lb⋅in 600 lb⋅ft = 7200 lb⋅in

350 lb-ft 600 lb-ft

sol’n: τmax 5400 + 4200 – 7200 = 2400

16TD π D4 − d 4 16(2400)(2.5) = π 2.5 4 − 2.34 = 2758.6 lb 2 in = 2.76 ksi

τB =

(

(

τB τB

)

7200 - 4200 = 3000

16TD π D4 − d 4 16(3000)(2.5) = π 2.5 4 − 2.34 = 3448.3 lb 2 in = 3.45 ksi

τA =

)

(

(

τA τB

)

)

5-10. The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear stress distribution over the cross section. 600 N 75mm 5mm 25mm 75mm

600 N

∑ M x = 0; 2(600)(0.075) = T T = 90 Nm π J= (D 4 − d 4 ) 32 π = (0.035 4 − 0.025 4 ) 32 J = 108 x10 −9 m 4 Tr τ max = J (90)(0.075) Nm 2 = (108 x10 −9 )m 4 = 14.45MPa Tr τ = J (90)(0.0125) = (108 x10 −9 ) = 10.40MPa

5-13. A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev/min. Determine the inner diameter d of the tube to nearest 1/8 in. if the allowable shear stress is τallow = 10 ksi.

d 2.5 in.

Given:

D = 2.5in

τ allow 10,000ld / in 2 Ρ = 3hp = 19,250lb. ft / s f = 2700rev / min = .75rev / s d =?

Solution:

Ρ = Tω 19250 = T [ 2π (0.75)] 19250 T= 2π (0.75) T = 4084.97lb.in 4084.97(1.25) τ max = π (1.25 4 − r 4 ) 2 (4084.97)(1.25)   4 π 4 π (1.25) 2 − ( r ) 2 =  10000   ( 4084097 )( 1 . 25 )( 2 ) πr 4 = (1.25 4 )π − 10000 6 . 649 r4 = π r = 1.206 d =r 2 d = 2.4in

1 d = 2 in 2

;

4

1 = 0.5 8

2

5-17. The steel shaft has a diameter of 1 in. and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft if the couple forces have a magnitude of F = 30 lb. 12in

8in. 8in.

F

F

sol’n:

∑M

A

AB D = 0.75 in d = 0.68 in BC D = 1 in d = 0.86 in

= 15(6 ) + 15(8) = T T = 210 lb ⋅ in

16TD π D4 − d 4 16(210)(0.75) = π 0.75 4 − 0.68 4 = 7818.72 lb 2 in = 7.8 ksi

τ AB =

(

(

τ AB τ AB

16TD π D4 − d 4 16(210)(1) = π 14 − 0.86 4 = 2361 lb 2 in = 2.36 ksi

τ BC =

)

)

(

(

τ BC τ BC

) )

5-21. The 20-mm-diameter steel shafts are connected using a brass coupling. If the yield point for the steel is (τY)st = 100 MPa, determine the applied torque T necessary to cause the steel to yield. If d = 40 mm, determine the maximum shear stress in the brass. The coupling has an inner diameter of 20 mm. d 20mm

T

T

To find the T; steel; τ =

2Ts πr 3

2Ts π (0.010) 3 Ts = 157.07 Nm

100 x10 6 =

To find the max.., τ; brass; τ =

2TR π (R − r 4 ) 2(157.07)(0.02) = π (0.02 4 − 0.014 ) 4

τ = 13.3MPa

5-22. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d. R

r

T T

Soln: P A πd 2 ⋅τ P = 4 T = PRn

τ =

 πd 2  ⋅ τ  Rn T =   4  T 4 τ = ; <1> π d 2 Rn 2T τ = ; < 2> πr 3 2T 4T = 3 πr π d 2 Rn 2r 3 n = d2

5-29. The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function N⋅m/m, where x is in meters. Determine the minimum torque T0 needed t = 25 x x to overcome friction and cause it to twist. Also, determine the absolute maximum stress in the shaft.

(

2

)

To 80mm

x 2m

(

t = 25 xe x

2

)

Solution: d=80mm=0.08 m

(

t = 25 xe x 16T τ = 3 πd

2

)

x2

T = ∫ 25 xe dx 2

0

2

T = 25∫ xe x dx 2

0 2

let u= x du=2x dx du/2=x dx x=0, u=0 x=2,u=4 T =∫

4

0

25e u du 2

4

= 12.5∫ e u du 0

[ ] = 12.5[e − 1] = 12.5 e u

4

0

4

T = 670 Nm

16T πd 3 16(670) = π (80 x10 −3 ) 3

τ =

τ = 6.66MPa

5-30. The solid shaft has a linear taper from rA at one end to rB at the other. Derive and equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis. T T

rB

x

L rA

ro

x L

Solution:

r = rB + r1 r1 r −r = A B L−x L (r − r )( L − x) r1 = A B L ( rA − rB )( L − x ) L rB L + ( rA − rB )( L − x) L rB L + ( L − x )rA − rB ( L − x ) L rB [ L − ( L − x)] + rA ( L − x) L rB ( x ) + rA ( L − x ) L xrB + rA ( L − x ) L

r = rB +

r= r=

r= r= r=

τ =

16T d ; r= 3 2 πd d = 2r

d 3 = 8r 3

τ =

16T π 8r 3

= τ =

τ =

2T πr 3 2T

 xr + rA ( L − x)  π B  L   3 2TL π [xrB + rA ( L − x)]

3

3

5-32. The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow = 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev/min, determine the minimum required thickness of the shaft’s wall.

Given: τ = 8ksi = 8000 psi D = 2.5in Ρ = 200hp = 100,000lbft / s f = 1140rev / min = 19rev / sec

t =?

Solutions:

Ρ 2πf 100,000lbft / s = 2π (19rev / s ) T = 837.66lbft or = 10,051.89lbin T=

16TD π (D 4 − d 4 ) 16(10,051.89)(2.5) 8,000 psi = π ( 2.5 4 − d 4 ) 402075.65 (2.5 4 − d 4 ) = π (8,000)

τ =

d 4 = 39 − 15.998 d = 4 23 d = 2.19in

Thickness: R=

D ; 2

r=

d 2

2.5 2.19 r= 2 2 R = 1.25 r = 1.1

R=

t = R−r t = 1.25 − 1.1 t = 0.15in

5-33. The drive shaft AB of an automobile is to be designed as a thin walled tube. The engine delivers 150 hp when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The material has an allowable shear stress τallow = 7 ksi.

Ρ 2πf 75,000lbft / s = 2π ( 25rev / s) T = 477.46lbft T = 5,729.58lbin T=

Given

Ρ = 150hp = 75,000lbft / s f = 1,500rev / min = 25rev / s D = 2.5in τ = 7 ksi = 7,000 psi t min = ?

Solution

16Td π (D 4 − d 4 ) 16(5729.58)(2.5) 7,000 = π (2.5 4 − d 4 ) 229,183 39 − d 4 = π (7,000)

τ =

d 4 = 39 − 10.42 d = 4 27.5968 d = 2.29in

Thickness: D 2 2.5 ; R= 2 R = 1.25in R=

t = R−r t = 1.25 − 1.15 t = 0.104in

d 2 2.29 r= 2 r = 1.146in r=

5-34. The drive shaft of a tractor is to be designed as a thin-walled tube. The engine delivers 200 hp when the shaft is turning at 1200 rev/min. Determine the minimum thickness of the wall of the shaft if the shaft’s outer diameter is 3 in. The material has an allowable shear stress of τallow = 7 ksi.

Given:

Ρ = 200hp = 100000lb. ft / s f = 1200rev / min = 20rev / s

D = 3in τ = 7 ksi = 7000 psi t max = ?

Solution: Ρ 2πf 100000lbft / s T= 2π ( 20rev / s ) T = 795.8lbft T=

T = 9549.3lbft

16TD π (D 4 − d 4 ) 16(9549.3)(3) 7000 = π (3 4 − d 4 ) 458366.4 81 − d 4 = π (7000)

τ =

d 4 = 81 − 20.84 d = 4 60.16 d = 2.785in

Max Thickness: R=

D 2

5-35. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. If it is rotating at 200 rad/s. determine its largest inner diameter to the nearest 1/8 in. If the allowable shear stress for the material is τallow = 25 ksi. A

B

6 in.

p ω 250000 T= 200 T = 15000lb ⋅ in 16TD τ = π D4 − d 4 16(15000)(2 ) 25000 = π 24 − d 4 T=

τ = 25ksi = 25000lb / in 2 1hp = 500 ft ⋅ lb / s  500 ft ⋅ lb / s   = 250000 ft / lb 500hp hp  

(

)

(

d 4 = (16 − 6.11) d = 1.75in

)

5-36. The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 6 ksi. If the outer diameter is 3 in. and the engine delivers 175 hp to the shaft when it is turning at 1250 rev/min. determine the minimum required thickness of the shaft’s wall. A

B

6 in.

5-37. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi. A

B

6 in.

5-38. The 0.75 in. diameter shaft for the electric motor develops 0.5 hp and runs at 1740 rev/min. Determine the torque produced and compute the maximum shear stress in the shaft. The shaft is supported by ball bearings at A and B. A

B

6 in.

5-42. The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

B A

5-46. The solid shaft of radius c is subjected to a torque T at its ends. Show that the maximum shear strain developed in the shaft is γmax = Tc/JG. What is the shear strain on an element located at point A, c/2 from the center of the shaft? Sketch the strain distortion of this element.

5-74. A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N⋅m. If the steel portion has a diameter of 30mm. determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa. C

B 680 N⋅m

1.60 m

0.75 m

sol’n: θAB = θBC

TL TL = JG AB JG BC

(680)(0.75) = (680)(1.6) 4 π d4 π (0.030) 9 (39 )(109 ) (75)(10 ) 32

32

d4 =

(1.6 )(0.030 4 )(75) (0.75)(39)

(

d = 4 3.3 10 −6

)

d = 0.0427 m d = 42.7 mm

m4

A

5-78. The composite shaft consists of a mid section that includes the 1-in-diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 100-lb.ft. The material is A36 steel. d 20mm

T

T

The solid shaft has a diameter of 0.75 in. If it is subjected to a torque shown, determine the maximum shear stress developed in regions BC and DE of the shaft. The bearings at A and B allow free rotation of the shaft.

T AB = 0 TBC = 35 lb. ft TDE = 25 lb. ft τ BC = 35 lb. ft τ BC =

(35)(0.358)(12) π (0.75)4

32 τ BC = 5.07 ksi τ DE = τ DE

(25)(0.375)(12) π (0.75)4

32 = 3.62 ksi

The solid shaft has a diameter of 0.75 in. If it is subjected to a torque shown, determine the maximum shear stress developed in regions CD and EF of the shaft. The bearings at A and F allow free rotation of the shaft.

TEF = 0 TCD = 15 lb. ft τ EF = 0 τ CD =

(15)(0.375)(12) π (0.75)4

32 τ CD = 2.17 ksi

5-49. The splined ends and gears attached to the A-36 steel shaft are subjected to the torque shown. Determine the angle of twist of end B with respect to the end A. The shaft has a diameter of 40 mm.

5-58. The engine of the helicopter is delivering 600 hp to he rotor shaft AB when the blade is rotating at 1200 rev/min. Determine the nearest 1/8 in. diameter of the shaft AB if the allowable shear stress is τ allow= 8 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft. long and made from 1.2 steel. 5-59.The engine of the helicopter is delivering 600 hp to he rotor shaft AB when the blade is rotating at 1200 rev/min. Determine the nearest 1/8 in. diameter of the shaft AB if the allowable shear stress is τ allow= 10.5 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft. long and made from 1.2 steel. 5-6. the solid 1.25 in diameter shaft is used to transmit the torques applied to the gears. If it is supported by smooth bearings at A and B, which do not resist torque, determine the shear stress developed in the shaft at points C and D. Indicate the shear stress on volume elements located at these points. 5-7. The shaft has an outer diameter of 1.25 in and an inner diameter of 1 in. If it is subjected to the applied torques as shown, determine the absolute maximum shear stress developed in the shaft. The smooth bearings at A and B do not resist torque.

5-50.The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Determine the angle of twist of gear C with respect to gear D. The shaft has a diameter of 40 mm.

6-1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exerts only vertical reactions on the shaft.

A

250 mm

B

800 mm

24 kN RB

ΣM B = 0

0.8 R A = 24(1.05) R A = 31.5kN RA 7.5 kN

ΣFY = 0 RB = −24 = 31.5 RB = 7.5kN

M A L = −24(0.25) M AL = −6kN .m

M BL = [(− 24)0.25 + 31.5(0.8)]kN .m M BL = 0

-24 kN

Vo = −24kN

V A = (− 24 + 31.5)kN

V A = 7.5kN -6 kN⋅m

VBL = 7.5kN

VB = [7.5 − 7.5]kN

VB = 0

6-2. The load binder is used to support a load. If the force applied to the handle is 50lb, determine the tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm ABC. T1

C

A B

ΣM c = 0

50 lb

50(15) = 3T1

12 in.

3 in.

T2

200 lb

T1 = 250lb ΣFY = 0 T2 = 250lb − 50lb T2 = 200lb V A = −50lb VBL = −50lb

VB = (− 50lb + 250lb ) VB = 200lb

-50 lb

VC L = 200lb

VC = (200 − 200)lb VC = 0

-600 lb⋅in

5- 2. The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-half of the applied torque (T/2).

r

T

r'

5-3. The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-quarter of the applied torque (T/4).

r

r'

T

5-5. The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and the three torques are applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe’s outer surface. Sketch the shear stress on the volume elements located at A and B.

B

C

450 lb-ft

A 350 lb-ft 600 lb-ft

5-10. The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear stress distribution over the cross section. 600 N 75mm 5mm 25mm 75mm

600 N

5-13. A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev/min. Determine the inner diameter d of the tube to nearest 1/8 in. if the allowable shear stress is τallow = 10 ksi.

d 2.5 in.

5-17. The steel shaft has a diameter of 1 in. and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft if the couple forces have a magnitude of F = 30 lb. 12in

8in. F

8in.

F

5-21. The 20-mm-diameter steel shafts are connected using a brass coupling. If the yield point for the steel is (τY)st = 100 MPa, determine the applied torque T necessary to cause the steel to yield. If d = 40 mm, determine the maximum shear stress in the brass. The coupling has an inner diameter of 20 mm. d 20mm

T

T

5-22. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d. R

r

T T

5-29. The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function N⋅m/m, where x is in meters. Determine the minimum torque T0 needed t = 25 x x to overcome friction and cause it to twist. Also, determine the absolute maximum stress in the shaft.

(

2

)

To 80mm

x 2m

(

t = 25 xe x

2

)

5-30. The solid shaft has a linear taper from rA at one end to rB at the other. Derive and equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis. T

T

rB B

A rA

x

L

5-35. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. If it is rotating at 200 rad/s. determine its largest inner diameter to the nearest 1/8 in. If the allowable shear stress for the material is τallow = 25 ksi. A

B

6 in.

5-36. The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 6 ksi. If the outer diameter is 3 in. and the engine delivers 175 hp to the shaft when it is turning at 1250 rev/min. determine the minimum required thickness of the shaft’s wall. A

B

6 in.

5-37. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi. A

B

6 in.

5-38. The 0.75 in. diameter shaft for the electric motor develops 0.5 hp and runs at 1740 rev/min. Determine the torque produced and compute the maximum shear stress in the shaft. The shaft is supported by ball bearings at A and B. A

B

6 in.

5-42. The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

B A

5-74. A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N⋅m. If the steel portion has a diameter of 30mm. determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa. C

B 680 N⋅m

1.60 m

A

0.75 m

5-78. The composite shaft consists of a mid section that includes the 1-in-diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 100-lb.ft. The material is A36 steel. d 20mm

T

T

6-3. Draw the shear and moment diagrams for the shaft. The bearings at A and D exert only vertical reactions on the shaft. The loading is applied to the pulleys at B and C and E. 14 in.

20 in.

15 in.

12 in.

A

E

B

C

80 lb

110 lb

D

35 lb

82.4 lb 35 lb 2.24 lb

ΣM A = 0

49 RD = 80(14) + 110(34) + 35(61) RD = 142.76lb ΣFY = 0

107.76 lb

R A = 80 + 110 + 35 − 142.76

1196 lb.ft

R A = 82.24lb V A = 82.24lb

1151.36 lb.ft

VB = 2.24lb VC = −107.76lb -420.24 lb.ft

VD = 35lb VE = 0 ΣM BL = (82.24)(14)lb.in M BL = 1151.36lb.in

ΣM CL = (82.24)(34) − 80(20)

6-4. Draw the shear and moment diagrams the beam.

