4_pure_bending1.ppt

CHAPTER

4

MECHANICS OF SOLIDS Pure Bending

GIK Institute of Engineering Sciences and Technology.

MECHANICS OF SOLIDS Pure Bending Pure Bending Other Loading Types Symmetric Member in Pure Bending Bending Deformations Strain Due to Bending Beam Section Properties Properties of American Standard Shapes Deformations in a Transverse Cross Section Sample Problem 4.2 Bending of Members Made of Several Materials Example 4.03

GIK Institute of Engineering Sciences and Technology

Eccentric Axial Loading in a Plane of Symmetry Example 4.07 Sample Problem 4.8 Unsymmetric Bending Example 4.08 General Case of Eccentric Axial Loading

4-2

MECHANICS OF SOLIDS Pure Bending

Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane GIK Institute of Engineering Sciences and Technology

4-3

MECHANICS OF SOLIDS Other Loading Types • Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple • Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple • Principle of Superposition: The normal stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress. GIK Institute of Engineering Sciences and Technology

4-4

MECHANICS OF SOLIDS Symmetric Member in Pure Bending • Internal forces in any cross section are equivalent to a couple. The moment of the couple is the section bending moment. • From statics, a couple M consists of two equal and opposite forces. • The sum of the components of the forces in any direction is zero. • The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane. • These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forces. Fx    x dA  0 M y   z x dA  0 M z    y x dA  M GIK Institute of Engineering Sciences and Technology

4-5

MECHANICS OF SOLIDS Bending Deformations Beam with a plane of symmetry in pure bending: • member remains symmetric

• bends uniformly to form a circular arc • cross-sectional plane passes through arc center and remains planar • length of top decreases and length of bottom increases • a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change • stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it GIK Institute of Engineering Sciences and Technology

4-6

MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology

4-7

MECHANICS OF SOLIDS Strain Due to Bending Consider a beam segment of length L. After deformation, the length of the neutral surface remains L. At other sections, L     y 

  L  L     y      y x  m 





L c



y



or

 ρ

y



(strain varies linearly)

c

m

y c

 x   m

GIK Institute of Engineering Sciences and Technology

4-8

MECHANICS OF SOLIDS Stress Due to Bending • For a linearly elastic material, y c

 x  E x   E m y    m (stress varies linearly) c

• For static equilibrium, y Fx  0    x dA     m dA c

 0   m  y dA c

First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid.

• For static equilibrium,  y  M    y x dA    y   m  dA  c 

  I M  m  y 2 dA  m c c m 

Mc M  I S

y Substituting  x    m c

x   GIK Institute of Engineering Sciences and Technology

My I 4-9

MECHANICS OF SOLIDS Beam Section Properties • The maximum normal stress due to bending, Mc M  I S I  section moment of inertia I S   section modulus c

m 

A beam section with a larger section modulus will have a lower maximum stress • Consider a rectangular beam cross section, 3 1 I 12 bh S   16 bh3  16 Ah c h2

Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending. • Structural steel beams are designed to have a large section modulus. GIK Institute of Engineering Sciences and Technology

4 - 10

MECHANICS OF SOLIDS Properties of American Standard Shapes

GIK Institute of Engineering Sciences and Technology

4 - 11

MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology

4 - 12

MECHANICS OF SOLIDS Deformations in a Transverse Cross Section • Deformation due to bending moment M is quantified by the curvature of the neutral surface   1 Mc  m  m   c Ec Ec I M  EI 1

• Although cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonzero,  y   x 

y 

 z   x 

y 

• Expansion above the neutral surface and contraction below it cause an in-plane curvature, 1    anticlastic curvature  

GIK Institute of Engineering Sciences and Technology

4 - 13

MECHANICS OF SOLIDS Sample Problem 4.2 SOLUTION: • Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. Y 

 yA A



I x   I  A d 2



• Apply the elastic flexural formula to find the maximum tensile and compressive stresses. m 

A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = 165 GPa and neglecting the effects of fillets, determine (a) the maximum tensile and compressive stresses, (b) the radius of curvature. GIK Institute of Engineering Sciences and Technology

Mc I

• Calculate the curvature 1





M EI

4 - 14

MECHANICS OF SOLIDS Sample Problem 4.2 SOLUTION:

Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. Area, mm2 1 20  90  1800 2 40  30  1200  A  3000

y , mm 50 20

yA, mm3 90 103 24 103 3  yA  114 10

3

 yA 114 10 Y    38 mm 3000 A



 

1 bh3  A d 2 I x   I  A d 2   12







1 90  203  1800  12 2  1 30  403  1200  182  12 12

I  868 103 mm  868 10-9 m 4 GIK Institute of Engineering Sciences and Technology

4 - 15



MECHANICS OF SOLIDS Sample Problem 4.2 • Apply the elastic flexural formula to find the maximum tensile and compressive stresses. Mc I M c A 3 kN  m  0.022 m A   I 868 10 9 mm4 M cB 3 kN  m  0.038 m B    I 868 10 9 mm4

m 

 A  76.0 MPa  B  131.3 MPa

• Calculate the curvature 1



 

