6_shearing_stresses1.pdf

CHAPTER

6

MECHANICS OF SOLIDS Shearing Stresses in Beams and ThinWalled Members

GIK Institute of Engineering Sciences and Technology.

MECHANICS OF SOLIDS Shearing Stresses in Beams and Thin-Walled Members Introduction Shear on the Horizontal Face of a Beam Element Example 6.01 Determination of the Shearing Stress in a Beam Shearing Stresses txy in Common Types of Beams Sample Problem 6.2 Longitudinal Shear on a Beam Element of Arbitrary Shape Example 6.04 Shearing Stresses in Thin-Walled Members

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS Introduction • Transverse loading applied to a beam results in normal and shearing stresses in transverse sections. • Distribution of normal and shearing stresses satisfies Fx    x dA  0 Fy  t xy dA  V Fz   t xz dA  0





M x   yt xz  z t xy dA  0 M y   z  x dA  0 M z    y  x   0

• When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces • Longitudinal shearing stresses must exist in any member subjected to transverse loading. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Shear on the Horizontal Face of a Beam Element • Consider prismatic beam • For equilibrium of beam element  Fx  0  H    D   D dA A

H 

M D  MC  y dA I A

• Note, Q   y dA A

M D  MC 

dM x  V x dx

• Substituting, VQ x I H VQ q   shear flow x I

H 

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MECHANICS OF SOLIDS Shear on the Horizontal Face of a Beam Element • Shear flow, q

H VQ   shear flow x I

• where Q   y dA A

 first moment of area above y1 I

2  y dA

A  A'

 second moment of full cross section

• Same result found for lower area

H  VQ    q x I Q  Q  0  first moment with respect to neutral axis H    H q 

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MECHANICS OF SOLIDS Example 6.01 SOLUTION:

• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank. • Calculate the corresponding shear force in each nail.

A beam is made of three planks, nailed together. Knowing that the spacing between nails is 25 mm and that the vertical shear in the beam is V = 500 N, determine the shear force in each nail.

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MECHANICS OF SOLIDS Example 6.01 SOLUTION:

• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank.

Q  Ay  0.020 m  0.100 m 0.060 m   120  10 6 m3 I

1 0.020 m 0.100 m 3  12 1 0.100 m 0.020 m 3  2[12

 0.020 m  0.100 m 0.060 m 2 ]  16.20  10 6 m 4 GIK Institute of Engineering Sciences and Technology

VQ (500 N )(120  10 6 m3 ) q  I 16.20  10 -6 m 4  3704 N m

• Calculate the corresponding shear force in each nail for a nail spacing of 25 mm. F  (0.025 m)q  (0.025 m)(3704 N m

F  92.6 N

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MECHANICS OF SOLIDS Determination of the Shearing Stress in a Beam • The average shearing stress on the horizontal face of the element is obtained by dividing the shearing force on the element by the area of the face. H q x VQ x   A A I t x VQ  It

t ave 

• On the upper and lower surfaces of the beam, tyx= 0. It follows that txy= 0 on the upper and lower edges of the transverse sections. • If the width of the beam is comparable or large relative to its depth, the shearing stresses at D1 and D2 are significantly higher than at D. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS

Shearing Stresses txy in Common Types of Beams • For a narrow rectangular beam, VQ 3 V  t xy   1  Ib 2 A 

t max 

y 2  c 2 

3V 2A

• For American Standard (S-beam) and wide-flange (W-beam) beams VQ It V t max  Aweb

t ave 

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MECHANICS OF SOLIDS • For a narrow rectangular beam, VQ 3 V  t xy   1  Ib 2 A 

t max 

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y 2  c 2 

3V 2A

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MECHANICS OF SOLIDS Sample Problem 6.2 SOLUTION: • Develop shear and bending moment diagrams. Identify the maximums. • Determine the beam depth based on allowable normal stress. A timber beam is to support the three concentrated loads shown. Knowing that for the grade of timber used,  all  1800 psi

t all  120 psi

• Determine the beam depth based on allowable shear stress. • Required beam depth is equal to the larger of the two depths found.

determine the minimum required depth d of the beam.