M C L = 1196.16lb.in

ΣM DL = (82.24)49 − 80(35) − 110(15) M DL = −420.24lb.in

ΣM E L = {82.24(61) − 80(47 ) − 110(27 ) M EL

+142.76(12)}(lb.in ) =0

for

2 kip

4 ft

2 kip

4 ft

2 kip

4 ft

2 kip

4 ft

4 ft

4 kip 2 kip -2 kip -4 kip 24 kip.ft 16 kip.ft

16 kip.ft

By symmetry : 2(4 ) RA = RF = (2) RA = RF = 4kip ΣM B L = 16kip. ft

ΣM C L = (4 )8 − 2(4 ) M C L = 24kip. ft

ΣM D L = 4(12) − 2(8) − 2(4 ) M D L = 24kip. ft

M E L = 4(16) − 2(12 ) − 2(8) − 2(4 ) M E L = 16kip. ft

ΣM FL = 4(20) − 2(16) − 2(12) − 2(8) − 2(4 ) M FL = 0

6-5. Draw the shear and moment diagrams for the rod. It is supported by a pin at A and a smooth plate at B. The plate slides w/in the groove and so it cannot support a vertical force, although it can support a moment. 15 kN

A B

4m

2m

15 kN

ΣFY = 0 R A = 1.5kN

60 kN.m 30 kN.m

ΣM B = 15(4 ) − 15(2 ) M B = 30kN .m ΣM C L = 60kN .m

6-6. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. Also, express the shear and moment in the shaft as a function of x w/in the region 125mm<x<725mm. 1500 N

800 N

B

A

x 125 mm

600 mm

75 mm

815.625 N

ΣM B = 0

15.625 N

800R A = 800(675) + 1500(75) R A = 815.625N ΣFY = 0 -1484.375 N

RB = 800 + 1500 − 815.625 RB = 1484.37 N

ΣM CL = 815.625(0.125) 111328 N.m 101953 N.m

M C L = 101.953N .m

ΣM DL = 815.625(0.725) − 800(0.6 ) M D = 111.328N .m

ΣM BL = 815.625(0.8) − 800(0.675) − 2500(0.075)

@ 125mm < x < 725mm

6-7. Draw the shear and moment diagrams for the beam.

V = 16.625 N

M = 815.625 x − 800(x − 0.125) M = (15.625 x + 100 )N .m

10 kN

8 kN

15 kN.m

2m

3m

18 kN 8 kN

ΣFY = 0 R A = 18kN

ΣM A = −10(2 ) − 8(5) − 15 M A = −75kN .m

-15 kN.m -39 kN.m

V A = 18kN

VC L = (18 − 10 )kN VC L = 8kN VB = 0

ΣM C L = −75 + 18(2 ) M CL = −39kN .m

ΣM BL = −39 + 8(3) M BL = −15kN .m ΣM B = 0

-75 kN.m

6-8. Draw the shear and moment for the pipe. The end screw is subjected to a horizontal force of 5kN. Hint: the reaction at the pin C must be replaced by equivalent loadings at point B on the axis of the pipe.

C

A 5 kN

80 mm B

400 mm

ΣM A = 0

0.4 RB = = 5(0.8) RB = 1kN ΣM B = 0 0.4 R A = (5 )(0.8)

-1 kN

R A = 1 kN V A = −1 kN -0.4 kN.m

VBL = −1 kN =0 MA =0

M BL = −1kN (0.4 m ) = −0.4kN ⋅ m M B = −0.4kN ⋅ m + 0.4kN ⋅ m =0

6-9. Draw the shear and moment diagrams for the beam. Hint: the 20kip load must be replaced by equivalent loadings at point C on the axis of the beam.

ΣM A = 0

12 RB = 15(4 ) − 20

ΣM B = 0

12 R A = 15(8) + 20

RB = 3.38 kip

R A = 11.67 kip

V A = 11.67 kip

VBL = −3.33 kip

VCL = 11.67 kip

VB = −3.33 + 3.33

VC = (11.67 − 15) kip = −3.33 kip MA =0

ΣM CL = 11.67(4 ) ΣM DL

= 46.67 kip ⋅ ft = 11.67(8) − 15(4 ) − 20

= 13.33 kip ⋅ ft M B = 13.33 − 3.33(4 ) =0

=0

6-10. The engine crane is used to support the engine, which has a weight of 1200 lb. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown.

3 ft

5 ft

A B

4 ft

C

1200 lb

1200

ΣM B = 0

5 R A = 2000lb(5 ft ) ΣFY = 0 RB = 2000lb + 1200lb

-2000

RB = 3200lb V A = −2000lb -6000

V AL = −2000lb VB = −2000lb + 3200lb = 1200lb VCL = 1200lb VCL = 1200lb − 1200lb =0 ΣM A = 0

2000(3) 2 = −6000lb ⋅ ft ΣM CL = (− 6000 − 6000)lb ⋅ ft

ΣM BL = −

=0

6-12. Draw the shear and moment diagrams for the compound beam w/c is pin connected at B. It is supported by a pin at A and a fix wall at C.

6 kip

8 kip

A C B

4 ft

6 ft

Mo = 0

M AL = (− 6 kip )(4 ft )

4 ft

4 ft

M DL

4

M CL

-4 -6 16

= −24 kip ⋅ ft = 40 kip ⋅ ft − 24 kip ⋅ ft = 16 kip ⋅ ft = 16 kip ⋅ ft − 16 kip ⋅ ft

=0 ΣM C = 0

14R A = 6(18) + 8(4 ) = 108 + 32 R A = 10 kip ΣFy = 0

-24

RC = 14 − 10 RC = 4 kip VO = −6 kip V AL = −6 kip V A = −6 kip + 10 kip = 4 kip VDL = 4 kip VD = 4 kip − 8 kip = −4kip VCL = −4 kip VC = −4 kip + 4 kip =0

6-15. Draw the shear and moment diagrams for the beam. Also, determine the shear and moment in the beam as a function of x, where 3ft<x<15ft. 1.5 kip/ft

50 kip.ft

A

B

x 3 ft

12 ft

ΣM B = 0

13.167

12 R A = 50 + (1.5)(12)(6) C

R A = 13.167kip -4.833

x

↑ ΣF = 0 RB = 1.5(12) − 13.167 RB = 4.833kip

7.79 kip.ft

By Similar ∆ : 18 13.167 = x 12 x = 8.778 ft M C = Area ∆ − 50

-50 kip.ft

13.167 (8.778) − 50 2 = 67.778 − 50 = 7.79 kip ⋅ ft =

@ 3 ft < x ≤ 15 ft : V = 13.167 − 1.5( x − 3) V = 17.7667 − 1.5 x M = 13.167(x − 3) −

1.5 ( x − 3)2 2 = 13.167 x − 39.5 − 0.75 x 2 − 6 x + 9

(

)

= 13.167 x − 39.5 − 0.75 x + 4.5 x − 6.75 2

M = −0.75x 2 + 17.667 x − 46.25

6-16. Draw the shear and moment diagrams for the beam.

800 lb/ft

A

B

800 lb/ft 8 ft

8 ft

-6400 51200 25600

ΣFy = 0

R A = 800(8) − 800(8) =0

VA = 0 VBL = −800(8) = −6400 lb

ΣM A = 800(8)(12 ) − 800(8)(4) = 51200 lb ⋅ ft ΣM B = 51200 − 800(8)(4 ) = 25600 lb ⋅ ft ΣM C = 51200 − 800(8)(12 ) − 800(8)(4 ) =0

VCL = −6400 + 6400 =0

6-17. The 50lb man sits in the center of the boat, w/c has a uniform width and weight per linear foot of 3lb/ft. Determine the maximum bending moment exerted on the boat. Assume that the water exerts a uniform distributed load upward on the bottom of the boat.

150 lb

7.5 ft

7.5 ft 75 lb

Σ Fy = 0 45 + 150 15 wo = 13 lb ft MA = 0 M B = Area ∆ w0 =

-75 lb

281.25 lb.ft 2°

= 75(0.75) 2°

1 lb ⋅ ft 2

MC = 0

VA = 0

VBL = 13(7.5) − 3(7.5) = 75 lb VB = (75 − 150) lb = −75 lb VCL = −75 + 75 =o

6-18. The footing supports the load transmitted by the two columns. Draw the shear and moment diagrams for the footing if the reaction of soil pressure on the footing is assumed to be uniform. 14 kip

14 kip

6 ft

12 ft

6 ft C

B

A

7 kip

D

7 kip 6 ft

ΣFy = 0

E

24wo = 14 kip +14 kip -7 kip

kip 24 ft wo = 7 kip 6 ft = 28

-7 kip

21 kip.ft

21 kip.ft

2°

2° 2°

7 kip (6 ft ) ft 6 = 7 kip VB = 7 kip −14 kip VBL =

= −7 kip VCL = −7 kip +

7 kip (12 ft ) ft 6

VCL = 7 kip VC = 7 kip −14 kip = −7 kip VDL = −7 kip + kip =0

6-19. Draw the shear and moment diagrams for the beam. 2 kip/ft 30 kip.ft A

5 ft

B

5 ft

5 ft

ΣM B = 0

10 R A = 2(5)(12.5) − 30

-0.5 kip

R A = 9.5 kip

-10 kip 2.5 kip.ft

VO = 0 V AL = −2(5 ) = −10 kip

2° -25 kip.ft

V A = (− 10 + 9.5) kip = −0.5 kip

1° -27.5 kip.ft

MO = 0 1 (10)(5) 2 = −25 kip ⋅ ft

M AL = −

M CL = (− 27.5 + 30) kip ⋅ ft = 2.5 kip ⋅ ft

M BL = [2.5 − 5(0.5)] kip ⋅ ft M BL = 0 ΣF y = 0 RB = 2(5) − 9.5 RB = 0.5 kip VB = (− 0.5 + 0.5 ) kip VB = 0

6-21. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as a function of x, where 4ft<x<10ft. by symmetry: 150 lb/ft

200 lb.ft

200 lb.ft B

A x 4 ft

6 ft

4 ft

450 lb

ΣFy = 0 -450 lb 200 lb.ft

2°

2°

R A + RB = 150(6 )

900 2 R A = 450 lb = RB R A = RB =

V = 450 − 150(x − 4 )

2

= 1050 − 150 x -200 lb.ft

-200 lb.ft

150( x − 4 ) 2 2 = −3200 + 1050x − 75 x

M = −200 + 450( x − 4 ) −

2

6-23. The T-beam is subjected to the loading shown. Draw the shear and moment diagrams for the beam. ΣF y = 0

R A + RB = 2000 + 100(18)

R A + RB = 3800 lb ; R A = 3566.67 lb ΣM A = 0

2000(6 ) + RB (18) = 100(18)(9 ) RB = 233.33

ΣFy = 0

V = R A − 2000 − 100( x − 6 ) 0 = 3566.67 − 2000 − 100 x + 600 x = 21.67 ft VMAX = −2 kip M MAX = −12 kip ⋅ ft

6-24. The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributed loading on the beam over its 2ft length. Draw the shear and moment diagrams for the beam if it supports a uniform loading of 2kip/ft. ΣM A = 0

2 wO (10) = 2(8)(5) wO = 4 kip

ft

ΣFy = 0 R A = 16 − wO = 16 − 8 R A = 8 kip V A = 8 kip

VE = (8 − 16) kip = −8 kip VB = (− 8 + 8) kip =0 16 8 = 8 x x = 4 ft MA = 0

ΣM C = 8(1) kip ⋅ ft

ΣM D = [8 + 2(4 )(2 )] kip ⋅ ft = 24 kip ⋅ ft ΣM D = [8(9 ) − 2(8)(4 )] kip ⋅ ft = 8 kip ⋅ ft ΣM B = 8(11) − 2(8)(6 ) + 4(2 )(1) =0

6-29. Draw the shear and moment diagrams for the beam. ΣM B = 0 1 Wo L  3 L  1 Wo L  1 L  ⋅  ⋅ +  +  2 2 2 2 2 2 6 2 W L2 W L2 LR A = o + o 12 6 WL RA = o 4 ΣFY = 0 LR A =

1 Wo L 1 Wo L 1 ⋅ + − ⋅ Wo L 2 2 2 2 4 Wo L RB = 4 RB =

Mmax L/2

Wo L L Wo L L ⋅ − ⋅ 4 2 4 6 W L = o 12

ΣM max =

6-30. Draw the shear and moment diagrams for the beam. ΣM B = 0 WL L 2 ⋅ LR A = o ⋅ 3 2 3 W L RA = o 4 ΣFY = 0 W L W L RB = o − o 2 4 W L RB = o 4 3 y A Wo = L L W L2 ΣM AL = − o 89 L 3

M RA x

WO y = L x ΣF y = 0 1 xy 2 W L W x2 0= O − O 4 2L L x 2 = − 2L 4 L x= 2 W  L L  1 L  WO L  WO L  L  ΣM CL = O  − − ⋅ ⋅     4  2 3 2 2  L 2  L 2  3 2  = 0.0934WO L2 − 0.11785WO L2

V = RA −

= −0.0244WO L2

6-31. Draw the shear and moment diagrams for the beam. ΣM B = 0 1  L  L 2 L  WO L  L L  1 WO L  2 L  LR A = WO   + +  + +   2  3  9 3  3  3 3 2 3 3 3 7WO L2 WO L2 WO L2 = + + 54 6 27 W L R A = O = RB 3 WO L = RA 3 W L 1 WO L VC = O − 3 2 3 WO L = 6 WO L WO L VD = − 6 3 W L =− O 6

VA =

by similar ∆ : WO L WO L 3 = 6 L x 3 W L WO = O 6x L x= 6

WO L WO L − 6 6 WO L =− 3 W L W L VB = R B = O − O 3 3

VB = −

VB = 0

ΣM C =

WO L  L  WO L  L   −   3  3 6 9

5WO L2 = 54 WO L  L L  WO L  L L  WO L  L   + −  + −   3 3 6 6 9 6 6  12  W L2 5W L2 W L2 = O − O − O 6 108 72 2 23WO L = 216

ΣM MAX =

ΣM D = =

23WO L2 1 WO L  L  −   216 2 6 6 5WO L2 54

WO L (L ) − WO L  L + 2L  − WO L  L + L  − WO L  2L  6 6 9 3  3 6 3 6  9  =0

ΣM B =

6-32. The ski supports the 180lb weight of the man. If the snow loading on its bottom surface is trapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for the ski. ΣFY = 0 1 1 180 = W (1.5) + W (1.5) + 3W 2 2 3W 180 = + 3W 2 (180)2 W= 9 = 40 lb ft 1 (40) 3  2 2 = 30 lb

3 (40) 2 = −30 lb

VB =

VC = −90 +

3 (40) − 180 2 = 90 − 180 VE = −90 lb

VD =

VE = 30 +

1 (40) 3  − 30 2 2 =0

1 (40) 3  3 ⋅ 1  2  2  2 3  = 15 lb ⋅ ft

ΣM B =

1 (40)(1.5) 1 + 3  + 3 (40) 3  2 2 2 2 4 = 105 lb ⋅ ft

ΣM E =

 90 + 30  ΣM C = 105 −  (1.5)  2  = 15 lb ⋅ ft ΣM D = 15 − =0

30 (1.5) 3

6-33. Draw the shear and moment diagrams for the beam. 50 kN/m

50 kN/m

B

A

4.5 m

4.5 m

112.5 kN 2°

2° -112.5 kN

168.75 kN.m

ΣM A = 0

(4.5) + 50 (4.5)(4.5) 2 + 4.5 1 9 R B = (50)   2 3 2 3 = 168.75 + 843.75 RB = 112.5 kN by symmetry : R A = RB = 112.5 kN 2

V A = 112.5 kN

VBL = (112.5 − 112.5) kN

VCL

=0 = 0 − 112.5 kN

= −112.5 kN VC = (− 112.5 + 112.5) kN =0 MA =0 112.5 (4.5) = 168.75 kN ⋅ m 3 112.5 (4.5) = 0 = 168.75 − 3

M BL = M CL

6-35. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4kN/m caused by bar C. Determine the intensity of the distributed load wo of the leaves on the pin and draw the shear and moment diagrams for the pin. ΣF y = 0 1  2  WO (0.2) = [0.4(0.06)] 2  WO = 1.2 kN m

0.4 kN/m

VA = 0 VBL

VCL

 1 =  (1.5)(0.02) kN  2 = 0.012 kN = [0.012 − 0.4(0.06)] kN = −0.012 kN

  1.5 VDL = − 0.012 +  (0.02)  kN   2  =0 MA = 0 M BL =

0.012 (0.02) = 0.00008 kN ⋅ m 3 0.012 (0.03) 2 = 0.00026 kN ⋅ m

M MAX = 0.00008 +

wo

wo 20 mm

60 mm

20 mm

0.012 kN 2°

30 mm

2° 0.012 kN

2.6x10-3 kN⋅m

6-36. Draw the shear and moment diagrams for the beam. 3 kip/ft 1500 lb.ft A B 12 ft

6 ft 900

ΣM A = 0 2°

3000 (6)(2 + 12) 2 127500 = 12 RB = 10625 lb RB = 1500 +

-1625

ΣFy = 0 1500

R A = 10625 −

3000 (6) 2

R A = 1625 lb 3° -18000

V A = −1625 lb V A = −1625 lb

VB = (− 1625 + 10625) lb VB = 9000 lb VCL = 9000 − [300(6 )(0.05)] =0 M A = 1500 lb ⋅ ft

M B = −1625(12 ) + 1500 M B = −18000 M C = −1800 + =0

9000 (6) 3

6-37. The compound beam consists of two segments that are pinned together at B. Draw the shear and moment diagrams if it support the distributed loading shown.

