Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 10.1, Problem 14P

Determine (a) the critical load for the square strut, (b) the radius of the round strut for which both struts have the same critical load. (c) Express the cross-sectional area of the square strut as a percentage of the cross-sectional area of the round strut. Use E = 200 GPa.

Fig. P10.13 and P10.14

Chapter 10.1, Problem 14P, Determine (a) the critical load for the square strut, (b) the radius of the round strut for which

(a)

Expert Solution
Check Mark
To determine

Find the critical load of the square strut.

Answer to Problem 14P

The critical load of the square strut is 64.3kN_.

Explanation of Solution

The modulus of elasticity of the strut is E=200GPa.

Determine the moment of inertia of the square strut (Isquare) using the relation.

Isquare=a412

Substitute 25 mm for a.

Isquare=(25mm×1m1000mm)412=3.2552×108m4

Determine the critical load (Pcr) for both the end pinned condition using the equation.

Pcr(square)=π2EIL2

Here, the modulus of elasticity is E and the length of the strut is L.

Substitute 200 GPa for E, 3.2552×108m4 for I, and 1 m for L.

Pcr(square)=π2×(200GPa×109N/m21GPa)×3.2552×10812=64.3×103N×1kN1000N=64.3kN

Therefore, the critical load of the square strut is 64.3kN_.

(b)

Expert Solution
Check Mark
To determine

Find the radius of the round strut when the critical load is same for square strut and round strut.

Answer to Problem 14P

The radius of the round strut is 14.27mm_.

Explanation of Solution

The modulus of elasticity of the strut is E=200GPa.

Determine the moment of inertia of the square strut (Isquare) using the relation.

Isquare=a412

Here, the size of the square strut is a.

Substitute 25 mm for a.

Isquare=(25mm×1m1000mm)412=3.2552×108m4

Determine the critical load (Pcr) for both the end pinned condition using the equation.

Pcr(square)=π2EIL2

Substitute 200 GPa for E, 3.2552×108m4 for I, and 1 m for L.

Pcr(square)=π2×(200GPa×109N/m21GPa)×3.2552×10812=64.3×103N×1kN1000N=64.3kN

The critical load of the square strut and the round strut is equal.

Pcr(square)=Pcr(round)=64.3kN

Determine the moment of inertia of the round strut (Iround) using the equation.

Pcr(round)=π2EIroundL2

Substitute 64.3 kN for Pcr(round), 200 GPa for E, and 1 m for L.

64.3kN×1000N1kN=π2×(200GPa×109N/m21GPa)×Iround12Iround=3.25748×108m4×(1000mm1m)4=3.25748×104mm4

Determine the radius of the round strut (c) using the relation.

Iround=πc44

Substitute 3.25748×104mm4 for Iround.

3.25748×104=π×c44c=14.27mm

Therefore, the radius of the round strut is 14.27mm_.

(c)

Expert Solution
Check Mark
To determine

Find the percentage of cross-sectional area of square strut to the cross-sectional area of round strut.

Answer to Problem 14P

The percentage of cross-sectional area of square strut to the cross-sectional area of round strut is 97.7%_.

Explanation of Solution

The modulus of elasticity of the strut is E=200GPa.

Find the cross sectional area of the square strut (Asquare)  using the equation.

Asquare=a2

Here, the size of the square strut is a.

Substitute 25 mm for a.

Asquare=252=625mm2

Find the cross sectional area of the round strut (Around) using the equation.

Around=πc2

Here, the radius of the round strut is c.

Substitute 14.27 mm for c.

Around=π×14.272=639.73mm2

Find the percentage of area of square strut to the area of round strut as follows;

Percentage=AsquareAround×100%

Substitute 625mm2 for Asquare and 639.73mm2 for Around.

Percentage=625639.73×100%=97.7%

Therefore, the percentage of cross-sectional area of square strut to the cross-sectional area of round strut is 97.7%_.

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Chapter 10 Solutions

Mechanics of Materials, 7th Edition

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