GIK Institute of Engineering Sciences and Technology

M EI 3 kN  m

165 GPa 868 10-9 m 4 

1

 20.95 10 3 m-1

   47.7 m

4 - 16

MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology

4 - 17

MECHANICS OF SOLIDS Bending of Members Made of Several Materials • Consider a composite beam formed from two materials with E1 and E2. • Normal strain varies linearly. x  

y



• Piecewise linear normal stress variation. 1  E1 x  

E1 y



 2  E2 x  

E2 y



Neutral axis does not pass through section centroid of composite section. • Elemental forces on the section are Ey E y dF1  1dA   1 dA dF2   2dA   2 dA



x  

My I

1   x



• Define a transformed section such that  2  n x

GIK Institute of Engineering Sciences and Technology

dF2  

nE1  y dA   E1 y n dA 



E n 2 E1 4 - 18

MECHANICS OF SOLIDS Example 4.03 SOLUTION: • Transform the bar to an equivalent cross section made entirely of brass • Evaluate the cross sectional properties of the transformed section • Calculate the maximum stress in the transformed section. This is the correct maximum stress for the brass pieces of the bar. Bar is made from bonded pieces of steel (Es = 29x106 psi) and brass (Eb = 15x106 psi). Determine the maximum stress in the steel and brass when a moment of 40 kip*in is applied. GIK Institute of Engineering Sciences and Technology

• Determine the maximum stress in the steel portion of the bar by multiplying the maximum stress for the transformed section by the ratio of the moduli of elasticity. 4 - 19

MECHANICS OF SOLIDS Example 4.03 SOLUTION: • Transform the bar to an equivalent cross section made entirely of brass. Es 29 106 psi n   1.933 Eb 15 106 psi bT  0.4 in  1.933  0.75 in  0.4 in  2.25 in

• Evaluate the transformed cross sectional properties 1 b h3  1 2.25 in.3 in 3 I  12 T 12

 5.063 in 4

• Calculate the maximum stresses m 

Mc 40 kip  in 1.5 in    11.85 ksi 4 I 5.063 in

 b max   m  s max  n m  1.933 11.85 ksi GIK Institute of Engineering Sciences and Technology

 b max  11.85 ksi  s max  22.9 ksi 4 - 20

MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology

4 - 21

MECHANICS OF SOLIDS Eccentric Axial Loading in a Plane of Symmetry • Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment  x   x centric   x bending 

• Eccentric loading FP M  Pd

P My  A I

• Validity requires stresses below proportional limit, deformations have negligible effect on geometry, and stresses not evaluated near points of load application.

GIK Institute of Engineering Sciences and Technology

4 - 22

MECHANICS OF SOLIDS Example 4.07 SOLUTION: • Find the equivalent centric load and bending moment

• Superpose the uniform stress due to the centric load and the linear stress due to the bending moment. • Evaluate the maximum tensile and compressive stresses at the inner and outer edges, respectively, of the superposed stress distribution.

An open-link chain is obtained by bending low-carbon steel rods into the shape shown. For 160 lb load, determine • Find the neutral axis by determining the location where the normal stress (a) maximum tensile and compressive is zero. stresses, (b) distance between section centroid and neutral axis GIK Institute of Engineering Sciences and Technology

4 - 23

MECHANICS OF SOLIDS Example 4.07 • Normal stress due to a centric load A  c 2   0.25 in 2  0.1963 in 2 P 160 lb 0   A 0.1963 in 2  815 psi

• Equivalent centric load and bending moment P  160 lb M  Pd  160 lb 0.6 in   104 lb  in

• Normal stress due to bending moment I  14 c 4  14  0.25 4  3.068  10 3 in 4 Mc 104 lb  in 0.25 in  m   I .068  10 3 in 4  8475 psi

GIK Institute of Engineering Sciences and Technology

4 - 24

MECHANICS OF SOLIDS Example 4.07

• Maximum tensile and compressive stresses t  0 m  815  8475 c  0  m  815  8475

 t  9260 psi

 c  7660 psi

• Neutral axis location 0

P My0  A I

P I 3.068 10 3 in 4 y0   815 psi AM 105 lb  in y0  0.0240 in

GIK Institute of Engineering Sciences and Technology

4 - 25

MECHANICS OF SOLIDS Sample Problem 4.8 The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MPa in compression. Determine the largest force P which can be applied to the link. SOLUTION: • Determine an equivalent centric load and bending moment. • Superpose the stress due to a centric load and the stress due to bending. From Sample Problem 2.4, A  3 10 3 m 2 Y  0.038 m I  868 10 9 m 4 GIK Institute of Engineering Sciences and Technology

• Evaluate the critical loads for the allowable tensile and compressive stresses. • The largest allowable load is the smallest of the two critical loads. 4 - 26

MECHANICS OF SOLIDS Sample Problem 4.8 • Determine an equivalent centric and bending loads. d  0.038  0.010  0.028 m P  centric load M  Pd  0.028 P  bending moment

• Superpose stresses due to centric and bending loads

0.028 P 0.022   377 P P Mc A P    A I 3 10 3 868 10 9 0.028 P 0.022   1559 P P Mc P B    A    A I 3 10 3 868 10 9

A  

• Evaluate critical loads for allowable stresses.  A  377 P  30 MPa

P  79.6 kN

 B  1559 P  120 MPa

P  79.6 kN

• The largest allowable load GIK Institute of Engineering Sciences and Technology

P  77.0 kN

4 - 27

MECHANICS OF SOLIDS Unsymmetric Bending • Analysis of pure bending has been limited to members subjected to bending couples acting in a plane of symmetry. • Members remain symmetric and bend in the plane of symmetry.