ΣM C = 0

WL(L ) 6 WL RA = 6 WL WL − RC = 2 6 WL RC = 3 @ V = 0; x? 2WL Wx = 6 L x= 3 LR A =

V=

WL Wx − 6 2

@ x = L/3 WL L ⋅ M MAX = 6 3 3 2 WL = 54 WL2 WL2 M CL = − 6 6 =0

6-38. Draw the shear and moment diagrams for the beam.

18 kN/m 12 kN/m

A

B 3m

ΣFy = 0  12 + 18  RB =  3  2  RB = 45 kN

2° -45 kN

ΣM B = 12(3)(1.5) +

1 (6)(3)(1) 2

= 54 kN ⋅ m 3°

V = −12 x − -54 kN.m

6 x 2

V = −15 x @ x=3 VB = −45 kN 12 x 2 6  x  − x  2 2 3 = −6 x 2 − x 2

M =−

M = −7 x 2 @ x=3 M B = −54 kN ⋅ m

6-39. Draw the shear and moment diagrams for the beam and determine the shear and moment as a function of x.

ΣM B = 0 6 R A = 200(3)(1.5) +

1 (200)(3)(1) 2 = 900 + 300 = 1200

R A = 200 N ΣFY = 0  600  RB =  3 − 1200  2  RB = 700 N 1 ( x − 3) 200 ( x − 33) 2  3  200 (x − 3)2 = 200 − 200 x + 600 − 6 100 2 = 500 − x =0 3 500 x2 = (3) 100 x = 15 x = 3.8

V = 200 − 200(x − 3) −

200 (x − 3)2 − 1 (x − 3)2  200  x − 3  2 2  3  3  100 2 ( x − 3)3 = 200 x − 100( x − 3) − 9 100 3 M =− x + 500 x − 600 9

M = 200 x −

@ x=0 V A = 200 N ; M A = 0 @ x = 3m; VCL = 200 N ; M CL = 600 N ⋅ m @ x = 3.8m; VDL = 0 ; M DL = 690.3 N ⋅ m @ x = 6m ; M BL = 0 VBL = −700 N VB = −700 + 700 =0

(Mindanao State University-General Santos City, Philippines) BSEE-(G.R.F) students

6-43. A member having the dimensions shown is to be used to resist an internal bending moment of M = 2 kip⋅ft. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b) about the y axis. Sketch the stress distribution for each case. y

z

x

6 in.

6 in.

a) about z axis 1 I z = ( )(6)(123 ) 12 I z = 864in 4 2(6)(12) 864 σ z = 0.167 kip / in 2 (1000lb / kip ) σz =

σ z = 167 psi (1)(12)(6 3 ) 12 I y = 216in 4 Iy =

( 2)(3)(12) 216 σ y = (0.333kip / in 2 )(1000lb / kip ) σy =

σ y = 333 psi

b) about y axis

6-45. A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (σallow)t = 22 ksi and (σallow)s = 15 ksi, respectively.

4 in. 4 in.

M

2 in. 2 in.

Mc I σ I M= t c I bh 2 = c 24

σ=

σcI c ( 4)( 2 3 ) 2 M = 15( ) 24 M = 30kip.in M =

bh 2 ) 24 ( 4)( 2 3 ) 2 M = 22( ) 24 M = 44kip.in M = 22(

We used the smaller value  ft  M = 30kip.in   12in  M = 2.5kip. ft

6-46. A member has the triangular cross section shown. If a moment of M = 800 lb⋅ft is applied to the cross section, determine the maximum tensile and compressive bending stresses in the member. Also, sketch a three dimensional view of the stress distribution acting over the cross section.

4 in. 4 in. M 2 in. 2 in.

Mc1 I 3 bh I= 36 ( 4)( 2 3 ) 3 I= 36 I = 4.62in 4 2h C1 = 3 22 3 C1 = 3 4 3 C1 = 3 C1 = 2.31in

σc =

( )

σc =

(9600(2.31))

4.62 σ c = 4800lb / in 2 σ c = 4.8ksi

h 3 2 3 C2 = 3 C2 = 1.15in C2 =

( )

σt =

(9600(1.15))

4.62 σ t = 2400lb / in 2 σ t = 2.4 ksi

6-49. A beam has the cross section shown. If it is made of steel that has an allowable stress of σallow =24 ksi, determine the largest internal moment the beam can resist if the moment is applied (a) about the z axis, (b) about the y axis. y 0.25 in. 3 in. 3 in.

0.25 in.

z

0.25 in. 3 in. 3 in.

a) about z-axis Mc I  (0.6 ) 0.253 I z = 2 + (0.6 )(0.25 ) 3.1252 12 

σ =

(

)

I z = 33.8125in

(

4

M z = 24(33.8125) / (3 + 0.25 ) M z = 249.7 kip.in M z = 20.8kip. ft

b) about y-axis

(

)

(

)

 (0.25 ) 0.6 3   (0.6 ) 0.25 3  I y = 2  + 12 12     4 I y = 9in

24(9 ) 3 M y = 72.06 kip.in

My =

M y = 6.0 kip. ft

) +  (0.2512)(0.6 ) 3

 



6-50. The beam is subjected to a moment of M = 40 kN⋅m. Determine the bending stress acting at points A and B. sketch the results on a volume element acting at each of these points. A B M = 40 kN⋅m

50 mm 50 mm 50 mm

(

50 mm 50 mm 50 mm

)

(

)

 (0.05 ) 0.153   (0.05 ) 0.053  I = I y = Iz =   + 2  12 12     −5 4 I = 1.51(10 ) m − Mz y M yz σA = + Iz Iy  (40 )(0.075 )(1000 ) σA = 0+   1.5(10 −5 )   σ A = 199 MPa

 (40 )(0.025 )(1000 )  σB = 0+   1.51(10 −5 )   σ B = 66.2 MPa

6-73. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 160 MPa. A

B

ΣM = 0 R A (1.2 ) = 600(2 ) − 400(0.6 ) 0.8 600

1.2

0.6

R A = 800 N 400

ΣM A = 0

RB (1.2 ) = 400(1.8) − 600(0.8) RB = 200 N

Mr πr 4 4 4M σ= 3 πr 4M r3 = πσ (4)(480) r =3 π 160(10 6 ) σ=

(

r = 0.0156m d = 31.3mm

)

6-81. The beam is subjected to the load P at its center. Determine the placement a of the supports so that the absolute maximum bending stress in the beam is as large as possible. What is this stress? P

d a

a

b L/2

L/2

+ ↑ ΣFy = 0 R AY + R BY = P R AY = R BY

For symmetrical loading

2 R AY = P

a = 0, L

P R AY = 2 ΣM c = 0

σ max =

MA = MB MA =

P  L    −a 2  2  

MB =

P  L    −a 2  2  

0≤a≤

L 2

@a = 0 PL   − 0 22  PL MA = 4 L @a = 2 P  L L   M A =  −   2  2 2   MA =

M max =

PL 4

(M max )(c ) I

3

bd 12 d c= 2 I=

σ max

σ max

PL  d    4 2 = bd 3 12  PL  12 =   2  8  bd

σ max =

( )

3 PL 2bd 2

  

6-89. The steel beam has the cross-sectional area shown. If w = 5 kip⋅ft, determine the absolute maximum bending stress in the beam. w

w

8 ft

8 ft

8 ft

8 in. 0.30 in. 0.30 in.

10 in. 0.30 in.

ΣM A = 0

24 R B = 40(4) + 40(2 ) RB = 40kip + ↑ ΣFy = 0 R A + R B = 80 R A = 40kip

(

)

( )

  (0.3) 10 3   (8) 0.30 3 I = 2 + 8(0.3) 5.15 2  +     12   12 I = 152.3in 4 (160)(12)(5.3) σ max = 152.3 σ max = 66.8 ksi

(

)

6-90. The beam has a rectangular cross section as shown. Determine the largest load P that can be supported on its overhanging ends so that the bending stress does not exceed σmax = 10 MPa. P

P 1.5 m

1.5 m

1.5 m

ΣM A = 0 1.5 RB + 1.5 P = 3P RB = P + ↑ ΣFy = 0 R A + R B = 2P RA = P Mc I (0.5) 0.253 I= 12 I = 1.953(10 −4 )m 4 1.5 P(0.125) 10(10 6 ) = 1.953(10 − 4 ) P = 10.4kN σ max =

(

)

250 mm 150 mm

6-91. The beam has the rectangular cross section shown. If P =12 kN, determine that absolute maximum bending stress in the beam. Sketch the stress distribution acting over the cross section. P

P 1.5 m

1.5 m

1.5 m

ΣM A = 0

RB (1.5) + 12(1.5) = 12(3) RB = 12kN

+ ↑ ΣFy = 0 R A + R B = 24 R A = 12kN I=

(0.15)(0.253 )

12 I = 1.953125(10 −4 ) m 4 18(0.125) σ max = 1.953125(10 −4 ) 18000(0.125) σ max = 1.953125(10 −4 ) σ max = 11.5MPa

250 mm 150 mm

6-93. A log that is 2 ft in diameter is to be cut into a rectangular section for use as a simply supported beam. If the allowable bending stress for the wood is σallow = 8 ksi, determine the largest load P that can be supported if the width of the beam is b = 8 in.

h b 2 ft P

8 ft

8 ft

6-102. The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 45°. z 125 mm 125 mm 250 mm

B

E M = 850 N⋅m

C

θ

A D

y

6-103. The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 30°. z 125 mm 125 mm 250 mm

B

E M = 850 N⋅m

C

θ

A D

y

6-105. The T-beam is subjected to a moment of M = 150 kip⋅in. directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis. The location y of the centroid, C, must be determined.

6-185. Determine the bending stress distribution in the beam at section a-a. sketch the distribution in three dimension acting over the cross section. 80 N

80 N

a a 400 mm

300 mm

300 mm

400 mm

80 N

80 N 15 mm 100 mm 15 mm

15 mm 75 mm

6-51. The aluminum machine part is subjected to a moment of M = 75 N⋅m. Determine the bending stress created at points B and C on the cross section. Sketch the results on a volume element located at each of these points. 20 mm 10 mm

10 mm 10 mm

20 mm

10 mm N B 10 mm C 40 mm

M = 75 N⋅m

6-53. A beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is M = 450 N⋅m, determine the resultant force that bending stress produces on the top board A and on the side board B. A

B 15 mm M = 40 kN⋅m 20 mm 200 mm 15 mm

20 mm 200 mm

6-47. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N⋅m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section. 25 mm

20 mm 200 mm 20 mm

M = 600 N⋅m

6-106. If the internal moment acting on the cross section of the strut has a magnitude of M = 800 N⋅m and is directed as shown, determine the bending stress at points A and B. The location z of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis. y 12 mm

12 mm

200 mm

z 12 mm

60°

150 mm

M = 800 N⋅m

z

200 mm

6-71. The axle of the freight car is subjected to wheel loadings of 20 kip. If it is supported by two journal bearings at C and D, determine that maximum bending stress developed at the center of the axle, where the diameter is 5.5 in.

C

A

10 in. 20 kip

B

60 in.

D

10 in. 20 kip

6-74. Determine the absolute maximum bending stress in the 1.5-in.-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces. 400 lb A 300 lb B

1 in. 18 in. 15 in.

6-75. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 22 ksi. 400 lb A 300 lb B

1 in. 18 in. 15 in.

6-72. Determine the absolute maximum bending stress in the 30-mm-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces. A

0.8 m 600 N

B

1.2 m

0.6 m 400 N

6-79. The steel shaft has a circular cross section with a diameter of 2 in. It is supported on smooth journal bearings A and B, which exert only vertical reactions on the shaft. Determine the absolute maximum bending stress if it is subjected to the pulley loadings shown.

A

B

20 in.

20 in.

500 lb

20 in.

300 lb

20 in.

500 lb

The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-half of the applied torque (T/2).

r

T

r'

by ratio and proportion: τ max τ = r r' T 2T = 4 4 πr πr ' 4 2r ' = r 4 r '4 =

r4 2

r4 2 r ' = 0.841r

r' = 4

r ' = r cos 32.7651

The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-quarter of the applied torque (T/4).

r

T

T 2( ) T τ = 23 = 3 πr ' πr ' 2T τ max = 3 πr b.) T 2πr '3 τ max τ = r r' 2T T = 4 4 πr πr ' 4 2r ' = r 4

τ =

r'4 =

r4 4

r4 4 r ' = 0.707r r ' = r sin 45 r' = 4

r'

The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and the three torques are applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe’s outer surface. Sketch the shear stress on the volume elements located at A and B.

B

A

C

450 lb-ft

450 lb⋅ft = 5400 lb⋅in 350 lb⋅ft = 4200 lb⋅in 600 lb⋅ft = 7200 lb⋅in

350 lb-ft 600 lb-ft

sol’n:

τmax 5400 + 4200 – 7200 = 2400

16TD π D4 − d 4 16(2400)(2.5) = π 2.5 4 − 2.34 = 2758.6 lb 2 in = 2.76 ksi

τB =

(

(

τB τB

)

7200 - 4200 = 3000

16TD π D4 − d 4 16(3000)(2.5) = π 2.5 4 − 2.34 = 3448.3 lb 2 in = 3.45 ksi

τA =

)

(

(

τA τB

)

)

The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear stress distribution over the cross section. 600 N 75mm 5mm 25mm 75mm

600 N

∑ M x = 0; 2(600)(0.075) = T T = 90 Nm π J= (D 4 − d 4 ) 32 π = (0.035 4 − 0.025 4 ) 32 J = 108 x10 −9 m 4 Tr τ max = J (90)(0.075) Nm 2 = (108 x10 −9 )m 4 = 14.45MPa Tr τ = J (90)(0.0125) = (108 x10 −9 ) = 10.40MPa

A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev/min. Determine the inner diameter d of the tube to nearest 1/8 in. if the allowable shear stress is τallow = 10 ksi.

d 2.5 in.

Given:

D = 2.5in

τ allow 10,000ld / in 2 Ρ = 3hp = 19,250lb. ft / s f = 2700rev / min = .75rev / s d =?

Solution:

Ρ = Tω 19250 = T [ 2π (0.75)] 19250 T= 2π (0.75) T = 4084.97lb.in 4084.97(1.25) τ max = π (1.25 4 − r 4 ) 2 (4084.97)(1.25)   4 π 4 π (1.25) 2 − ( r ) 2 =  10000   ( 4084097 )( 1 . 25 )( 2 ) πr 4 = (1.25 4 )π − 10000 6 . 649 r4 = π r = 1.206 d =r 2 d = 2.4in

1 d = 2 in 2

;

4

1 = 0.5 8

2

The steel shaft has a diameter of 1 in. and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft if the couple forces have a magnitude of F = 30 lb. 12in

8in. 8in.

F

F

sol’n:

∑M

A

AB D = 0.75 in d = 0.68 in BC D = 1 in d = 0.86 in

= 15(6 ) + 15(8) = T T = 210 lb ⋅ in

16TD π D4 − d 4 16(210)(0.75) = π 0.75 4 − 0.68 4 = 7818.72 lb 2 in = 7.8 ksi

τ AB =

(

(

τ AB τ AB

16TD π D4 − d 4 16(210)(1) = π 14 − 0.86 4 = 2361 lb 2 in = 2.36 ksi

τ BC =

)

)

(

(

τ BC τ BC

) )

The 20-mm-diameter steel shafts are connected using a brass coupling. If the yield point for the steel is (τY)st = 100 MPa, determine the applied torque T necessary to cause the steel to yield. If d = 40 mm, determine the maximum shear stress in the brass. The coupling has an inner diameter of 20 mm. d 20mm

T

T

To find the T; steel; τ =

2Ts πr 3

2Ts π (0.010) 3 Ts = 157.07 Nm

100 x10 6 =

To find the max.., τ; brass; τ =

2TR π (R − r 4 ) 2(157.07)(0.02) = π (0.02 4 − 0.014 ) 4

τ = 13.3MPa

The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d. R

r

T T

Soln: P A πd 2 ⋅τ P = 4 T = PRn

τ =

 πd 2  ⋅ τ  Rn T =   4  T 4 τ = ; <1> π d 2 Rn 2T τ = ; < 2> πr 3 2T 4T = 3 πr π d 2 Rn 2r 3 n = d2

The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function N⋅m/m, where x is in meters. Determine the minimum torque T0 needed t = 25 x x to overcome friction and cause it to twist. Also, determine the absolute maximum stress in the shaft.

