MECHANICS OF MATERIALS

Third Edition

CHAPTER

MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University

Shearing Stresses in Beams and ThinWalled Members

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

Third Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

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 τxy in Common Types of Beams Further Discussion of the Distribution of Stresses in a ... Sample Problem 6.2 Longitudinal Shear on a Beam Element of Arbitrary Shape Example 6.04 Shearing Stresses in Thin-Walled Members Plastic Deformations Sample Problem 6.3 Unsymmetric Loading of Thin-Walled Members Example 6.05 Example 6.06 © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Introduction • Transverse loading applied to a beam results in normal and shearing stresses in transverse sections. • Distribution of normal and shearing stresses satisfies Fx = ∫ σ x dA = 0 Fy = ∫ τ xy dA = −V Fz = ∫ τ xz dA = 0

(

)

M x = ∫ y τ xz − z τ xy dA = 0 M y = ∫ z σ x dA = 0 M z = ∫ (− y σ 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Shear on the Horizontal Face of a Beam Element • Consider prismatic beam • For equilibrium of beam element ∑ Fx = 0 = ∆H + ∫ (σ D − σ 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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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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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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Example 6.01 SOLUTION: • Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank. VQ (500 N )(120 × 10− 6 m3 ) q= = I 16.20 × 10-6 m 4 = 3704 N m

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

• Calculate the corresponding shear force in each nail for a nail spacing of 25 mm.

+ (0.020 m × 0.100 m )(0.060 m )2 ] = 16.20 × 10− 6 m 4 © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

F = (0.025 m)q = (0.025 m)(3704 N m

F = 92.6 N

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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

τ ave =

• On the upper and lower surfaces of the beam, τyx= 0. It follows that τxy= 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Shearing Stresses τxy in Common Types of Beams • For a narrow rectangular beam, VQ 3 V ⎛⎜ 1− τ xy = = ⎜ Ib 2 A ⎝

τ max =

y 2 ⎞⎟ c 2 ⎟⎠

3V 2A

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

τ ave =


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Further Discussion of the Distribution of Stresses in a Narrow Rectangular Beam • Consider a narrow rectangular cantilever beam subjected to load P at its free end: 3 P ⎛⎜ τ xy = 1− 2 A ⎝⎜

y 2 ⎞⎟ c 2 ⎟⎠

σx = +

Pxy I

• Shearing stresses are independent of the distance from the point of application of the load. • Normal strains and normal stresses are unaffected by the shearing stresses. • From Saint-Venant’s principle, effects of the load application mode are negligible except in immediate vicinity of load application points. • Stress/strain deviations for distributed loads are negligible for typical beam sections of interest. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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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, σ all = 1800 psi

τ 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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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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Sample Problem 6.2 • Determine the beam depth based on allowable normal stress. σ 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.

τ all =

• Required beam depth is equal to the larger of the two. d = 10.71in. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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Longitudinal Shear on a Beam Element of Arbitrary Shape • We have examined the distribution of the vertical components τxy on a transverse section of a beam. We now wish to consider the horizontal components τxz of the stresses. • Consider prismatic beam with an element defined by the curved surface CDD’C’. ∑ Fx = 0 = ∆H + ∫ (σ D − σ C )dA a

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


VQ ∆x I

q=

∆H VQ = ∆x I 6 - 14

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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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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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 in 2 = 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 © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

• Based on the spacing between nails, determine the shear force in each nail. lb ⎞ ⎛ F = f A = ⎜ 46.15 ⎟(1.75 in ) in ⎠ ⎝

F = 80.8 lb 6 - 16

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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 τ zx = τ xz ≈

∆H VQ = t ∆x It

• Previously found a similar expression for the shearing stress in the web τ xy =

VQ It

• NOTE: τ xy ≈ 0 τ xz ≈ 0 © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

in the flanges in the web 6 - 17

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Shearing Stresses in Thin-Walled Members • The variation of shear flow across the section depends only on the variation of the first moment. q =τt =

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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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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Plastic Deformations I c

• Recall: M Y = σ Y = maximum elastic moment • For M = PL < MY , the normal stress does not exceed the yield stress anywhere along the beam. • For PL > MY , yield is initiated at B and B’. For an elastoplastic material, the half-thickness of the elastic core is found from ⎛ 1 yY2 ⎞ 3 Px = M Y ⎜1 − 2 ⎟ ⎜ 3c ⎟ 2 ⎝ ⎠

• The section becomes fully plastic (yY = 0) at the wall when 3 PL = M Y = M p 2

• Maximum load which the beam can support is Pmax = © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

Mp L 6 - 20

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Plastic Deformations • Preceding discussion was based on normal stresses only • Consider horizontal shear force on an element within the plastic zone, ∆H = −(σ C − σ D )dA = −(σ Y − σ Y )dA = 0

Therefore, the shear stress is zero in the plastic zone. • Shear load is carried by the elastic core, 3 P ⎛⎜ τ xy = 1− 2 A′ ⎜⎝

τ max =

y 2 ⎞⎟ where A′ = 2byY 2⎟ yY ⎠

3P 2 A′

• As A’ decreases, τmax increases and may exceed τY © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

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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.


( )

)

VQ (50 kips ) 15.98 in 3 τ= = It 394 in 4 (0.770 in )

(

τ = 2.63 ksi

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Unsymmetric Loading of Thin-Walled Members • Beam loaded in a vertical plane of symmetry deforms in the symmetry plane without twisting. σx = −

My I

τ ave =

VQ It

• Beam without a vertical plane of symmetry bends and twists under loading. σx = −


My I

τ ave ≠

VQ It

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Unsymmetric Loading of Thin-Walled Members • If the shear load is applied such that the beam does not twist, then the shear stress distribution satisfies D VQ τ ave = V = ∫ q ds It B

B

E

A

D

F = ∫ q ds = − ∫ q ds = − F ′

• F and F’ indicate a couple Fh and the need for the application of a torque as well as the shear load. F h = Ve

• When the force P is applied at a distance e to the left of the web centerline, the member bends in a vertical plane without twisting.


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Example 6.05 • Determine the location for the shear center of the channel section with b = 4 in., h = 6 in., and t = 0.15 in. e=

Fh I

• where b

b VQ

Vb h F = ∫ q ds = ∫ ds = ∫ st ds I0 2 0 0 I Vthb 2 = 4I

2 ⎡1 3 1 3 ⎛ h⎞ ⎤ I = I web + 2 I flange = th + 2 ⎢ bt + bt ⎜ ⎟ ⎥ 12 ⎝ 2 ⎠ ⎥⎦ ⎢⎣12 1 th 2 (6b + h ) ≅ 12

• Combining, e=

b h 2+ 3b

=

4 in. 6 in. 2+ 3(4 in.)


e = 1.6 in.

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Example 6.06 • Determine the shear stress distribution for V = 2.5 kips. τ=

q VQ = t It

• Shearing stresses in the flanges, VQ V h Vh = (st ) = s It It 2 2I Vhb 6Vb = τB = 2 1 th 2 (6b + h ) th(6b + h )

τ=

(12 )

=

6(2.5 kips )(4 in ) = 2.22 ksi (0.15 in )(6 in )(6 × 4 in + 6 in )

• Shearing stress in the web,

( )

1 VQ V 8 ht (4b + h ) 3V (4b + h ) = = τ max = 2 1 It th (6b + h )t 2th(6b + h ) 12

= © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

3(2.5 kips )(4 × 4 in + 6 in ) = 3.06 ksi 2(0.15 in )(6 in )(6 × 6 in + 6 in ) 6 - 26