Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
10th Edition
ISBN: 9780134319650
Author: Russell C. Hibbeler
Publisher: PEARSON
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Chapter 9, Problem 9.1RP
To determine

The principal stresses at point A.

Expert Solution & Answer
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Answer to Problem 9.1RP

The principal stresses at point A (σ1) and (σ2) are 119psi_ and 119psi_.

Explanation of Solution

Calculate the normal stress (σA) acting at point A using the relation:

σA=MyzIy (1)

Here, My is the moment in y-direction, z is the distance in z-direction from centroid to point A, and Iy is the moment of inertia.

Sketch the internal forces and moment in free body diagram at point A as shown in Figure 1.

Mechanics of Materials (10th Edition), Chapter 9, Problem 9.1RP , additional homework tip  1

Apply Equilibrium equations to find the value of moment at point A.

Sum of moments in y-direction is equal to 0.

ΣMy=0My(20×10)=0My=200lbin

Sum of moments in x-direction is equal to 0.

ΣMx=0Tx+(20×12)=0Tx=240lbin

Sum of forces in z-direction is equal to 0.

ΣVz=0Vz20=0Vz=20lb

Find the moment of inertia of the section (I):

Outer radius of the pipe is 1.5 in. and the inner radius of the pipe is 1.375 in.

I=π4[(1.5in)4(1.375in)4]=1.1687in4

Find the polar moment of inertia of the section (J):

J=π2[(1.5in)4(1.375in)4]=2.3374in4

Sketch the cross section at point A as shown in Figure 2.

Mechanics of Materials (10th Edition), Chapter 9, Problem 9.1RP , additional homework tip  2

Find the first moment of area at point A (QA)z using the relation:

(QA)z=Σy¯'A' (2)

Here, y¯' is the centroid distance to point A and A' is the area from centroid to point A.

Refer to Figure 2.

(QA)z=4×1.5in.3π[12×π×(1.5in.)2]4×1.375in.3π[12×π×(1.375in.)2]=0.51693in3.

Substitute 200lbin for My, 0 for z, and 1.1687in4 for Iy in Equation (1).

σA=(200lbin)(0)1.1687in4=0

Find the shear stress (τA) at point A using the relation:

τA=(VQIt)zTρJ (3)

Substitute 20lb for Vz, 0.51693in3 for (QA)z, 1.1687in4 for I, 2×0.125in. for t, 240lbin for T, 1.5in. for ρ, and 2.3374in4 for J in Equation (3).

τA=(20lb)(0.51693in3)(1.1687in4)2(0.125in.)(240lbin)(1.5in.)(2.3374in4)=118.6psi

Sketch the state of stress at point A as shown in Figure 3.

Mechanics of Materials (10th Edition), Chapter 9, Problem 9.1RP , additional homework tip  3

Refer to Figure 3.

The value of normal stresses are σx=0 and σz=0.

The value of shear stress is τxz=118.6psi.

Find the principal stresses (σ1) and (σ2) at point A:

σ1,2=σx+σz2±(σxσz2)2+τxz2 (4)

Substitute 0 for σx, 0 for σz, and 118.6psi for τxz in Equation (4).

σ1,2=0±0+(118.6)2=0±118.6σ1=0+118.6=118.6psi

σ1119psiσ2=0118.6=118.6psi119psi

Therefore, the principal stresses (σ1) and (σ2) at point A are 119psi_ and 119psi_.

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Chapter 9 Solutions

Mechanics of Materials (10th Edition)

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