(

2

)

To 80mm

x 2m

(

t = 25 xe x

2

)

Solution: d=80mm=0.08 m

(

t = 25 xe x 16T τ = 3 πd

2

)

x2

T = ∫ 25 xe dx 2

0

2

T = 25∫ xe x dx 2

0 2

let u= x du=2x dx du/2=x dx x=0, u=0 x=2,u=4 T =∫

4

0

25e u du 2

4

= 12.5∫ e u du 0

[ ] = 12.5[e − 1] = 12.5 e u

4

0

4

T = 670 Nm

16T πd 3 16(670) = π (80 x10 −3 ) 3

τ =

τ = 6.66MPa

The solid shaft has a linear taper from rA at one end to rB at the other. Derive and equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis. T

T

rB B

A rA

x

L

Solution:

r = rB + r1 r1 r −r = A B L−x L (r − r )( L − x) r1 = A B L ( rA − rB )( L − x ) L rB L + ( rA − rB )( L − x) L rB L + ( L − x )rA − rB ( L − x ) L rB [ L − ( L − x)] + rA ( L − x) L rB ( x ) + rA ( L − x ) L xrB + rA ( L − x ) L

r = rB +

r= r=

r= r= r=

τ =

16T d ; r= 3 2 πd d = 2r

d 3 = 8r 3

τ =

16T π 8r 3

= τ =

τ =

2T πr 3 2T

 xr + rA ( L − x)  π B  L   3 2TL π [xrB + rA ( L − x)]

3

3

The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow = 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev/min, determine the minimum required thickness of the shaft’s wall.

Given: τ = 8ksi = 8000 psi D = 2.5in Ρ = 200hp = 100,000lbft / s f = 1140rev / min = 19rev / sec

t =?

Solutions:

Ρ 2πf 100,000lbft / s = 2π (19rev / s ) T = 837.66lbft or = 10,051.89lbin T=

16TD π (D 4 − d 4 ) 16(10,051.89)(2.5) 8,000 psi = π ( 2.5 4 − d 4 ) 402075.65 (2.5 4 − d 4 ) = π (8,000)

τ =

d 4 = 39 − 15.998 d = 4 23 d = 2.19in

Thickness: R=

D ; 2

r=

d 2

2.5 2.19 r= 2 2 R = 1.25 r = 1.1

R=

t = R−r t = 1.25 − 1.1 t = 0.15in

The drive shaft AB of an automobile is to be designed as a thin walled tube. The engine delivers 150 hp when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The material has an allowable shear stress τallow = 7 ksi.

Ρ 2πf 75,000lbft / s = 2π ( 25rev / s) T = 477.46lbft T = 5,729.58lbin T=

Given

Ρ = 150hp = 75,000lbft / s f = 1,500rev / min = 25rev / s D = 2.5in τ = 7 ksi = 7,000 psi t min = ?

Solution

16Td π (D 4 − d 4 ) 16(5729.58)(2.5) 7,000 = π (2.5 4 − d 4 ) 229,183 39 − d 4 = π (7,000)

τ =

d 4 = 39 − 10.42 d = 4 27.5968 d = 2.29in

Thickness: D 2 2.5 ; R= 2 R = 1.25in R=

t = R−r t = 1.25 − 1.15 t = 0.104in

d 2 2.29 r= 2 r = 1.146in r=

The drive shaft of a tractor is to be designed as a thin-walled tube. The engine delivers 200 hp when the shaft is turning at 1200 rev/min. Determine the minimum thickness of the wall of the shaft if the shaft’s outer diameter is 3 in. The material has an allowable shear stress of τallow = 7 ksi.

Given:

Ρ = 200hp = 100000lb. ft / s f = 1200rev / min = 20rev / s

D = 3in τ = 7 ksi = 7000 psi t max = ?

Solution: Ρ 2πf 100000lbft / s T= 2π ( 20rev / s ) T = 795.8lbft T=

T = 9549.3lbft

16TD π (D 4 − d 4 ) 16(9549.3)(3) 7000 = π (3 4 − d 4 )

τ =

81 − d 4 =

458366.4 π (7000)

d 4 = 81 − 20.84 d = 4 60.16 d = 2.785in

Max Thickness:

t = R−r t = 0.12in

D 2 R = 1.5 d r = = 1.3925in 2 R=

A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. If it is rotating at 200 rad/s. determine its largest inner diameter to the nearest 1/8 in. If the allowable shear stress for the material is τallow = 25 ksi. A

B

6 in.

p ω 250000 T= 200 T = 15000lb ⋅ in 16TD τ = π D4 − d 4 16(15000)(2 ) 25000 = π 24 − d 4 T=

τ = 25ksi = 25000lb / in 2 1hp = 500 ft ⋅ lb / s  500 ft ⋅ lb / s   = 250000 ft / lb 500hp hp  

(

)

(

d 4 = (16 − 6.11) d = 1.75in

)

The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 6 ksi. If the outer diameter is 3 in. and the engine delivers 175 hp to the shaft when it is turning at 1250 rev/min. determine the minimum required thickness of the shaft’s wall. A

B

6 in.

P 2πf 87500 = 2π (20.833) T = 668.45 ft.lb T = 8021.4lb.in 16TD τ = π D4 − d 4 16(8021.4 )(2.5) 6000 = π 2.5 4 − d 4 320856.87 39 − d 4 = π (6000) T=

(

)

(

)

d 4 = 21.98in d = 2.165in thickness; D 2.5 = 1.25 R= = 2 2 d 2.165 r= = = 1.08 2 2 t = R−r t = 0.167in

16TD π D4 − d 4 16(1322.2 )(4.5) = π 4.5 4 − 2 4

τ =

(

)

(

τ = 483.17lb / in 2

)

A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

A

B

6 in.

16TD π D4 − d 4 16(2T ) ; T = 11137.2lb.in 25000 = π 2 4 − 1.84 4 P ω= T 3 10 6 = 11137.2 ω = 296rad / s

τ =

(

)

(

( )

)

The 0.75 in. diameter shaft for the electric motor develops 0.5 hp and runs at 1740 rev/min. Determine the torque produced and compute the maximum shear stress in the shaft. The shaft is supported by ball bearings at A and B. A

P 2πf 260 = 2 π (29 ) T = 1 . 51 ft .lb T = 18 . 12 lb .in

T =

16 T πd 3 16 (18 . 12 = π 0 . 75 3 τ = 219 psi τ =

(

)

)

B

6 in.

The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

B A

(

16TD τπ D 4 − d 4 ; T = 16 D π D4 − d 4 P T= ω P τπ D 4 − d 4 = ω 16 D 6 3 10 (2500π ) 2 4 − 1.84 4 = ω 32 6 3 10 ω= 11137.22 ω = 296.4rad / s

τ =

(

)

(

( )

)

(

( )

)

)

The solid shaft of radius c is subjected to a torque T at its ends. Show that the maximum shear strain developed in the shaft is γmax = Tc/JG. What is the shear strain on an element located at point A, c/2 from the center of the shaft? Sketch the strain distortion of this element.

τ max

τ max γ max γ max

Tr = ;r=c J Tc = J = γG τ = G Tc = JG

Tρ c ; ρ= 2 j Tc = 2J

τ =

τ = γG γ =

τ G

γ =

Tc 2 JG

A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N⋅m. If the steel portion has a diameter of 30mm. determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa. C

B 680 N⋅m

1.60 m

0.75 m

sol’n: θAB = θBC

TL TL = JG AB JG BC

(680)(0.75) = (680)(1.6) 4 π d4 π (0.030) 9 (39 )(109 ) (75)(10 ) 32

32

d4 =

(1.6 )(0.030 4 )(75) (0.75)(39)

(

d = 4 3.3 10 −6

)

d = 0.0427 m d = 42.7 mm

m4

A

The composite shaft consists of a mid section that includes the 1-in-diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 100-lb.ft. The material is A-36 steel. d 20mm

T

T

5-14. The solid shaft has a diameter of 0.75 in. If it is subjected to a torque shown, determine the maximum shear stress developed in regions BC and DE of the shaft. The bearings at A and B allow free rotation of the shaft.

T AB = 0 TBC = 35lb. ft TDE = 25lb. ft

35(0.375)(12 ) π 0.75 4 32 = 8.07 ksi

τ BC = τ BC

τ DE = τ DE

(

)

(40)(0.375)(12)

(

π 0.75 4 32 = 3.62ksi

)

5-15. The solid shaft has a diameter of 0.75 in. If it is subjected to a torque shown, determine the maximum shear stress developed in regions CD and EF of the shaft. The bearings at A and F allow free rotation of the shaft.

TEF = 0 TCD = 15lb τ EF = 0

15(0.375)(12) π 0.75 4 32 = 2.17ksi

τ CD = τ CD

(

)

5-49. The splined ends and gears attached to the A-36 steel shaft are subjected to the torque shown. Determine the angle of twist of end B with respect to the end A. The shaft has a diameter of 40 mm.

( )

G = 75 10 9 N / m 2 TBD = 400 N .m TCD = 200 N .m T AC = 300 N .m d = 0.04m r = 0.02m TL θ =Σ JG 400(0.5)(0.2 ) 200(0.4)(2 ) 300(0.3)(2 ) + − θB/ A = 4 9 4 9 π 0.02 75 10 π 0.02 75 10 π 0.02 4 75 10 9 = 0.010079813rad (57.3) θ B / A = 0.0578°

(

)( ( ))

(

)( ( ))

(

)( ( ))

5-58. The engine of the helicopter is delivering 600 hp to he rotor shaft AB when the blade is rotating at 1200 rev/min. Determine the nearest 1/8 in. diameter of the shaft AB if the allowable shear stress is τ allow= 8 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft. long and made from 1.2 steel.

P 2πf 300000 = 2π (20) T = 2387.3lb. ft T = 28648lb.in

T=

16T πd 3 16(28648) d3 = π (8000)

τ =

1 d = 2.6in ; 0. 6  = 0.75 8 d = 2.75in

5-59.The engine of the helicopter is delivering 600 hp to he rotor shaft AB when the blade is rotating at 1200 rev/min. Determine the nearest 1/8 in. diameter of the

shaft AB if the allowable shear stress is τ allow= 10.5 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft. long and made from 1.2 steel.

P 2πf 300000 = 2π (20) T = 2387lb. ft

T=

T = 28648lb.in 16T τ = 3 πd 16(28648) d3 = π (10500) 1 d = 2.4in ; 0 .4  = 0.5 8 d = 2.5in

5-6. the solid 1.25 in diameter shaft is used to transmit the torques applied to the gears. If it is supported by smooth bearings at A and B, which do not resist torque, determine the shear stress developed in the shaft at points C and D. Indicate the shear stress on volume elements located at these points.

TC = 1500lb.in TD = 600lb.in Tr τ = J 1500(1625)(32) τC = π 1.25 4 τ C = 3.91ksi

(

)

600(0.625)(32) π 1.25 4 τ D = 1.56ksi

τD =

(

)

5-7. The shaft has an outer diameter of 1.25 in and an inner diameter of 1 in. If it is subjected to the applied torques as shown, determine the absolute maximum shear stress developed in the shaft. The smooth bearings at A and B do not resist torque.

16TD ; Tmax = 1500lb.in π D4 − d 4 16(1500)(1.25) = π 1.25 4 − 14 τ max = 6.62ksi τ max =

(

(

)

)

5-50.The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Determine the angle of twist of gear C with respect to gear D. The shaft has a diameter of 40 mm.

G = 75(10 9 ) N / m 2 TBD = 400 N .m TCD = 200 N .m T AC = 300 N .m d = 0.84m θ =Σ

TL JG

400(0.5)(2 ) 200(0.4 )(2 ) + 4 9 π 0.02 75 10 π 0.02 4 75 10 9 = 0.243°

θC / D = θC / D

(

)( ( ))

(

)( ( ))

T T

rB

x

L rA

ro

x L

The simply supported shaft has a moment of inertia of 2I for given region BC and a moment of inertia I for regions AB and CD. Determine the maximum deflection P

of the beam due to the load P. A

B

I

C

I

2I

L/4

L/4

L/4 PL

2 PL

PL

2

8

L/4

2 PL

8

16

D

2 2 PL

16

8

δ TP/A

∑M

D

= 0:

RA L = P

L 2

R A = RD =

TD / A TD / A

TB / A TB / A

P 2

5PL3 7 PL3 5 PL3 PL3 3PL3 5 PL3 = + + + + + 384 1536 512 384 512 1536 5PL3 = 128 PL3 PL3 PL3 = + + 192 512 1536 PL3 = 128

PL3 5 PL3 +δ 128 = 128 L L 2 3 3 PL δ = 256

TD/A

Determine the equation of the elastic curve for the beam using the x coordinate that is valid for 0 ≤ x ≤ L/2. Specify the slope at A and the beam’s maximum deflection. EI is constant. P

A

B x

L 2

ΣFy = 0 R A + RB = P ΣM A = 0 L RB L = P  2 P RB = 2 P RA = 2

L 2

ΣM = 0 p M = x 2 p EIy' ' = x 2 p EIy' = x 2 + C1 4 p 3 EIy = x + C1 x + C 2 12 @ y = 0, x = 0 C2 = 0 @ y' = 0, x =

L 2

2

0=

PL   + C1 42

C1 = −

PL2 16

P 3 PL2 x − x 12 16 P y= x 4 x 2 − 3L2 48EI L ymax @ x = 2 PL ymax = − 48EI

EIy =

(

)

@x =0 y =θ = 0+ θA =

− PL2 16 EI

C1 EI

Determine the equations of the elastic curve using the x1 and x2 coordinates. Specify the slope at A and the maximum deflection. EI is constant. P

P a

a A

B x1 x2 L

The beam is made of two rods and is subjected to the concentrated load P. Determine the slope at C. The moments of inertia of the rods are IAB and IAC, and the modulus of elasticity is E. P B C

A l L

Determine the elastic curve for the cantilevered beam, which is subjected to the couple moment Mo. Also compute the maximum slope and maximum deflection of the beam. EI is constant. A

B Mo x L

ΣFy = 0 RA = 0 ΣM A = 0 MO = MO

ΣM = 0 M = MO EIy ' = M O x + C1 EIy =

MO 2 x + C1 x + C2 2

@ y = 0, x = 0 C2 = 0 @ y ' = 0, x = L C1 = − M O L MO 2 x − M O Lx 2 ymax @ x = L

EIy =

M O L2 − M O L2 2 M L2 =− O 2 EI

EIymax = ymax

θ max @ x = 0 C1 EI M L =− O EI

y ' = θ max = 0 + θ max

Determine the deflection at the center of the beam and the slope at B. EI is constant.

M0

B A

x L

Σ M B = 0; RA L = −M O M RA = − O L Σ M A = 0; MO L M = −RA x + M O RB =

− M O x2 EIy ' = + M O x + C1 2L M x3 M x2 EIy = − O + O + C1 x + C 2 6L 2 @ x = 0, y = 0 C2 = 0 @ x = L, y = 0 − M O L3 M O L2 + + C1 L + 0 6L 2 M L C1 = − O 3 @x =0 0=

MOL 3 M L y' = − O 3 EI L @x = 2 M L3 M L2 M L2 EIy max = − O + O − O 48 L 8 6 2 3M O L y max = − 48 EI EIy ' = −

− M O x 3 M O x 2 M O Lx + − EIy = 6L 2 3 MO − x 3 + 3Lx − 2 L2 x y= 6 LEI

[

]

Determine the elastic curve for the simply supported beam, which is subjected to the couple moments Mo. Also, compute the maximum slope and the maximum deflection of the beam. EI is constant. M0

M0 x L

Determine the equations of the elastic curve using the coordinates x1 and x2, and specify the slope at A. EI is constant. A

B

C M0

x1

x2 L

L

ΣFy = 0; R A + RB = 0 R A = − RB ΣM A = 0; RB L = M O MO L − MO RA = L RB =

ΣM = 0; M M = O x L MO 2 x + C1 EIy ' = 2L M EIy = O x 3 + C1 x + C 2 6L @ y = 0, x = 0 C2 = 0 @ y = 0, x = L C1 = −

MOL 6

θA @ x = 0 EI θ = −C1 MOL ; 6 EI negative sign indicates below

θ =−

The floor beam of the airplane is subjected to the loading shown. Assuming that the fuselage exerts only vertical reactions on the ends of the beam, determine the maximum deflection of the beam. EI is constant.

