EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 6.5, Problem 53P
To determine

To provide: The expression of maximum shearing stress as τmax=k(VA).

The constant k for each orientation.

Expert Solution & Answer
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Answer to Problem 53P

The constant k for orientation (a) is 2.08_.

The constant k for orientation (b) is 2.10_.

Explanation of Solution

Given information:

The beam is a hollow square of side a and thickness t.

The beam is subjected to a vertical shear of V.

Calculation:

Orientation (a)

Sketch the cross section of an extruded beam as shown in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 6.5, Problem 53P , additional homework tip  1

Refer to Figure 1.

The height of the section above neutral axis is h=32a.

Area of the cross member is A1=A2=at.

Calculate the area of the member A as shown below.

A=6at

Calculate the moment of inertia of the beam I as shown below.

I=2ath2+4×13ath2=10ath23

Substitute 32a for h.

I=10at3×(32a)2=10at3×34a2=52a3t

Calculate the first moment of area as shown below.

Q=Ay¯

Here, A is the area of the section and y¯ is the centroid of the section.

Calculate the first moment of area along the neutral axis as shown below.

Qmax=at×h+(2×at×h2)=ath+ath=2ath

Substitute 32a for h.

Qmax=2at×32a=3a2t

Calculate the shear stress τ as shown below.

τ=VQIt

Here V is the vertical shear.

Substitute 3a2t for Q, 2t for t, and 52a3t for I in Equation (2).

τ=V×3a2t52a3t×2t=3V5at

Substitute 16A for at.

τ=3V516A=635VA

Hence, the expression for maximum shearing stress is 635VA.

Calculate the constant as shown below.

τ=2.08VA

Therefore, the constant k is 2.08_.

(b)

Sketch the cross section of an extruded beam as shown in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 6.5, Problem 53P , additional homework tip  2

Refer to Figure 2.

The height of the section above neutral axis is h=a2.

Area of the cross member is A1=A2=at.

Calculate the area of the member A as shown below.

A=6at

Calculate the moment of inertia I of the beam as shown below.

I=4×(112ath2+at(a2+h2)2)+4×13t×(a2)3=4×[112ath2+at(a2+h2)2+124a3t]

Substitute a2 for h.

I=4(112at(a2)2+at(a2+a22)2+124a3t)=4×(148a3t+at(34a)2+124a3t)=4×(348a3t+916a3t)=4×(3+2748)a3t

=3012a3t=52a3t

Calculate the first moment of area as shown below.

Q=Ay¯

Here, A is the area of the section and y¯ is the centroid of the section.

Calculate the first moment of area along the neutral axis as shown below.

Qmax=2×at×(a2+h2)+2×12at×a4=2at(a2+h2)+a2t4

Substitute a2 for h.

Qmax=2at(a2+a22)+a2t4=2at×3a4+a2t4=3a2t2+a2t4=74a2t

Calculate the shear stress as shown below.

τ=VQIt

Here V is the vertical shear.

Substitute 74a2t for Q, 2t for t, and 52a3t for I in Equation (2).

τ=V×74a2t52a3t×2t=7V4×5at=720Vat

Substitute 16A for at.

τ=720V16A=2110VA

Hence, the expression for maximum shearing stress is 2110VA_.

Calculate the constant as shown below.

τ=2.1VA

Therefore, the constant k is 2.1_.

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Chapter 6 Solutions

EBK MECHANICS OF MATERIALS

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