EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 4.9, Problem 129P

4.127 through 4.134 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

Chapter 4.9, Problem 129P, 4.127 through 4.134 The couple M is applied to a beam of the cross section shown in a plane forming

Flg. P4.129

(a)

Expert Solution
Check Mark
To determine

Find the stress at point A.

Answer to Problem 129P

The stress at point A is σA=29.3MPa_.

Explanation of Solution

Given information:

The couple acts in a vertical plane M=25kNm.

The angle is θ=15°.

Calculation:

Sketch the beam cross section as shown in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 4.9, Problem 129P

Refer to Figure 1.

Calculate the moment along y direction (My) as shown below.

My=Msinθ

Substitute 25kNm for M and 15° for θ.

My=25sin15°=6.4705kNm

Calculate the moment along z direction (Mz) as shown below.

Mz=Mcosθ

Substitute 25kNm for M and 15° for θ.

Mz=25cos15°=24.148kNm

Calculate the moment of inertia along y axis (Iy) as shown below.

Iy=112×80×903+112×80×303=5.04×106mm4×(1m1,000mm)4=5.04×106m4

Calculate the moment of inertia along z axis (Iz) as shown below.

Iz=13×90×603+13×60×203+13×30×1003=16.64×106mm4×(1m1,000mm)4=16.64×106m4

Calculate the stress (σ) as shown below.

σ=MyzIyMzyIz (1)

The location of point A along z axis z=45mm.

The location of point A along y axis y=60mm.

Calculate the stress at point A (σA) as shown below.

Substitute 6.4705kNm for My, 24.148kNm for Mz, 5.04×106m4 for Iy, 16.64×106m4 for Iz, 45mm for z, and 60mm for y in Equation (1).

σA=(6.4705kNm×(45mm×1m1,000mm)5.04×106m424.148kNm×(60mm×1m1,000mm)16.64×106m4)=57.772×10387.072×103=29.3×103kN/mm2×1MPa1,000kN/mm2=29.3MPa

Therefore, the stress at point A is σA=29.3MPa_.

(b)

Expert Solution
Check Mark
To determine

The stress at point B.

Answer to Problem 129P

The stress at point B is σB=144.8MPa_.

Explanation of Solution

Given information:

The couple acts in a vertical plane M=25kNm.

The angle is θ=15°.

Calculation:

Refer to part (a).

The moment along y axis My=6.4705kNm.

The moment along z axis Mz=24.148kNm.

The moment of inertia along y axis Iy=5.04×106m4.

The moment of inertia along z axis Iz=16.64×106m4.

Refer to Figure 1 in part (a).

The location of point B along z axis z=45mm.

The location of point B along y axis y=60mm.

Calculate the stress at point B (σB) as shown below.

Substitute 6.4705kNm for My, 24.148kNm for Mz, 5.04×106m4 for Iy, 16.64×106m4 for Iz, 45mm for z, and 60mm for y in Equation (1).

σB=(6.4705kNm×(45mm×1m1,000mm)5.04×106m424.148kNm×(60mm×1m1,000mm)16.64×106m4)=57.772×10387.072×103=144.844×103kN/mm2×1MPa1,000kN/mm2=144.8MPa

Therefore, the stress at point B is σB=144.8MPa_.

(c)

Expert Solution
Check Mark
To determine

The stress at point D.

Answer to Problem 129P

The stress at point D is σD=125.9MPa_.

Explanation of Solution

Given information:

The couple acts in a vertical plane M=25kNm.

The angle is θ=15°.

Calculation:

Refer to part (a).

The moment along y axis is My=6.4705kNm.

The moment along z axis is Mz=24.148kNm.

The moment of inertia along y axis is Iy=5.04×106m4.

The moment of inertia along z axis is Iz=16.64×106m4.

Refer to Figure 1 in part (a).

The location of point D along z axis z=15mm.

The location of point D along y axis y=100mm.

Calculate the stress at point D (σD) as shown below.

Substitute 6.4705kNm for My, 24.148kNm for Mz, 5.04×106m4 for Iy, 16.64×106m4 for Iz, 15mm for z, and 100mm for y in Equation (1).

σD=(6.4705kNm×(15mm×1m1,000mm)5.04×106m424.148kNm×(100mm×1m1,000mm)16.64×106m4)=19.257×103+145.12×103=125.863×103kN/mm2×1MPa1,000kN/mm2=125.9MPa

Therefore, the stress at point D is σD=125.9MPa_.

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Chapter 4 Solutions

EBK MECHANICS OF MATERIALS

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