STANDALONE CODE MECHANICS OF MATERIALS-M
STANDALONE CODE MECHANICS OF MATERIALS-M
11th Edition
ISBN: 9780137605200
Author: HIBBELER
Publisher: PEARSON
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Chapter 8, Problem 1RP

If it supports a cable loading of 800 lb, determine the maximum normal stress at section a–a and sketch the stress distribution acting over the cross section. Use the curved-beam formula to calculate the bending stress.

Chapter 8, Problem 1RP, If it supports a cable loading of 800 lb, determine the maximum normal stress at section aa and

Expert Solution & Answer
Check Mark
To determine

The maximum tensile stress (σt)max at section a-a.

The maximum compressive stress (σc)max .

To sketch:

The stress distribution over the cross section.

Answer to Problem 1RP

The maximum tensile stress (σt)max is 15.8 ksi.

The maximum compressive stress (σc)max is 10.5ksi_ .

Explanation of Solution

Given information:

The force in the cable is 800 lb.

Diameter of the circular is 1.25 in.

Calculation:

Expression to find the location of neutral (R) surface from the center of curvature of the hook:

R=AAdAr (1)

Here, R is the location of neutral axis, A is the cross sectional area of the member, r is the arbitrary position, and dA is the area element on the cross section.

Determine the radius (r) of the circular cross section as shown below:

r=d2 (2)

Here, d is the diameter of the circular cross section.

Substitute 1.25 in. for d in Equation (2).

r=1.252=0.625in.

Determine the area (A) of circular cross section as shown below:

A=πr2 (3)

Here, r is the radius of the circular cross section.

Substitute 0.625 in. for r in Equation (3).

A=π(0.625)2=1.227184in2

Determine the value of dAr using the relation as shown below:

dAr=2π(r¯r¯2c2) (4)

Here, c is the radius of cross section and r¯ is the centroid of the section.

Find the distance measured from the center of curvature to the centroid of the cross section r¯ value as shown in below:

r¯=3.75+2.52=3.125in.

Substitute 0.625 in. for c and 3.125 in. for r¯ in Equation (4).

dAr=2π(3.1253.12520.6252)=2π(0.0631)=0.39670

Substitute 1.227184in2 for A and 0.39670 for dAr in Equation (1).

R=1.2271840.39670=3.0934in.

Sketch the cross section of eye hook as shown in Figure 1.

STANDALONE CODE MECHANICS OF MATERIALS-M, Chapter 8, Problem 1RP , additional homework tip  1

Let the moment acting at the section be M.

Express to the value of M as shown below:

M=F×R (5)

Here, F is the load and R is the radius.

M=800(3.09343)=2.474×103

Determine the bending stress (σ) using the curve beam formula as shown below.

σ=M(Rr)Ar(r¯R)+PA (6)

Here, M is the applied moment and P is the applied load.

Substitute 2.474×103 for M, 3.0934in. for R, 0.625 in. for r, 3.125in. for r¯ , 2.5 in. for r, 1.227184in2 for A, and 800 lb for P in Equation (5).

Determine the maximum tensile stress (σt)max using the curve beam formula:

(σt)max=2.475(103)(3.093432.5)1.227184(2.5)(3.1253.09343)+8001.227184=1468.7390.096855+651.89898=15,816psi×1ksi103psi=15.8ksi

Hence, the maximum tensile stress (σt)max is 15.8 ksi.

Determine the maximum compressive stress (σc)max using the curve beam formula:

Substitute 2.474×103 for M, 3.0934in. for R, 0.625 in. for r, 3.125in. for r¯ , 3.75 in. for r, 1.227184in2 for A, and 800 lb for P in Equation (5).

(σc)max=2.475(103)(3.093433.75)1.227184(3.75)(3.1253.09343)+8001.227184=1625.0100.14528+651.89898=11,185.366+651.89898=10,533psi(1ksi103psi)

(σc)max=10.5ksi

Hence, the maximum compressive stress (σc)max is 10.5ksi_

Sketch the stress distribution (tensile and compressive stress) along the cross section as shown in Figure 2.

STANDALONE CODE MECHANICS OF MATERIALS-M, Chapter 8, Problem 1RP , additional homework tip  2

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Chapter 8 Solutions

STANDALONE CODE MECHANICS OF MATERIALS-M

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