80 lb/ft

2 ft

8 ft

2 ft

ΣFy = 0 R A + R B = 80(8) = 640 ΣM B = 0;

R A (12) = 80(8)(6)

R A = 320lb

18.77 kip. ft 3 EI 18.77 = kip. ft 3 EI

RB = 320lb

y max = −

 x − 2  x − 10  M = R A x + 80( x − 10)   − 80(x − 2)  2   2 

∆ max

EIy' ' = 320x + 40(x − 10) − 40( x − 2) 40 40 3 3 EIy' = 160x 2 + ( x − 10) − ( x − 2) + C1 3 3 160 3 10 10 4 4 EIy = x + ( x − 10) − (x − 2) + C1 x + C 2 3 3 3 @ x = 0, y = 0 C2 = 0 2

2

@ x = 12, y = 0 160 0= (12)3 + 10 (2)4 − 10 (10)4 + C112 3 3 3 14720 C1 = − lb. ft 3 3 160 3 10 10 14720 4 4 EIy = x + ( x − 10) − (x − 2) − x 3 3 3 3 y max @ x = 6 160 3 10 4 14720 (6) + (4) − (6) 3 3 3 56320 lb. ft 3 =− 3EI

EIy = y max

The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 3 kip/ft 5 kip-ft

5 kip-ft B

A 4 ft

8 ft

4 ft

ΣM B = 0

R A (8) − 3(8)(4) = 0 R A = 12kips

ΣFy = 0 R A + RB = 24 RB = 12kips  x − 12   x − 4 M = −5 + R A (x − 4) − 3( x − 4)   + RB ( x − 12) + 3( x − 12)  2   2  3 12 3 12 2 3 2 3 EIy' = −5 x + ( x − 4) − (x − 4) + ( x − 12) + (x − 12) + C1 6 2 6 2 2 1 1 5x 3 4 3 4 + 2( x − 4) − (x − 4) + 2(x − 12) + (x − 12 ) + C1 x + C 2 EIy = − 8 8 2 @ x = 8, y' = 0 C1 = −24 1 1 5x 2 3 4 3 4 + 2( x − 4) − (x − 4) + 2(x − 12) + (x − 12 ) − 24 x + C 2 8 8 2 @ x = 4, y = 0 C 2 = 136 EIy = −

y=

1 EI

 5x 2  1 1 3 4 3 4 + 2(x − 4) − ( x − 4) + 2(x − 12) + ( x − 12) − 24 x + 136 kip. ft 3 − 8 8  2 

The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft, and at C by a thrust bearing that exerts horizontal and vertical reactions on the shaft. Determine the equation of the elastic curve. EI is constant. C

A

B P x a

ΣM = 0 RB (a ) = P(a + b )

P(a + b ) a ΣFy = 0 RB =

R A = RB − P RA =

Pb a

M = − R A x + RB ( x − a ) Pbx P (a + b )(x − a ) EIy' ' = − + a a 2 2 Pbx P(a + b )( x − a ) EIy' = − + + C1 2a 2 3 Pbx 3 P(a + b )( x − a ) EIy = − + + C1 x + C 2 6a 6a @ x = 0, y = 0 C2 = 0 @ x = a, y = 0 − Pba 3 + C1 a 6a Pba C1 = 6 3 Pbx 3 P(a + b )( x − a ) Pba EIy = − x + + 6a 6a 6 3 1  Pbx 3 P(a + b )(x − a ) Pba  x y= − + +  EI  6a 6a 6  0=

b

The shafts support the two pulley loads shown. Determine the equation of the elastic curve. The bearing at A and B exerts only vertical reactions on the shaft. EI is constant.

B A x 20 in.

20 in.

20 in. 40 lb

60 lb

ΣM A = 0 R B (40 ) = 40(20 ) + 60(60 ) R B = 110lb ↑ ΣFy = 0 R B = R A + 60 + 40 R A = 10lb ↓ M = − R A x − 40( x − 20 ) + R B (x − 40 )

EIy ' ' = −10 x − 40(x − 20 ) + 110 (x − 40 )

− 10 x 2 40( x − 20 ) 110(x − 40 ) EIy ' = − + + C1 2 2 2 3 3 EIy = −1.67 x 3 − 6.67( x − 20 ) + 18.33( x − 40 ) + C1 x + C 2 2

2

@ x = 0, y = 0 C2 = 0 @ x = 40, y = 0

( )

( )

0 = −1.67 40 3 − 6.67 20 3 + 40C1 C1 = 4006 EIy = −1.67 x 3 − 6.67( x − 20 ) + 18.33( x − 40 ) + 4006 x 1 3 3 y= − 1.67 x 3 − 6.67 (x − 20 ) + 18.33( x − 40 ) + 4006 x lb.in 3 EI 3

[

3

]

The shafts support the two pulley loads shown. Determine the equation of the elastic curve. The bearing at A and B exerts only vertical reactions on the shaft. EI is constant. x A

12 in.

B

24 in.

24 in. 80 lb

50 lb

ΣM A = 0

50(12 ) + RB (48) = 80(24) RB = 27.5lb ΣFy = 0 R A + RB = 50 + 80 R A = 102.5lb M = −50 x + 102.5( x − 12) − 80(x − 36)

50 x 2 102.5( x − 12) 80( x − 36) EIy' = − + − + C1 2 2 2 3 3 EIy = −8.33x 3 + 17.1( x − 12) − 13.33(x − 36) + C1 x + C 2 2

2

@ x = 12, y = 0 0 = −14400 + 12C1 + C 2 @ x = 60, y = 0 0 = −95040 + 60C 2 equating eqn1 & 2 C1 = 1680 C 2 = −5760

EIy = −8.33x 3 + 17.1( x − 12) − 13.33(x − 36) + 1680x − 5760 3

y=

[

3

]

1 3 3 − 8.33x 3 + 17.1( x − 12) − 13.33(x − 36) + 1680x − 5760 lb.in 3 EI

The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 6 kip/ft

B

A x l 9 ft.

15 ft.

ΣM B = 0 R A (15 ) −

1 (6 )(9 ) 1 (9 ) + 15 − 6(15 )(7.5 ) = 0 2  3  

R A = 77.4 kips ΣFy = 0 1 (9 )(6 ) − 6(15) 2 R B = 39.6 kips R A + RB −

equating eqn.1 & 2 C1 = −256.5 C 2 = 2636 .55  x 5 77.4  (x − 9)3 + 1 (x − 9)5  1 − + y= 180 6 180  EI  − 256.5 x + 2636 .55  

y 6 = x 9 2 y= x 3 1 1 1  1 M = − (xy ) x  + R A ( x − 9 ) + ( y − 6 )(x − 9 ) (x − 9) 2 2 3  3 1  2  1  12  1 x x  x  + 77 .4( x − 9 ) +  x − 6 (x − 9 ) (x − 9 ) 2  3  3  23  3 3 x 1 3 =− + 77 .4( x − 9 ) + (x − 9 ) 9 9 4 x 77 .4 EIy ' = − + ( x − 9 ) 2 + 1 ( x − 9 ) 4 + C1 36 36 2 5 77 .4 x EIy = − + (x − 9)3 + 1 (x − 9 )5 + C1 x + C 2 180 6 180 @x = 9:y = 0 EIy ' ' = −

( )

1 5 9 + C1 (9) + C 2 180 @ x = 24 : y = 0 0=_

0=

(eqn.1)

−1 77 .4 24 5 + (24 − 9 )3 + 1 (24 − 9 )5 + C1 24 + C 2 180 6 180

( )

(eqn.2 )

Determine the slope and deflection at C. EI is constant. 15 kip A C B 30 ft.

15 ft.

Determine the slope and deflection at B. EI is constant. P A B L

1 PL (L ) 2 PL2 θB = 2 EI

EIθ B =

EI∆B =

2L 2   PL 3 2

PL3 ∆B = 3EI

Determine the maximum slope and the maximum deflection of the beam. EI is constant. M0

M0

B

A L

ΣFy = 0; R A = − RB ΣM A = 0; RB L = 0 RB = 0

ΣM = 0; M = MO

θ max @ x = 0

EIy' = M O x + C1 M EIy = O x 2 + C1 x + C 2 2 @ y = 0, x = 0 C2 = 0 @ y ' = 0, x = C1 = −

L 2

MOL 2

M O x 2 M O Lx − 2 2 L y max @ x = 2 M L2 M L2 EIy max = O − O 8 4 2 M L y max = − O 8 EI EIy =

EIθ = 0 + C1 θ =−

MOL 2 EI

The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 4 kip 2 kip

A

4 kip/fit x

B 8 ft.

8 ft.

8 ft.

The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant. 20 kN

20 kN

B

A x 3m

1.5 m

1.5 m

ΣFy = 0; R A + RB = 40kN ΣM B = 0;

R A (3) = 20(4.5) − 20(1.5)

R A = 20kN RB = 20kN

M = R A ( x − 1.5) + RB (x − 4.5) − 20x M = 20( x − 1.5) + 20(x − 4.5) − 20x

EIy' = 10(x − 1.5) + 10( x − 4.5) − 10 x 2 + C1 2

2

10 (x − 1.5)3 + 10 (x − 4.5)3 − 10 x 3 + C1 x + C 2 3 3 3 @ x = 0, y = 0 C2 = 0 EIy =

@ x = 6, y = 0 10 10 10 3 3 3 0 = (4.5) + (1.5) − (6) + C1 (6) 3 3 3 2 C1 = 67.5kN .m 10 (x − 1.5)3 + 10 (x − 4.5)3 − 10 x 3 + 67.5 x + C 2 3 3 3 @ x = 4.5, y = 0 10 10 10 0 = (3)3 + (0)3 − (4.5)3 + 67.5(4.5) + C 2 3 3 3 2 C 2 = −90kN .m EIy =

y=

1 10 (x − 1.5)3 + 10 (x − 4.5)3 − 10 x 3 + 67.5x − 90kN .m 3  EI  3 3 3 

Determine the equation of the elastic curve. Specify the slope at A. EI is constant. w

B

C A x L

L

The beam is subjected to the linearly varying distributed load. Determine the maximum slope of the beam. EI is constant. w B A x

ΣFy = 0

L

R A + RB =

wO L 2

ΣM B = 0

(

)

1 wO L2 6 w L RA = o 6 wo L RB = 3 by ratio and proportion RA L =

@x = L

(

y wO = x L w x y= O L 1 (xy ) x  2 3 1 1 EIy ' ' = (wO Lx ) − wO x 3 6 6 1 1  w x4 EIy ' = wO Lx 2 −  O 12 24  L

M = R A (x) −

(

)

  + C1  4 1 1 w x  EIy = wO Lx 3 −  O  + C1 x + C 2 36 20  L  @ x = 0, y = 0 C2 = 0 @ x = L, y = 0

(

)

(

)

(

)

w L4 1 wO L4 − O + C1 L 36 120 L 3 7w L C1 = − O 360 1 1  w x4 EIy ' = wO Lx 2 −  O 12 24  L 0=

(

)

)

(

)

7 wO L3 1 1 3 3 EIy ' = wO L − wO L − 12 24 360 3 w L EIy ' = O 45 w L3 θ max = O 45 EI

 7 wO L3  − 360 

The beam is subjected to the load shown. Determine the equation of the slope and elastic curve. EI is constant. 3 kN/m

A

15 kN⋅m B

C

x 3m

5m

ΣM B = 0;

R A (5) + 3(5)(2.5 ) − 15 = 0 R A = 4.5kN

ΣM A = 0 R B (5) = (3)(5)(2.5) + 15 R B = 10.5kN  x  x −5 M = R A x − 3(x )  + R B ( x − 5) + 3(x − 5 )  2  2 

(x − 5 ) + C x3 2 + 5.25( x − 5) + 1 2 2 x4 ( x − 5)4 + C x + C 3 EIy = 0.75 x 3 − + 1.75( x − 5) + 1 2 8 8 @ x = 0, y = 0 3

EIy ' = 2.25 x 2 −

C2 = 0 @ x = 5, y = 0

( )

0 = 0.75 5 3 −

54 + C1 5 8

C1 = −3.125  ( x − 5)3 − 3.125 x3 2 1  2.25 x 2 − + 5.25(x − 5) +  y' = 2 2  EI    y=

1 EI

 ( x − 5)4 − 3.125 x  x4 3 3 0 . 75 x 1 . 75 x 5 ( ) − + − +   8 8  

The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. EW=1.6(103) ksi. 1.5 kip

0.8 kip/ft A

12 in.

C B

x

6 in. 9 ft

9 ft

ΣM A = 0 RB (9 ) = 1.5(18) +

1 (0.8)(9) 2 (9) 2 3 

RB = 5.4kips ΣFy = 0 1 (0.8)(9) + 5.4 = 0 2 R A = −0.3kips

R A − 1.5 −

( )

1 1  2 2 M = −0.3 x −  x 2 y  + 5.4(x − 9 ) + 0.4( x − 9 ) + ( y − 0.8)(x − 9 ) 6 6   by ratio and proportion y 0.8 = 9 x 0.8 x y= 9 1  0.8 x  0.8 2   2 2 EIy' ' = −0.3x −  x  + 5.4( x − 9 ) + 0.4(x − 9 ) +  − 0.8 ( x − 9 ) 6 9  54  

( )

0.4(x − 9 ) 0.8( x − 9 ) − 0.3 x 2 0.8 x 4 5.4( x − 9 ) EIy' = − + + + + C1 2 54(4 ) 2 3 54(4 ) 2

3

4

0.8(x − 9 ) − 0.3 x 3 0.8 x 5 5.4(x − 9 ) 0.4( x − 9 ) EIy = − + + + + C1 x + C 2 6 1080 6 12 1080 @ x = 0, y = 0; C 2 = 0 3

@x =9 y = 0

( )

0.8 9 5 + 9C1 = 0; C1 = 8.9 1080 @ x = 18, EI = 1.6 10 3 − 0.05(9 ) − 3

( )

y = −0.765 in.

4

5

Determine the deflection and slope at C. EI is constant.

C

A

M0 B x L

L

Determine the deflection at C and the slope of the beam at A, B, and C. EI is constant.

A

8 kN⋅m B

C

3m

6m

ΣFy = 0 R A + RB = 0 ΣM B = 0

R A (6) = −8

4 4 R A = − kN ; RB = kN 3 3 M = R A x + RB (x − 6 ) 4  4 EIy' ' = − x +  (x − 6) 3  3 4 4 2 EIy' = − x 2 + ( x − 6) + C1 6 6 2 2 2 = − x 2 + (x − 6) + C1 3 3 2 2 3 EIy = − x 3 + (x − 6) + C1 x + C 2 9 9 @ y = 0, x = 0 C2 = 0

( )

@ x = 6, y = 0 2 2 0 = − 6 3 + (0) + C1 6 9 9 2 C1 = 8kN .m

( )

( )

2 2 2 2 x + (x − 6) + 8 3 3 slope @ A; x = 0 8 EIy' = 8, θ A = EI slope @ B; x = 6 − 16 16 EIy' = −16, θ B = or θ B = EI EI slope @ C ; x = 9 40 40 EIy' = −40, θ C = − or θ C = EI EI Eiy' = −

deflection@ C; @ x = 9 2 2 3 EIy = − x 3 + (x − 6) + 8 x 9 9 EIy = −84 84 EI 84 kN .m 3 ∆C = EI yC = −

Determine the slope at B and deflection at C. EI is constant. P

P 2

P 2

A

B C a

a

a

a

The beam is subjected to the load shown. Determine the slope at B and deflection at C. EI is constant. w w

A

B C

ΣFy = 0 R A + R B = 2wa ΣM B = 0;

a 2

a 2

a

a

x−a M = R A x + w (x − a )  2 w 2 EIy ' ' = wax + ( x − a ) − 2 3 wax 2 w (x − a ) EIy ' = + − 2 6 4 wax 3 w (x − a ) EIy = + − 6 24 @ y = 0, x = 0

a  R A (3a ) = wa  + 2a  + 2  2 wa 2 5wa 2 wa 2 R A (3a ) = + 2 2 R A = wa

C2 = 0

RB = wa

@ y = 0, x = 3a 27 wa 4 w (2 a )4 − + 6 24 7 C1 = − wa 3 12 3 wax 2 w (x − a ) EIy ' = + 2 6 4 3 wax w (x − a ) EIy = + 6 24 slope @ B ; x = 3 a 0 =

(

2  wx  x − 2a  − w ( x − 2 a ) −  2 2    wx 2 w 2 − (x − 2 a ) 2 2 wx 3 w 3 − (x − 2 a ) + C 1 6 6 wx 4 w (x − 2 a )4 + C 1 x + C 2 − 24 24

w (3 a )4 − w (a )4 + C 1 (3 a ) 24 24

)

(

)

7 wx 3 w 3 wa 3 − (x − 2 a ) − 6 6 12 wx 4 w (x − 2 a )4 − 7 wa 3 x − − 24 24 12 −

(

)

wa (3 a )2 + w (2 a )3 − w (3 a )3 − w (a )3 − 7 2 6 6 6 12 3 7 wa θB = 12 EI 3 deflection @ C ; x = a 2 EI θ =

3

4

4

wa  3  w 1  w 3  7 wa  a −  a −  a + 6 2  24  2  24  2  24 4 25 wa EIy = − 48 25 wa 4 25 wa 4 yC = − ; ∆C = 48 EI 48 EI

EIy =

3

3   a 2 

Determine the elastic curve for the simply supported beam the x coordinate 0 ≤x ≤ L/2. Also, determine the slope at A, and the maximum deflection of the beam. EI is constant. ΣFy = 0 1  wo L  1  wo L   +   2 2  2 2  w L R A + RB = o 2 ΣM A = 0 R A + RB =

wo L  2  L  wo L  L 1  L  +   = RB L    + 4  3  2  4  2 3  2  w L2 w L2 RB L = o + o 12 6 w L RB = o 4 w L RA = o 4 BY RATIO AND PROPORTION : y wo = x L 2 wo x y= L 2 ΣM = 0 1  2w Lx 2  x wo L  − (x ) = 0 M +  o 2 L  3 4 w x 3 w Lx M =− o + o 3L 4 4 wo x wo Lx 2 EIy' = − + + C1 12 L 8 w x 5 w Lx 3 EIy = − o + o + C1 x + C 2 60 L 24 @ x = 0; y = 0 C2 = 0 L ; y' = 0 2 4 2 w L w LL 0 = − o   + o   + C1 12 L  2  8 2

@x =

C1 = −

5 wo L3 192

wo

A

B x L

y max @ x =

L : 2

EIy MAX = −

wo L  L  wo L  L  5wo L3 +     − 60 L  2  24  2  192

5

y MAX = −

wo L4 24 EI

θ MAX @ x = 0 EI θ MAX = 0 + 0 + C 1 θ MAX

w o L3 = 192 EI

3

L   2

Draw the shear and moment diagrams for the shaft. The bearings at A and B exerts only vertical reactions on the shaft.