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MECHANICS OF SOLIDS Sample Problem 6.2 SOLUTION: Develop shear and bending moment diagrams. Identify the maximums. Vmax  3 kips M max  7.5 kip  ft  90 kip  in

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MECHANICS OF SOLIDS Sample Problem 6.2 • Determine the beam depth based on allowable normal stress.  all 

M max S

1800 psi 

90  103 lb  in.

0.5833 in. d 2

d  9.26 in. 1 bd3 I  12 I S   16 b d 2 c

 16 3.5 in.d 2  0.5833 in.d 2

• Determine the beam depth based on allowable shear stress. 3 Vmax 2 A 3 3000 lb 120 psi  2 3.5 in. d d  10.71in.

t all 

• Required beam depth is equal to the larger of the two. d  10.71in. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS

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MECHANICS OF SOLIDS Longitudinal Shear on a Beam Element of Arbitrary Shape • We have examined the distribution of the vertical components txy on a transverse section of a beam. We now wish to consider the horizontal components txz of the stresses. • Consider prismatic beam with an element defined by the curved surface CDD’C’.  Fx  0  H    D   C dA a

• Except for the differences in integration areas, this is the same result obtained before which led to H 

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VQ x I

q

H VQ  x I 6 - 17

MECHANICS OF SOLIDS Example 6.04 SOLUTION: • Determine the shear force per unit length along each edge of the upper plank. • Based on the spacing between nails, determine the shear force in each nail. A square box beam is constructed from four planks as shown. Knowing that the spacing between nails is 1.5 in. and the beam is subjected to a vertical shear of magnitude V = 600 lb, determine the shearing force in each nail. GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Example 6.04 SOLUTION: • Determine the shear force per unit length along each edge of the upper plank.





VQ 600 lb  4.22 in 3 lb q   92 . 3 I in 27.42 in 4 q lb  46.15 2 in  edge force per unit length

f 

For the upper plank, Q  Ay  0.75in.3 in.1.875 in.  4.22 in 3

For the overall beam cross-section, 1 4.5 in   1 3 in  I  12 12 3

3

 27.42 in 4 GIK Institute of Engineering Sciences and Technology

• Based on the spacing between nails, determine the shear force in each nail. lb   F  f    46.15 1.75 in  in   F  80.8 lb 6 - 19

MECHANICS OF SOLIDS Shearing Stresses in Thin-Walled Members • Consider a segment of a wide-flange beam subjected to the vertical shear V. • The longitudinal shear force on the element is H 

VQ x I

• The corresponding shear stress is t zx  t xz 

H VQ  t x It

• Previously found a similar expression for the shearing stress in the web t xy 

VQ It

• NOTE: t xy  0 t xz  0 GIK Institute of Engineering Sciences and Technology

in the flanges in the web 6 - 20

MECHANICS OF SOLIDS Shearing Stresses in Thin-Walled Members • The variation of shear flow across the section depends only on the variation of the first moment. q tt 

VQ I

• For a box beam, q grows smoothly from zero at A to a maximum at C and C’ and then decreases back to zero at E. • The sense of q in the horizontal portions of the section may be deduced from the sense in the vertical portions or the sense of the shear V.

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MECHANICS OF SOLIDS Shearing Stresses in Thin-Walled Members • For a wide-flange beam, the shear flow increases symmetrically from zero at A and A’, reaches a maximum at C and the decreases to zero at E and E’. • The continuity of the variation in q and the merging of q from section branches suggests an analogy to fluid flow.

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MECHANICS OF SOLIDS Sample Problem 6.3 SOLUTION: • For the shaded area, Q  4.31in 0.770 in 4.815 in   15.98 in 3

• The shear stress at a, Knowing that the vertical shear is 50 kips in a W10x68 rolled-steel beam, determine the horizontal shearing stress in the top flange at the point a.

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 



VQ 50 kips 15.98 in 3 t  It 394 in 4 0.770 in 



t  2.63 ksi

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