EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 8.3, Problem 47P

Three forces are applied to the bar shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c.

Fig. P8.47

Chapter 8.3, Problem 47P, Three forces are applied to the bar shown. Determine the normal and shearing stresses at (a) point

(a)

Expert Solution
Check Mark
To determine

The normal and shearing stress at point a.

Answer to Problem 47P

The normal stress at point a is 18.39MPa_.

The shear stress at point a is 0.391MPa_.

Explanation of Solution

Given information:

The centric force p is 10kN.

Calculation:

At point A:

Find the area of cross section (A) using the equation:

A=bh (1)

Here, b is the width of the bar and h is the height of the bar.

Substitute 60mm for b and 32mm for h in Equation (1).

A=60×32=1.920mm2(1m103mm)2=1920×106m2

Find the moment of inertia (Iz) of section using the relation:

Iz=bh312 (2)

Substitute 60mm for b and 32mm for h in Equation (2).

Iz=60×32312=1,966,08012=163.84×103mm4(1012m41mm4)=163.84×109m4

Find the moment of inertia (Iy) of section using the relation:

Iy=bh312 (3)

Substitute 32mm for b and 60mm for h in Equation (3).

Iy=32(60)312=6,912,00012=576×103mm4(1012m41mm4)=576×109m4

Sketch the side view of bar as shown in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 8.3, Problem 47P , additional homework tip  1

At the section containing point a, b, and c.

Refer to Figure 1.

P=10kNVy=750NVz=500N

Find the moment about z axis as follows:

MZ=(180×103)(750)=135Nm

Sketch the side view of bar as shown in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 8.3, Problem 47P , additional homework tip  2

Find the moment about y axis as follows:

My=(220×103)(500)=110Nm

T=0

Find the normal stress (σ) at point a using the formula as follows:

σ=PA+MzyIzMyzIy (4)

Here, P is the centric force, A is the area of rectangular cross section, Iy moment of inertia about y axis, Iz is moment of inertia about z axis, y is the distance from point a, My and Mz is bending couples, and z is distance from z axis.

Substitute 10kN for P, 1920×106m2 for A, 110Nm for My, 135Nm for Mz, 576×109m4 for Iy, 163.84×109m4 for Iz, 16mm for y, and 0 for z in Equation (4).

σ=10kN(103N1kN)1920×106+135×16mm(1m103mm)163.84×109+110×0576×109=10×1031920×106+135×0.016163.84×109=5,208,333.333+13,183,593.75+0=18,391,927.08Pa(1MPa106Pa)

σ=18.39MPa

Thus, the normal stress at point a is 18.39MPa_.

Sketch the cross section at point a as shown in figure 3.

EBK MECHANICS OF MATERIALS, Chapter 8.3, Problem 47P , additional homework tip  3

Determine the first moment area (Q) as follows:

Q=Az¯ (5)

Here, A is the cross section area and z¯ is the distance from neutral axis.

Refer to Figure 2.

Substitute (32×30)mm2 for A and 15mm for y¯ in Equation (5).

Q=(32×30)×15=14,400mm3

Find the shear stress (τ) at point a using the formula as follows:

τ=VzQIyt (6)

Here, Vz is shear about z axis, Q is the first moment area, and t is the thickness of the bar.

Substitute 500N for Vz, 14,400mm3 for Q, 576×109m4 for Iy, and 32mm for t in Equation (6).

τ=500×14.4×103mm3(1m103mm)3576×109×32mm(1m103mm)=7.2×1031.8432×109=390,625Pa(1MPa106Pa)

τ=0.391MPa

Thus, the shear stress at point a is 0.391MPa_.

(b)

Expert Solution
Check Mark
To determine

The normal and shearing stresses at point b.

Answer to Problem 47P

The normal stress at point b is 21.3MPa_.

The shear stress at point b is 0.293MPa_.

Explanation of Solution

Calculation:

At point b:

Find the normal stress (σ) at point b using the formula as follows:

σ=PA+MzyIzMyzIy (7)

Substitute 10kN for P, 1920×106m2 for A, 110Nm for My, 135Nm for Mz, 576×109m4 for Iy, 163.84×109m4 for Iz, 16mm for y, and 15mm for z in Equation (7).

σ=10kN(103N1kN)1920×106+135×16mm(1m103mm)163.84×109110×(15mm(1m103mm))576×109=10×1031920×106+135×0.016163.84×109+2,864,583.33=5,208,333.333+13,183,593.75+2,864,583.33=21,256,510.42Pa(1MPa106Pa)

σ=21.3MPa

Thus, the normal stress at point b is 21.3MPa_.

Sketch the cross section at point b as shown in figure 4.

EBK MECHANICS OF MATERIALS, Chapter 8.3, Problem 47P , additional homework tip  4

Determine the first moment area (Q) as follows:

Q=Az¯ (8)

Here, A is the cross section area and z¯ is the distance from neutral axis.

Refer to Figure 2.

Substitute (32×15)mm2 for A and 22.5mm for y¯ in Equation (8).

Q=(32×15)22.5=10,800mm3

Find the shear stress (τ) at point b using the formula as follows:

τ=VzQIyt (9)

Here, Vz is shear about z axis, Q is the first moment area, and t is the thickness of the bar.

Substitute 500N for Vz,  10,800mm3 for Q, 576×109m4 for Iy, and 32mm for t in Equation (9).

τ=500×10.8×103mm3(1m103mm)3576×109×32mm(1m103mm)=5.4×1031.8432×109=292,968.75Pa(1MPa106Pa)

τ=0.293MPa

Thus, the shear stress at point b is 0.293MPa_.

(c)

Expert Solution
Check Mark
To determine

The normal and shearing stresses at point c.

Answer to Problem 47P

The normal stress at point c is 24.1MPa_.

The shear stress at point c is 0_.

Explanation of Solution

Calculation:

Find the normal stress (σ) at point b using the formula as follows:

σ=PA+MzyIzMyzIy (10)

Substitute 10kN for P, 1920×106m2 for A, 110Nm for My, 135Nm for Mz, 576×109m4 for Iy, 163.84×109m4 for Iz, 16mm for y, and 30mm for z in Equation (10).

σ=10kN(103N1kN)1920×106+135×16mm(1m103mm)163.84×109110×(30mm(1m103mm))576×109=10×1031920×106+135×0.016163.84×109+5,729,166.67=5,208,333.333+13,183,593.75+5,729,166.67=24,121,093.75Pa(1MPa106Pa)

σ=24.1MPa

Thus, the normal stress at point c is 24.1MPa_.

Find the shear stress (τ) at point c using the formula as follows:

τ=VzQIyt (11)

The point c is edge on the cross section. Since Q is zero.

Substitute 500N for Vz, 0 for Q, 576×109m4 for Iy, and 32mm for t in Equation (11).

τ=500×0576×109×32=0

Thus, the shear stress at point c is 0_.

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Chapter 8 Solutions

EBK MECHANICS OF MATERIALS

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