A

250 mm

B

800 mm

24 kN RB

ΣM B = 0

0.8 R A = 24(1.05) R A = 31.5kN RA 7.5 kN

ΣFY = 0 RB = −24 = 31.5 RB = 7.5kN

M A L = −24(0.25) M AL = −6kN .m

M BL = [(− 24)0.25 + 31.5(0.8)]kN .m M BL = 0

-24 kN

Vo = −24kN

V A = (− 24 + 31.5)kN

V A = 7.5kN -6 kN⋅m

VBL = 7.5kN

VB = [7.5 − 7.5]kN VB = 0

The load binder is used to support a load. If the force applied to the handle is 50lb, determine the tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm ABC. T1

C

A B

ΣM c = 0

50 lb

50(15) = 3T1

12 in.

3 in.

T2

200 lb

T1 = 250lb ΣFY = 0 T2 = 250lb − 50lb T2 = 200lb V A = −50lb VBL = −50lb

VB = (− 50lb + 250lb ) VB = 200lb

-50 lb

VC L = 200lb

VC = (200 − 200)lb VC = 0

-600 lb⋅in

Draw the shear and moment diagrams for the shaft. The bearings at A and D exert only vertical reactions on the shaft. The loading is applied to the pulleys at B and C and E. 14 in.

20 in.

15 in.

12 in.

A

E

B

C

80 lb

110 lb

D

35 lb

82.4 lb 35 lb 2.24 lb

ΣM A = 0

49 RD = 80(14) + 110(34) + 35(61) RD = 142.76lb ΣFY = 0

107.76 lb

R A = 80 + 110 + 35 − 142.76

1196 lb.ft

R A = 82.24lb V A = 82.24lb

1151.36 lb.ft

VB = 2.24lb VC = −107.76lb -420.24 lb.ft

VD = 35lb VE = 0 ΣM BL = (82.24)(14)lb.in M BL = 1151.36lb.in

ΣM CL = (82.24)(34) − 80(20)

Draw the shear and moment diagrams for beam.

M C L = 1196.16lb.in

ΣM DL = (82.24)49 − 80(35) − 110(15) M DL = −420.24lb.in

ΣM E L = {82.24(61) − 80(47 ) − 110(27 ) M EL

+142.76(12)}(lb.in ) =0

the

2 kip

4 ft

2 kip

4 ft

2 kip

4 ft

2 kip

4 ft

4 ft

4 kip 2 kip -2 kip -4 kip 24 kip.ft 16 kip.ft

16 kip.ft

By symmetry :

2(4 ) (2) RA = RF = 4kip ΣM B L = 16kip. ft RA = RF =

ΣM C L = (4 )8 − 2(4 ) M C L = 24kip. ft

ΣM D L = 4(12) − 2(8) − 2(4 ) M D L = 24kip. ft

M E L = 4(16) − 2(12 ) − 2(8) − 2(4 ) M E L = 16kip. ft

ΣM FL = 4(20) − 2(16) − 2(12) − 2(8) − 2(4 ) M FL = 0

Draw the shear and moment diagrams for the rod. It is supported by a pin at A and a smooth plate at B. The plate slides w/in the groove and so it cannot support a vertical force, although it can support a moment. 15 kN

A B

4m

2m

15 kN

ΣFY = 0 R A = 1.5kN

60 kN.m 30 kN.m

ΣM B = 15(4 ) − 15(2 ) M B = 30kN .m ΣM C L = 60kN .m

Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. Also, express the shear and moment in the shaft as a function of x w/in the region 125mm<x<725mm. 1500 N

800 N

B

A

x 125 mm

600 mm

75 mm

815.625 N

15.625 N

-1484.375 N

111328 N.m 101953 N.m

ΣM B = 0

800R A = 800(675) + 1500(75) R A = 815.625N ΣFY = 0 RB = 800 + 1500 − 815.625 RB = 1484.37 N

ΣM CL = 815.625(0.125) M C L = 101.953N .m

ΣM DL = 815.625(0.725) − 800(0.6 ) M D = 111.328N .m

ΣM BL = 815.625(0.8) − 800(0.675) − 2500(0.075)

Draw the shear and moment diagrams for @ 125mm < x < 725mm

the beam.

V = 16.625 N 10 kN

8 kN

M = 815.625 x − 800( x − 0.125) M = (15.625 x + 100)N .m

15 kN.m

2m

3m

18 kN 8 kN

ΣFY = 0 R A = 18kN

ΣM A = −10(2 ) − 8(5) − 15 -15 kN.m -39 kN.m

-75 kN.m

M A = −75kN .m V A = 18kN

VC L = (18 − 10 )kN VC L = 8kN VB = 0

ΣM C L = −75 + 18(2 ) M CL = −39kN .m

ΣM BL = −39 + 8(3) M BL = −15kN .m ΣM B = 0

Draw the shear and moment for the pipe. The end screw is subjected to a horizontal force of 5kN. Hint: the reaction at the pin C must be replaced by equivalent loadings at point B on the axis of the pipe.

C

A 5 kN

80 mm B

400 mm

ΣM A = 0

0.4 RB = = 5(0.8) RB = 1kN ΣM B = 0 0.4 R A = (5 )(0.8)

-1 kN

R A = 1 kN V A = −1 kN -0.4 kN.m

VBL = −1 kN =0 MA =0

M BL = −1kN (0.4 m ) = −0.4kN ⋅ m M B = −0.4kN ⋅ m + 0.4kN ⋅ m =0

The engine crane is used to support the engine, which has a weight of 1200 lb. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown.

3 ft

5 ft

A B

4 ft

C

1200 lb

1200

ΣM B = 0

5 R A = 2000lb(5 ft ) ΣFY = 0 RB = 2000lb + 1200lb

-2000

RB = 3200lb V A = −2000lb -6000

V AL = −2000lb VB = −2000lb + 3200lb = 1200lb VCL = 1200lb VCL = 1200lb − 1200lb =0 ΣM A = 0

2000(3) 2 = −6000lb ⋅ ft ΣM CL = (− 6000 − 6000)lb ⋅ ft

ΣM BL = −

=0

Draw the shear and moment diagrams for the compound beam w/c is pin connected at B. It is supported by a pin at A and a fix wall at C.

6 kip

8 kip

A C B

4 ft

6 ft

Mo = 0

M AL = (− 6 kip )(4 ft )

4 ft

4 ft

M DL

4

M CL

-4 -6 16

= −24 kip ⋅ ft = 40 kip ⋅ ft − 24 kip ⋅ ft = 16 kip ⋅ ft = 16 kip ⋅ ft − 16 kip ⋅ ft

=0 ΣM C = 0

14R A = 6(18) + 8(4 ) = 108 + 32 R A = 10 kip ΣFy = 0

-24

RC = 14 − 10 RC = 4 kip VO = −6 kip V AL = −6 kip V A = −6 kip + 10 kip = 4 kip VDL = 4 kip VD = 4 kip − 8 kip = −4kip VCL = −4 kip VC = −4 kip + 4 kip =0

Draw the shear and moment diagrams for the beam. Also, determine the shear and moment in the beam as a function of x, where 3ft<x<15ft. 1.5 kip/ft

50 kip.ft

A

B

x 3 ft

12 ft

ΣM B = 0

13.167

12 R A = 50 + (1.5)(12)(6) C

R A = 13.167kip -4.833

x

↑ ΣF = 0 RB = 1.5(12) − 13.167 RB = 4.833kip

7.79 kip.ft

By Similar ∆ : 18 13.167 = x 12 x = 8.778 ft M C = Area ∆ − 50

-50 kip.ft

13.167 (8.778) − 50 2 = 67.778 − 50 = 7.79 kip ⋅ ft =

@ 3 ft < x ≤ 15 ft : V = 13.167 − 1.5( x − 3) V = 17.7667 − 1.5 x M = 13.167(x − 3) −

1.5 ( x − 3)2 2 = 13.167 x − 39.5 − 0.75 x 2 − 6 x + 9

(

)

= 13.167 x − 39.5 − 0.75 x + 4.5 x − 6.75 2

M = −0.75x 2 + 17.667 x − 46.25

Draw the shear and moment diagrams for the beam.

800 lb/ft

A

B

800 lb/ft 8 ft

8 ft

-6400 51200 25600

ΣFy = 0

R A = 800(8) − 800(8) =0

VA = 0 VBL = −800(8) = −6400 lb

ΣM A = 800(8)(12 ) − 800(8)(4) = 51200 lb ⋅ ft ΣM B = 51200 − 800(8)(4 ) = 25600 lb ⋅ ft ΣM C = 51200 − 800(8)(12 ) − 800(8)(4 ) =0

VCL = −6400 + 6400 =0

The 50lb man sits in the center of the boat, w/c has a uniform width and weight per linear foot of 3lb/ft. Determine the maximum bending moment exerted on the boat. Assume that the water exerts a uniform distributed load upward on the bottom of the boat.

150 lb

7.5 ft

7.5 ft 75 lb

Σ Fy = 0 45 + 150 15 wo = 13 lb ft MA = 0 M B = Area ∆ w0 =

-75 lb

281.25 lb.ft 2°

= 75(0.75) 2°

1 lb ⋅ ft 2

MC = 0

VA = 0

VBL = 13(7.5) − 3(7.5) = 75 lb VB = (75 − 150) lb = −75 lb VCL = −75 + 75 =o

The footing supports the load transmitted by the two columns. Draw the shear and moment diagrams for the footing if the reaction of soil pressure on the footing is assumed to be uniform. 14 kip

14 kip

6 ft

12 ft

6 ft C

B

A

7 kip

D

7 kip 6 ft

ΣFy = 0

E

24wo = 14 kip +14 kip -7 kip

kip 24 ft wo = 7 kip 6 ft = 28

-7 kip

21 kip.ft

21 kip.ft

2°

2° 2°

7 kip (6 ft ) ft 6 = 7 kip VB = 7 kip −14 kip VBL =

= −7 kip VCL = −7 kip +

7 kip (12 ft ) ft 6

VCL = 7 kip VC = 7 kip −14 kip = −7 kip VDL = −7 kip + kip =0

Draw the shear and moment diagrams for the beam. 2 kip/ft 30 kip.ft A

5 ft

B

5 ft

5 ft

ΣM B = 0

10 R A = 2(5)(12.5) − 30

-0.5 kip

R A = 9.5 kip

-10 kip 2.5 kip.ft

VO = 0 V AL = −2(5 ) = −10 kip

2° -25 kip.ft

V A = (− 10 + 9.5) kip = −0.5 kip

1° -27.5 kip.ft

MO = 0 1 (10)(5) 2 = −25 kip ⋅ ft

M AL = −

M CL = (− 27.5 + 30) kip ⋅ ft = 2.5 kip ⋅ ft

M BL = [2.5 − 5(0.5)] kip ⋅ ft M BL = 0 ΣF y = 0 RB = 2(5) − 9.5 RB = 0.5 kip VB = (− 0.5 + 0.5 ) kip VB = 0

Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as a function of x, where 4ft<x<10ft. by symmetry: 150 lb/ft

200 lb.ft

200 lb.ft B

A x 4 ft

6 ft

4 ft

450 lb

ΣFy = 0 -450 lb 200 lb.ft

2°

2°

R A + RB = 150(6 )

900 2 R A = 450 lb = RB R A = RB =

V = 450 − 150(x − 4 )

2

= 1050 − 150 x -200 lb.ft

-200 lb.ft

150( x − 4 ) 2 2 = −3200 + 1050x − 75 x

M = −200 + 450( x − 4 ) −

2

Draw the shear and moment diagrams for the beam. 50 kN/m

50 kN/m

B

A

4.5 m

4.5 m

112.5 kN 2°

2° -112.5 kN

168.75 kN.m

ΣM A = 0

(4.5) + 50 (4.5)(4.5) 2 + 4.5 1 9 R B = (50)   2 3 2 3 = 168.75 + 843.75 RB = 112.5 kN by symmetry : R A = RB = 112.5 kN 2

V A = 112.5 kN

VBL = (112.5 − 112.5) kN

VCL

=0 = 0 − 112.5 kN

= −112.5 kN VC = (− 112.5 + 112.5) kN =0 MA =0 112.5 (4.5) = 168.75 kN ⋅ m 3 112.5 (4.5) = 0 = 168.75 − 3

M BL = M CL

The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4kN/m caused by bar C. Determine the intensity of the distributed load wo of the leaves on the pin and draw the shear and moment diagrams for the pin. ΣF y = 0 1  2  WO (0.2 ) = [0.4(0.06)] 2  WO = 1.2 kN m

0.4 kN/m

VA = 0 VBL

 1 =  (1.5)(0.02) kN  2 = 0.012 kN

VCL = [0.012 − 0.4(0.06)] kN = −0.012 kN   1.5 VDL = − 0.012 +  (0.02)  kN   2  =0 MA = 0 M BL =

0.012 (0.02) = 0.00008 kN ⋅ m 3 0.012 (0.03) 2 = 0.00026 kN ⋅ m

M MAX = 0.00008 +

wo 20 mm

wo 60 mm

20 mm

0.012 kN 2°

30 mm

2° 0.012 kN

2.6x10-3 kN⋅m

Draw the shear and moment diagrams for the beam. 3 kip/ft 1500 lb.ft A B 12 ft

6 ft 900

ΣM A = 0 2°

3000 (6)(2 + 12) 2 127500 = 12 RB = 10625 lb RB = 1500 +

-1625

ΣFy = 0 1500

R A = 10625 −

3000 (6) 2

R A = 1625 lb 3° -18000

V A = −1625 lb V A = −1625 lb

VB = (− 1625 + 10625) lb VB = 9000 lb VCL = 9000 − [300(6 )(0.05)] =0 M A = 1500 lb ⋅ ft

M B = −1625(12 ) + 1500 M B = −18000 M C = −1800 + =0

9000 (6) 3

Draw the shear and moment diagrams for the beam and determine the shear and moment as a function of x. 18 kN/m 12 kN/m

A

B 3m

ΣM B = 0 2°

6 R A = 200(3)(1.5) + -45 kN

1 (200)(3)(1) 2 = 900 + 300 = 1200

R A = 200 N 3°

ΣFY = 0 -54 kN.m

 600  RB =  3 − 1200  2  RB = 700 N 1 ( x − 3) 200 ( x − 33) 2  3  200 (x − 3)2 = 200 − 200 x + 600 − 6 100 2 = 500 − x =0 3 500 x2 = (3) 100 x = 15 x = 3.8

V = 200 − 200(x − 3) −

200 (x − 3)2 − 1 (x − 3)2  200  x − 3  2 2  3  3  100 2 ( x − 3)3 = 200 x − 100( x − 3) − 9 100 3 M =− x + 500 x − 600 9

M = 200 x −

A member having the dimensions shown is to be used to resist an internal bending moment of M = 2 kip⋅ft. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b) about the y axis. Sketch the stress distribution for each case. y

z

x

6 in.