• The neutral axis of the cross section coincides with the axis of the couple • Will now consider situations in which the bending couples do not act in a plane of symmetry. • Cannot assume that the member will bend in the plane of the couples. • In general, the neutral axis of the section will not coincide with the axis of the couple. GIK Institute of Engineering Sciences and Technology

4 - 28

MECHANICS OF SOLIDS Unsymmetric Bending • 0  Fx    x dA      m dA y  c



or 0   y dA

neutral axis passes through centroid

Wish to determine the conditions under which the neutral axis of a cross section of arbitrary shape coincides with the axis of the couple as shown. • The resultant force and moment from the distribution of elementary forces in the section must satisfy Fx  0  M y

M z  M  applied couple

GIK Institute of Engineering Sciences and Technology

 y  M  M   y     m dA • z  c  σ I or M  m I  I z  moment of inertia c

defines stress distribution • 0  M y   z x dA   z   m dA

y  c  or 0   yz dA  I yz  product of inertia

couple vector must be directed along a principal centroidal axis 4 - 29

MECHANICS OF SOLIDS Unsymmetric Bending Superposition is applied to determine stresses in the most general case of unsymmetric bending. • Resolve the couple vector into components along the principle centroidal axes. M z  M cos

M y  M sin 

• Superpose the component stress distributions x  

Mzy Myy  Iz Iy

• Along the neutral axis, x  0   tan  

GIK Institute of Engineering Sciences and Technology

M cos  y  M sin   y Mzy Myy   Iz Iy Iz Iy

y Iz  tan  z Iy

4 - 30

MECHANICS OF SOLIDS Example 4.08 SOLUTION: • Resolve the couple vector into components along the principle centroidal axes and calculate the corresponding maximum stresses. M z  M cos 

M y  M sin 

• Combine the stresses from the component stress distributions. x  

Mzy Myy  Iz Iy

A 1600 lb-in couple is applied to a rectangular wooden beam in a plane • Determine the angle of the neutral forming an angle of 30 deg. with the axis. vertical. Determine (a) the maximum y Iz tan    tan  stress in the beam, (b) the angle that the z Iy neutral axis forms with the horizontal plane. GIK Institute of Engineering Sciences and Technology

4 - 31

MECHANICS OF SOLIDS Example 4.08 • Resolve the couple vector into components and calculate the corresponding maximum stresses. M z  1600 lb  in  cos 30  1386 lb  in M y  1600 lb  in sin 30  800 lb  in 1 1.5 in 3.5 in 3  5.359 in 4 I z  12 1 3.5 in 1.5 in 3  0.9844 in 4 I y  12 The largest tensile stress due to M z occurs along AB

1 

M z y 1386 lb  in 1.75 in   452.6 psi  4 Iz 5.359 in

The largest tensile stress due to M z occurs along AD

2 

M yz Iy



800 lb  in 0.75 in   609.5 psi 0.9844 in 4

• The largest tensile stress due to the combined loading occurs at A.  max   1   2  452.6  609.5 GIK Institute of Engineering Sciences and Technology

 max  1062 psi 4 - 32

MECHANICS OF SOLIDS Example 4.08 • Determine the angle of the neutral axis. Iz 5.359 in 4 tan   tan   tan 30 4 Iy 0.9844 in  3.143

  72.4o

GIK Institute of Engineering Sciences and Technology

4 - 33

MECHANICS OF SOLIDS General Case of Eccentric Axial Loading • Consider a straight member subject to equal and opposite eccentric forces. • The eccentric force is equivalent to the system of a centric force and two couples. P  centric force M y  Pa M z  Pb

• By the principle of superposition, the combined stress distribution is P Mz y M yz x    A Iz Iy

• If the neutral axis lies on the section, it may be found from My Mz P y z Iz Iy A GIK Institute of Engineering Sciences and Technology

4 - 34

MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology

4 - 35

MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology

4 - 36

MECHANICS OF SOLIDS

GIK Institute of Engineering Sciences and Technology

4 - 37

MECHANICS OF SOLIDS Practice Problems

4.1, 4.4, 4.5, 4.6, 4.9, 4.17, 4.20, 4.25, 4.34, 4.38, 4.55, 4.100, 4.102, 4.105, 4.106, 4.107, 4.112, 4.113, 4.127, 4.129, 4.132, 4.134, 4.136, 4.138, 4.140, 4.146, 4.148 GIK Institute of Engineering Sciences and Technology

4 - 38