6 in.

a) about z axis 1 I z = ( )(6)(123 ) 12 I z = 864in 4 2(6)(12) 864 σ z = 0.167 kip / in 2 (1000lb / kip ) σz =

σ z = 167 psi

(1)(12)(6 3 ) 12 I y = 216in 4 Iy =

( 2)(3)(12) 216 σ y = (0.333kip / in 2 )(1000lb / kip ) σy =

σ y = 333 psi

b) about y axis

A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (σallow)t = 22 ksi and (σallow)s = 15 ksi, respectively.

4 in. 4 in.

M

2 in. 2 in.

Mc I σ I M= t c I bh 2 = c 24

σ=

σcI c ( 4)( 2 3 ) 2 M = 15( ) 24 M = 30kip.in M =

bh 2 ) 24 ( 4)( 2 3 ) 2 M = 22( ) 24 M = 44kip.in M = 22(

We used the smaller value  ft  M = 30kip.in   12in  M = 2.5kip. ft

A member has the triangular cross section shown. If a moment of M = 800 lb⋅ft is applied to the cross section, determine the maximum tensile and compressive bending stresses in the member. Also, sketch a three dimensional view of the stress distribution acting over the cross section.

4 in. 4 in. M 2 in. 2 in.

Mc1 I 3 bh I= 36 ( 4)( 2 3 ) 3 I= 36 I = 4.62in 4 2h C1 = 3 22 3 C1 = 3 4 3 C1 = 3 C1 = 2.31in

σc =

( )

σc =

(9600(2.31))

4.62 σ c = 4800lb / in 2 σ c = 4.8ksi

h 3 2 3 C2 = 3 C2 = 1.15in C2 =

( )

σt =

(9600(1.15))

4.62 σ t = 2400lb / in 2 σ t = 2.4 ksi

A beam has the cross section shown. If it is made of steel that has an allowable stress of σallow =24 ksi, determine the largest internal moment the beam can resist if the moment is applied (a) about the z axis, (b) about the y axis. y 0.25 in. 3 in. 3 in.

0.25 in.

z

0.25 in. 3 in. 3 in.

a) about z-axis Mc I  (0.6 ) 0.253 I z = 2 + (0.6 )(0.25 ) 3.1252 12 

σ =

(

)

I z = 33.8125in

(

4

M z = 24(33.8125) / (3 + 0.25 ) M z = 249.7 kip.in M z = 20.8kip. ft

b) about y-axis

(

)

(

)

 (0.25 ) 0.6 3   (0.6 ) 0.25 3  I y = 2  + 12 12     4 I y = 9in

24(9 ) 3 M y = 72.06 kip.in

My =

M y = 6.0 kip. ft

) +  (0.2512)(0.6 ) 3

 



The beam is subjected to a moment of M = 40 kN⋅m. Determine the bending stress acting at points A and B. sketch the results on a volume element acting at each of these points.

A B M = 40 kN⋅m

50 mm 50 mm 50 mm

(

50 mm 50 mm 50 mm

)

(

)

 (0.05 ) 0.153   (0.05 ) 0.053  I = I y = Iz =   + 2  12 12     −5 4 I = 1.51(10 ) m σA =

− Mz y M yz + Iz Iy

 (40 )(0.075 )(1000 )  σA = 0+   1.5(10 −5 )   σ A = 199 MPa  (40 )(0.025 )(1000 )  σB = 0+   1.51(10 −5 )   σ B = 66.2 MPa

Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 160 MPa.

A

0.8 600

B

1.2

ΣM = 0 R A (1.2 ) = 600(2 ) − 400(0.6 ) R A = 800 N

0.6 400

ΣM A = 0

RB (1.2 ) = 400(1.8) − 600(0.8) RB = 200 N

Mr πr 4 4 4M σ= 3 πr 4M r3 = πσ (4)(480) r =3 π 160(10 6 )

σ=

(

r = 0.0156m d = 31.3mm

)

The beam is subjected to the load P at its center. Determine the placement a of the supports so that the absolute maximum bending stress in the beam is as large as possible. What is this stress? P

d a

a

b L/2

L/2

+ ↑ ΣFy = 0 R AY + R BY = P R AY = R BY

For symmetrical loading

2 R AY = P

a = 0, L

P R AY = 2 ΣM c = 0

σ max =

MA = MB MA =

P  L    −a 2  2  

MB =

P  L    −a 2  2  

0≤a≤

L 2

@a = 0 PL   − 0 22  PL MA = 4 L @a = 2 P  L L   M A =  −   2  2 2   MA =

M max =

PL 4

(M max )(c ) I

3

bd 12 d c= 2 I=

σ max

σ max

PL  d    4 2 = bd 3 12  PL  12 =   2  8  bd

σ max =

( )

3 PL 2bd 2

  

The beam has a rectangular cross section as shown. Determine the largest load P that can be supported on its overhanging ends so that the bending stress does not exceed σmax = 10 MPa.

P

P 1.5 m

1.5 m

1.5 m

ΣM A = 0 1.5 RB + 1.5 P = 3P RB = P

250 mm 150 mm

+ ↑ ΣFy = 0 R A + RB = 2P RA = P Mc I (0.5) 0.253 I= 12 I = 1.953(10 −4 )m 4 1.5 P(0.125) 10(10 6 ) = 1.953(10 − 4 ) σ max =

(

P = 10.4kN

)

The beam has the rectangular cross section shown. If P =12 kN, determine that absolute maximum bending stress in the beam. Sketch the stress distribution acting over the cross section. P

P 1.5 m

1.5 m

1.5 m

ΣM A = 0

RB (1.5) + 12(1.5) = 12(3) RB = 12kN

+ ↑ ΣFy = 0 R A + R B = 24 R A = 12kN I=

(0.15)(0.253 )

12 I = 1.953125(10 −4 ) m 4 18(0.125) σ max = 1.953125(10 −4 ) 18000(0.125) σ max = 1.953125(10 −4 ) σ max = 11.5MPa

250 mm 150 mm

A log that is 2 ft in diameter is to be cut into a rectangular section for use as a simply supported beam. If the allowable bending stress for the wood is σallow = 8 ksi, determine the largest load P that can be supported if the width of the beam is b = 8 in. h b 2 ft P

8 ft

h=

(24

2

− 82

)

= 22.6 in ΣM B = 0 RA (16 ) = 8 P P 2 P RB = 2 bh 2 S= 6 (8)(22.6)2 = 6 = 681in 4 M , M = (96 )RA σ= S (96)(RA ) 8= 681 RA = 56.75 kip RA =

M = 96(56.75)

= 5448 kip ⋅ ft P = 2 RA

= 2(56.75) = 113.5 kip P = 114 kip

8 ft

The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 45°. z 125 mm 125 mm 250 mm

B

E M = 850 N⋅m

C

θ

A D

θ = 45° M Y = 850 cos 45 = 601.04 Nm M Z = 850 sin 45 = 601.04 Nm

(

)

(

)

1 3 ⋅ (0.25)(0.25) = 3.2552 10 −4 m 4 12 1 3 I Z = ⋅ (0.25)(0.25) = 3.2552 10 −4 m 4 12 − M Z y MY z σA = + IZ IY IY =

− 601.04(− 0.125) 601.04(− 0.125) + 3.2552 10 −4 3.2552 10 − 4 =0 =

(

)

(

)

− 601.04(− 0.125) (0.125)601.04 + 3.2552 10 − 4 3.2552 10 − 4 = 462 kPa

σB =

(

)

(

)

− 601.04(0.125) 601.04(0.125) + 3.2552 10 −4 3.2552 10 −4 = −462 kPa

σD =

(

)

(

)

− 601.04(0.125) 601.04(0.125) + 3.2552 10 −4 3.2552 10 −4 =0

σE =

(

)

(

)

y

The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 30°. z 125 mm 125 mm 250 mm

B

E M = 850 N⋅m

C

θ

A D

θ = 30° M Y = 850 cos 30 = 736.12 Nm M Z = 850 sin 30 = 425 Nm

(

)

(

)

1 3 ⋅ (0.25)(0.25) = 3.2552 10− 4 m4 12 1 3 I Z = ⋅ (0.25)(0.25) = 3.2552 10− 4 m4 12 − M Z y MY z σA = + IZ IY IY =

− 425(− 0.125) 736.12(− 0.125) + 3.2552 10− 4 3.2552 10− 4 = −119 kPa − 425(− 0.125) 736.12(0.125) σB = + 3.2552 10− 4 3.2552 10− 4 = 446 kPa − 425(0.125) 736.12(0.125) σD = + 3.2552 10− 4 3.2552 10− 4 = −446 kPa =

(

)

(

)

(

)

(

)

(

(

)

)

− 425(0.125) 736.12(0.125) + 3.2552 10− 4 3.2552 10− 4 = 119 kPa

σE =

(

)

(

)

y

The T-beam is subjected to a moment of M = 150 kip⋅in. directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis. The location y of the centroid, C, must be determined. y 150 kip.in 6in.

6in.

A 2in. y 60° z

8in.

B

M Y = 150 sin 60 = 129.9038 kip ⋅ in

2in.

M Z = −150 cos 60 = −75 kip ⋅ in Σy A ΣA [(4)(8)(2) + (9)(2)(12)] = (8)(2) + (2)(12) = 7 in

yo =

  1  1 I Y =  (8) 2 3 + (2 ) 12 3  12  12   4 = 293.33 in

( )

( )

 1  1 2 3 2 I Z =  (2 ) 8 3 + 8(2)(7 − 4 )  +  (12)(2 ) + 12(2 )(9 − 7 )    12   12 4 = 333.33 in

( )

− 75(3) 129.9(6) + 333.333 293.333 = 3.33 ksi

σA =−

− 75(− 7 ) 129.9(− 1) + 333.33 293.333 = −2.0178 ksi

σB = −

∴ σ max = 3.33 ksi

tan α = =

IZ (tan θ ) IY

(333.333) (tan 30)

293.33 α = −33.2 − 30 = −63.2°

The aluminum machine part is subjected to a moment of M = 75 N⋅m. Determine the bending stress created at points B and C on the cross section. Sketch the results on a volume element located at each of these points. 20 mm 10 mm

10 mm

10 mm

20 mm

10 mm B N 10 mm C

M = 75 N⋅m

40 mm

0.005(0.08)(0.01) + 2(0.04)(0.01)(0.03) 0.01(0.08) + (0.04)(0.01) = 0.0175 m

y=

 1   1 2 I Z =  (0.08) 0.013 + (0.08)(0.01)(0.0175 − 0.005) 2  + 2  (0.01) 0.04 3 + (0.03 − 0.0175)   12  12  

(

)

( )

= 0.3633 10 6 m 4 σB = −

(75)(− 0.0175)

(

0.3633 10 −6 σ B = 3.61MPa

)

− (75)(− 0.0075) 0.3633 10 −6 σ c = 1.55MPa σC =

(

)

(

)

A beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is M = 450 N⋅m, determine the resultant force that bending stress produces on the top board A and on the side board B. A

B 15 mm M = 40 kN⋅m 20 mm 200 mm 20 mm 200 mm

15 mm

( ( )

)

(

)

(

)

1  1  I y = 2  (0.2) 0.02 3 + (0.2 )(0.02) 0.112  + 2  (0.015) 0.24 3  12  12  −4 4 = 1.316 10 m − MZ y MY z σA = + IZ Iy σA = 0+0 =0 FRA = 0

450(0.12 ) 1.316 10 −4 = 0.410MPa

σB = 0+

FRB =

(

)

(410334.34)(0.24)(0.015)

= 1.5kN

1000

The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N⋅m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section. 25 mm

20 mm 200 mm 20 mm

Σy A ΣA ( 0.0125)(0.025)(0.24 ) + (2 )(0.1)(0.15)(0.02) = 0.025(0.24 ) + 2(0.15)(0.02) y = 0.05625m C = 0.175 − 0.05625 = 0.11875m y=

(

)

1 2 I =  (0.24) 0.0253 + (0.24)(0.025)(0.05625 − 0.0125)  + 12  1 2 2  (0.02) 0.153 + (0.02 )(0.15)(0.1 − 0.05625)  12 

( ) = 34.53(10 )m −6

4

Mc I 600(0.11875) = 24.53 10 −6 = 2.06MPa

σ max =

(

)

M = 600 N⋅m

If the internal moment acting on the cross section of the strut has a magnitude of M = 800 N⋅m and is directed as shown, determine the bending stress at points A and B. The location z of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis. y 12 mm

12 mm

200 mm

z 12 mm

60°

150 mm

M = 800 N⋅m

z

z=

(0.006)(0.012)(0.4) + (2)(0.081)(0.012)(0.138) (0.012)(0.4) + (2)(0.012)(0.138)

= 0.366m C = 0.15 − 0.0366 = 0.1134m

(

)

1 2 I Y =  (0.4 ) 0.012 3 + (0.4 )(0.012)(0.081 − 0.0366)   12 −6 4 = 16.3374 10 m

(

IZ

)  1 = 2  (0.138)(0.012 ) + (0.138)(0.012)(0.194 ) + 12 3

2

 1 3  12 (0.012) 0.4  = 0.18869 10 −3 m 4 − 400(0.2 ) (− 692.82)(− 0.1134) σA = + 0.1886 10 −3 16.3374 10 −6 = 4.38MPa

(

)

(

)

(

)

(

)

− 400(− 0.2 ) − 692.82(0.0366) + 0.11869 10 −3 16.3374 10 −6 = −1.13MPa I tan α = Z tan θ IY

σB =

(

)

( (

(

) )

 0.18869 10 −3  tan α =  tan 30 −6  16.3374 10  α = −87.1°

)



200 mm

The beam has a rectangular cross section as shown. Determine the largest load P that can be subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces. A

0.8 m

1.2 m

0.6 m 400 N

600 N

3

d  π  πd 3 Mc 2 σ = ; I=   = 4 32 I Mr = 2 πr 4 32M = πd 3 32(480) = 3 π (0.030) σ = 181 N / m 2

B

A member having the dimensions shown is to be used to resist an internal bending moment of M = 2 kip⋅ft. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b) about the y axis. Sketch the stress distribution for each case. y

z

x

6 in.

6 in.

a) about z axis 1 I z = ( )(6)(123 ) 12 I z = 864in 4 2(6)(12) 864 σ z = 0.167 kip / in 2 (1000lb / kip ) σz =

σ z = 167 psi

b) about y axis

(1)(12)(6 3 ) 12 I y = 216in 4 Iy =

( 2)(3)(12) 216 σ y = (0.333kip / in 2 )(1000lb / kip ) σy =

σ y = 333 psi

A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (σallow)t = 22 ksi and (σallow)s = 15 ksi, respectively.

4 in. 4 in.

M

2 in. 2 in.

Mc I σ I M= t c I bh 2 = c 24

σ=

σcI c ( 4)( 2 3 ) 2 M = 15( ) 24 M = 30kip.in M =

bh 2 ) 24 ( 4)( 2 3 ) 2 M = 22( ) 24 M = 44kip.in M = 22(

We used the smaller value  ft  M = 30kip.in   12in  M = 2.5kip. ft

A member has the triangular cross section shown. If a moment of M = 800 lb⋅ft is applied to the cross section, determine the maximum tensile and compressive bending stresses in the member. Also, sketch a three dimensional view of the stress distribution acting over the cross section.

4 in. 4 in. M 2 in. 2 in.

Mc1 I 3 bh I= 36 ( 4)( 2 3 ) 3 I= 36 I = 4.62in 4 2h C1 = 3 22 3 C1 = 3 4 3 C1 = 3 C1 = 2.31in

σc =

( )

σc =

(9600(2.31))

4.62 σ c = 4800lb / in 2 σ c = 4.8ksi

h 3 2 3 C2 = 3 C2 = 1.15in C2 =

( )

σt =

(9600(1.15))

4.62 σ t = 2400lb / in 2 σ t = 2.4 ksi

A beam has the cross section shown. If it is made of steel that has an allowable stress of σallow =24 ksi, determine the largest internal moment the beam can resist if the moment is applied (a) about the z axis, (b) about the y axis. y 0.25 in. 3 in. 3 in.

0.25 in.

z

0.25 in. 3 in. 3 in.

a) about z-axis Mc I  (0.6 ) 0.253 I z = 2 + (0.6 )(0.25 ) 3.1252 12 

σ =

(

)

I z = 33.8125in

(

4

M z = 24(33.8125) / (3 + 0.25 ) M z = 249.7 kip.in M z = 20.8kip. ft

b) about y-axis

(

)

(

)

 (0.25 ) 0.6 3   (0.6 ) 0.25 3  I y = 2  + 12 12     4 I y = 9in

24(9 ) 3 M y = 72.06 kip.in

My =

M y = 6.0 kip. ft

) +  (0.2512)(0.6 ) 3

 



The beam is subjected to a moment of M = 40 kN⋅m. Determine the bending stress acting at points A and B. sketch the results on a volume element acting at each of these points. A B M = 40 kN⋅m

50 mm 50 mm 50 mm

(

50 mm 50 mm 50 mm

)

(

)

 (0.05 ) 0.153   (0.05 ) 0.053  I = I y = Iz =   + 2  12 12     −5 4 I = 1.51(10 ) m − Mz y M yz σA = + Iz Iy  (40 )(0.075 )(1000 ) σA = 0+   1.5(10 −5 )   σ A = 199 MPa

 (40 )(0.025 )(1000 )  σB = 0+   1.51(10 −5 )   σ B = 66.2 MPa

Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 160 MPa. A

B

ΣM = 0 R A (1.2 ) = 600(2 ) − 400(0.6 ) 0.8 600

1.2

0.6

R A = 800 N 400

ΣM A = 0

RB (1.2 ) = 400(1.8) − 600(0.8) RB = 200 N

Mr πr 4 4 4M σ= 3 πr 4M r3 = πσ (4)(480) r =3 π 160(10 6 ) σ=

(

r = 0.0156m d = 31.3mm

)

The beam is subjected to the load P at its center. Determine the placement a of the supports so that the absolute maximum bending stress in the beam is as large as possible. What is this stress? P

d a

a

b L/2

L/2

+ ↑ ΣFy = 0 R AY + R BY = P R AY = R BY

For symmetrical loading

2 R AY = P

a = 0, L

P R AY = 2 ΣM c = 0

σ max =

MA = MB MA =

P  L    −a 2  2  

MB =

P  L    −a 2  2  

0≤a≤

L 2

@a = 0 PL   − 0 22  PL MA = 4 L @a = 2 P  L L   M A =  −   2  2 2   MA =

M max =

PL 4

(M max )(c ) I

3

bd 12 d c= 2 I=

σ max

σ max

PL  d    4 2 = bd 3 12  PL  12 =   2  8  bd

σ max =

( )

3 PL 2bd 2

  

The steel beam has the cross-sectional area shown. If w = 5 kip⋅ft, determine the absolute maximum bending stress in the beam.

wo

wo

8 in 0.3 in

0.30 in 10 in 0.30 in

8 ft

8 ft

8 ft

40 kip

-40 kip 160 kip.ft

ΣM A = 0

24 R B = 40(4 ) + 40(2 ) RB = 40kip

+ ↑ ΣFy = 0 R A + RB = 80 R A = 40kip

(

)

( )

  (0.3) 10 3   (8 ) 0.30 3 I = 2 + 8(0.3) 5.15 2  +     12   12 I = 152.3in 4 (160)(12)(5.3) σ max = 152.3 σ max = 66.8 ksi

(

)

The beam has a rectangular cross section as shown. Determine the largest load P that can be supported on its overhanging ends so that the bending stress does not exceed σmax = 10 MPa. P

P 1.5 m

1.5 m

1.5 m

ΣM A = 0 1.5 RB + 1.5 P = 3P RB = P + ↑ ΣFy = 0 R A + R B = 2P RA = P Mc I (0.5) 0.253 I= 12 I = 1.953(10 −4 )m 4 1.5 P(0.125) 10(10 6 ) = 1.953(10 − 4 ) P = 10.4kN σ max =

(

)

250 mm 150 mm

The beam has the rectangular cross section shown. If P =12 kN, determine that absolute maximum bending stress in the beam. Sketch the stress distribution acting over the cross section. P

P 1.5 m

1.5 m

1.5 m

ΣM A = 0

RB (1.5) + 12(1.5) = 12(3) RB = 12kN

+ ↑ ΣFy = 0 R A + R B = 24 R A = 12kN I=

(0.15)(0.253 )

12 I = 1.953125(10 −4 ) m 4 18(0.125) σ max = 1.953125(10 −4 ) 18000(0.125) σ max = 1.953125(10 −4 ) σ max = 11.5MPa

250 mm 150 mm

. A log that is 2 ft in diameter is to be cut into a rectangular section for use as a simply supported beam. If the allowable bending stress for the wood is σallow = 8 ksi, determine the largest load P that can be supported if the width of the beam is b = 8 in.

h b 2 ft P

8 ft

h=

(24

2

− 82

)

= 22.6 in ΣM B = 0 RA (16 ) = 8 P P 2 P RB = 2 bh 2 S= 6 (8)(22.6)2 = 6 = 681in 4 M , M = (96 )RA σ= S (96)(RA ) 8= 681 RA = 56.75 kip RA =

M = 96(56.75) = 5448 kip ⋅ ft P = 2 RA = 2(56.75) = 113.5 kip P = 114 kip

8 ft

The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 45°. z 125 mm 125 mm 250 mm

B

E M = 850 N⋅m

C

θ

A D

θ = 45° M Y = 850 cos 45 = 601.04 Nm M Z = 850 sin 45 = 601.04 Nm

(

)

(

)

1 3 ⋅ (0.25)(0.25) = 3.2552 10 −4 m 4 12 1 3 I Z = ⋅ (0.25)(0.25) = 3.2552 10 −4 m 4 12 − M Z y MY z σA = + IZ IY IY =

− 601.04(− 0.125) 601.04(− 0.125) + 3.2552 10 −4 3.2552 10 − 4 =0 =

(

)

(

)

− 601.04(− 0.125) (0.125)601.04 + 3.2552 10 − 4 3.2552 10 − 4 = 462 kPa

σB =

(

)

(

)

− 601.04(0.125) 601.04(0.125) + 3.2552 10 −4 3.2552 10 −4 = −462 kPa

σD =

(

)

(

)

− 601.04(0.125) 601.04(0.125) + 3.2552 10 −4 3.2552 10 −4 =0

σE =

(

)

(

)

y

The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 30°. z 125 mm 125 mm 250 mm

B

E M = 850 N⋅m

C

θ

A D

θ = 30° M Y = 850 cos 30 = 736.12 Nm M Z = 850 sin 30 = 425 Nm

(

)

(

)

1 3 ⋅ (0.25)(0.25) = 3.2552 10− 4 m4 12 1 3 I Z = ⋅ (0.25)(0.25) = 3.2552 10− 4 m4 12 − M Z y MY z σA = + IZ IY IY =

− 425(− 0.125) 736.12(− 0.125) + 3.2552 10− 4 3.2552 10− 4 = −119 kPa − 425(− 0.125) 736.12(0.125) σB = + 3.2552 10− 4 3.2552 10− 4 = 446 kPa − 425(0.125) 736.12(0.125) σD = + 3.2552 10− 4 3.2552 10− 4 = −446 kPa =

(

)

(

)

(

)

(

)

(

(

)

)

− 425(0.125) 736.12(0.125) + 3.2552 10− 4 3.2552 10− 4 = 119 kPa

σE =

(

)

(

)

y

The T-beam is subjected to a moment of M = 150 kip⋅in. directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis. The location y of the centroid, C, must be determined.

M Y = 150 sin 60 = 129.9038 kip ⋅ in M Z = −150 cos 60 = −75 kip ⋅ in Σy A ΣA [(4)(8)(2) + (9)(2)(12)] = (8)(2) + (2)(12) = 7 in

yo =

 1   1 I Y =  (8) 2 3 + (2 ) 12 3  12  12   4 = 293.33 in

( )

( )

 1  1 2 3 2 I Z =  (2 ) 8 3 + 8(2)(7 − 4 )  +  (12)(2 ) + 12(2 )(9 − 7 )    12   12 = 333.33 in 4

( )

− 75(3) 129.9(6) + 333.333 293.333 = 3.33 ksi

σA =−

− 75(− 7 ) 129.9(− 1) + 333.33 293.333 = −2.0178 ksi

σB = −

∴ σ max = 3.33 ksi

tan α = =

IZ (tan θ ) IY

(333.333) (tan 30)

293.33 α = −33.2 − 30 = −63.2°

6-185. Determine the bending stress distribution in the beam at section a-a. Sketch the distribution in three dimension acting over the cross section.

( )

(

 (0.15) 0.13  (0.75) 0.153 I= + 12 12   = 2.52 10 −6 m 4 Mc σ = I (32)(0.050)Nm 2 = 2.52 = 634.6 kPa

(

)

)

The aluminum machine part is subjected to a moment of M = 75 N⋅m. Determine the bending stress created at points B and C on the cross section. Sketch the results on a volume element located at each of these points. 20 mm 10 mm

10 mm

10 mm

20 mm

10 mm B N 10 mm C

M = 75 N⋅m

40 mm

0.005(0.08)(0.01) + 2(0.04)(0.01)(0.03) 0.01(0.08) + (0.04)(0.01) = 0.0175 m

y=

 1  1  2 I Z =  (0.08) 0.013 + (0.08)(0.01)(0.0175 − 0.005) 2  + 2  (0.01) 0.04 3 + (0.03 − 0.0175)  12   12  

(

)

( )

= 0.3633 10 6 m 4 (75)(− 0.0175) σB = − 0.3633 10 −6 σ B = 3.61MPa

(

)

− (75)(− 0.0075) 0.3633 10 −6 σ c = 1.55MPa σC =

(

)

(

)

A beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is M = 450 N⋅m, determine the resultant force that bending stress produces on the top board A and on the side board B. A

B 15 mm M = 40 kN⋅m 20 mm 200 mm 20 mm 200 mm

15 mm

( ( )

)

(

)

(

)

  1 1 I y = 2  (0.2) 0.02 3 + (0.2 )(0.02) 0.112  + 2  (0.015) 0.24 3    12 12 −4 4 = 1.316 10 m − MZ y MY z σA = + IZ Iy σA = 0+0 =0 FRA = 0

450(0.12 ) 1.316 10 −4 = 0.410MPa (410334.34)(0.24)(0.015) FRB = 1000 = 1.5kN

σB = 0+

(

)

The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N⋅m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section. 25 mm

20 mm 200 mm 20 mm

Σy A ΣA ( 0.0125)(0.025)(0.24 ) + (2 )(0.1)(0.15)(0.02) = 0.025(0.24 ) + 2(0.15)(0.02) y = 0.05625m C = 0.175 − 0.05625 = 0.11875m y=

(

)

1 2 I =  (0.24) 0.0253 + (0.24)(0.025)(0.05625 − 0.0125)  + 12  1 2 2  (0.02) 0.153 + (0.02 )(0.15)(0.1 − 0.05625)  12 

( ) = 34.53(10 )m −6

4

Mc I 600(0.11875) = 24.53 10 −6 = 2.06MPa

σ max =

(

)

M = 600 N⋅m

If the internal moment acting on the cross section of the strut has a magnitude of M = 800 N⋅m and is directed as shown, determine the bending stress at points A and B. The location z of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis. y 12 mm

12 mm

200 mm

z 12 mm

60°

150 mm

M = 800 N⋅m

z

z=

(0.006)(0.012)(0.4) + (2)(0.081)(0.012)(0.138) (0.012)(0.4) + (2)(0.012)(0.138)

= 0.366m C = 0.15 − 0.0366 = 0.1134m

(

)

1 2 I Y =  (0.4 ) 0.012 3 + (0.4 )(0.012)(0.081 − 0.0366)   12 −6 4 = 16.3374 10 m

(

IZ

)  1 = 2  (0.138)(0.012 ) + (0.138)(0.012)(0.194 ) + 12 3

2

 1 3  12 (0.012) 0.4  = 0.18869 10 −3 m 4 − 400(0.2 ) (− 692.82)(− 0.1134) σA = + 0.1886 10 −3 16.3374 10 −6 = 4.38MPa

(

)

(

)

(

)

(

)

− 400(− 0.2 ) − 692.82(0.0366) + 0.11869 10 −3 16.3374 10 −6 = −1.13MPa I tan α = Z tan θ IY

σB =

(

)

( (

(

) )

 0.18869 10 −3  tan α =  tan 30 −6  16.3374 10  α = −87.1°

)



200 mm

The axle of the freight car is subjected to wheel loadings of 20 kip. If it is supported by two journal bearings at C and D, determine that maximum bending stress developed at the center of the axle, where the diameter is 5.5 in.

ΣM D = 0

20(70) + 20 = RC (80) RC = 20kip

ΣFy = 0 20 + 20 − 20 = RD R

D

= 20 kip

Mc I Mr = 4 πr 4 (4)(200) = π 2.75 3 σ = 12.24ksi σ =

(

)

6-74. Determine the absolute maximum bending stress in the 1.5-in.-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces.

ΣM B = 0

R A (30) = 400(18) − 300(15)

R A = 90 lb ΣFY = 0 RB = 400 + 300 − 90 RB = 610 lb

Mc I 64Mr = πd 4

σ =

=

64(4500)(7.5)

π (1.5 ) = 13.6 Ksi

4

d  π  2 I=   4 4 πd I= 64

4

6-75. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 22 ksi.

Given: σ = 22 Ksi d =?

Solution: ΣM B = 0

R A (30) = 400(18) − 300(15)

R A = 90 lb ΣFY = 0 RB = 400 + 300 − 90 RB = 610 lb σ =

Mc I 3

σ =

32M πd 3

32(4500) πd 3 144000 d3 = π (22000)

22 =

d = 3 2.1 d = 1.28 in

;

d  π  πd 3 Mr 2 I= 4 =   = 4 32 πr 4

Determine the absolute maximum bending stress in the 30-mm-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces. A

0.8 m

1.2 m

0.6 m 400 N

600 N

3

d  π  πd 3 Mc 2  σ = ; I= = 4 32 I Mr = 2 πr 4 32M = πd 3 32(480) = 3 π (0.030) σ = 181 N / m 2

B

6-79. The steel shaft has a circular cross section with a diameter of 2 in. It is supported on smooth journal bearings A and B, which exert only vertical reactions on the shaft. Determine the absolute maximum bending stress if it is subjected to the pulley loadings shown.

ΣM B = 0 ; R A (80) = 500(60) + 300(40) + 500(20) R A = 650lb ΣM A = 0 RB (80) = 800(60) + 300(40) + 500(20) RB = 650lb Mc Mr 4 M = 4 = 3 I πr πr 4 4(16000) = π 13 σ = 20.4ksi

σ =

( )

Determine the absolute maximum bending stress in the 20mm diameter pin. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4kN/m caused by bar C. Determine the intensity of the distributed load wo of the leaves on the pin and draw the shear and moment diagrams for the pin.

0.4 kN/m

Mc I π 90.014 I= 4 = 7.853 10 −9 m 4

σ= wo 20 mm

wo 60 mm

20 mm

(

0.00026(0.01) 7.853 10 −9 = 331kPa

σ max = σ max 2.6x10-3 kN⋅m

)

(

)

If the beam ABC has square cross-section 6in by 6in, determine the absolute bending stress in the beam. The 20kip load must be replaced by equivalent loadings at point C on the axis of the beam.

( )

1 (6) 63 12 I = 108in 4 I=

Mc I 46.68(3)(12 ) = 108 = 15.6ksi

σ max = σ max

6-67. If the crane boom ABC has a rectangular cross-section with the base of 2in a height of 3in, determine the absolute maximum bending stress in the boom. The engine crane is used to support the engine, which has a weight of 1200 lb.

( )

1 (2) 33 12 = 4.5in 4

I=

 kip  12in   M = 6000lb. ft    1000lb  ft  = 72kip.in 72(1.5) 4.5 = 24ksi

σ max =

6-86. The simply supported beam is made from four ¾ in diameter rod w/c are bounded as shown determine the maximum bending stress in the beam due to the loading shown.

ΣM B = 0

R A (10) = 80(8) + 80(2 )

R A = 80lb ΣM A = 0 RB = 80lb 3 (4)in − 1 (4)in 4 4 = 2in Mr M σ = 3 = 32 3 πr πd 4 160lb. ft = 1920lb.in 32M σ = πd 3 32(1920) = π 23 = 2.4ksi d=

6-48. The beam is made from three boards nailed together as shown. If the moment acting on the cross-section is M=600N.m, determine the resultant force bending stress produces on the top board. y 945 = 25 93.75 y = 252kPa 945 + 252 = 598.5kPa 2 FR = (598.5)(25)(240) / 1000 Ave. stress =

FR = 3.591N

5-59. The beam is subjected to a moment M=30lb.ft. Determine the bending stress at points A and B. Also, sketch a three dimensional view of the stress distribution acting over the entire cross-sectional area.

y=

2(1)(0.5) + (2)(0.5)(3)(1)(2 ) 2(1) + 2(0.5)(3)(1)

= 1.4in 2  3 3  1 2(1) 2 + (2) 0.9 2 + 2   + (1)(3)(2 − 1.4 )  12   36  2 4 = 4.367in 30(12 )(4 − 1.4 ) σA = 4.367 = 214.35 psi 30(12 )(0.4 ) σD = 4.367 = 32.977 psi I=

(

)

(Mindanao State University-General Santos City, Philippines) BSEE-(G.R.F) students 2004