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 8.3, Problem 54P

Three steel plates, each 13 mm thick, are welded together to form a cantilever beam. For the loading shown, determine the normal and shearing stresses at points d and e.

Fig. P8.53 and P8.54

Chapter 8.3, Problem 54P, Three steel plates, each 13 mm thick, are welded together to form a cantilever beam. For the loading

Expert Solution & Answer
Check Mark
To determine

The normal and shearing stress at point a and b.

Answer to Problem 54P

The normal stress at point d is 42.2MPa_.

The shear stress at point d is 2.83MPa_.

The normal stress at point e is 12.74MPa_.

The shear stress at point e is 0_.

Explanation of Solution

Given information:

The thickness (t) of the steel plate is 13mm.

Calculation:

Refer to Figure P8.51 in the textbook.

Forces at H are as follows:

The force in x direction, Fx=9kN.

The force in y direction, Fy=13kN

The force in z direction, Fz=0

Moments at H are as follows:

Find the moment about x axis as follows:

Mx=(0.400×13kN(103N1kN))=0.400×13,000=5,200Nm

Find the moment about y axis as follows:

My=(0.400×9kN(103N1kN))=3,600Nm

Mz=0

At point A:

Sketch the I section as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 8.3, Problem 54P , additional homework tip  1

Refer to Figure 1.

Find the centroid section x¯ at point a as follows:

x¯=752=37.5mm

Find the centroid section y¯ at point a as follows:

y¯=132=6.5mm

Find the area of the section using the relation:

A=2(b×d)flange+(b×d)web (1)

Here, b is the width and d is the depth of the flange and web respectively.

Refer to Figure 1.

Substitute 150mm for bf, 13mm for df, 13mm for bw, and 49mm for dw in Equation (1).

A=2×150(13)+13×49=3,900+637=4,357mm2(1m103mm)2=4537×106m2

Find the moment of inertia (Ix) about x axis as follows:

Ix=2[bfdf312+bfdf(x¯y¯)2]+bwdw312 (2)

Substitute 150mm for bf, 13mm for df, 13mm for bw, 37.5mm for x¯, 6.5mm for y¯ and 49mm for dw in Equation (2).

Ix=2[150×13312+150×13(37.56.5)2]+13×49312=2[27,462.5+1,873,950]+127,453.08=3,930,278.08mm2(1m103mm)2=3.9303×106m4

Find the moment of inertia (Iy) about y axis as follows:

Iy=2[dfbf312]+bw3dw12 (3)

Substitute 150mm for bf, 13mm for df, 13mm for bw, and 49mm for dw in Equation (3).

Iy=2[13×150312]+49×13312=7,312,500+8,971.08=7,321,471.083mm4(1m103mm)4=7.3215×106m4

At point d:

Sketch the I section for point b as shown in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 8.3, Problem 54P , additional homework tip  2

Refer to Figure 2.

Find the area of section (A) as follows:

A=bd (4)

Substitute 60mm for b and 13mm for d in Equation (4).

A=60×13=780mm2

Refer to Figure 2.

The centroid of the x¯ is 45mm.

The centroid of the y¯ is 31mm.

Find the first moment area (Qx) at point b as follows:

Qx=Ay¯ (5)

Substitute 780mm2 for A and 31mm for y¯ in Equation (5).

Qx=780×31=24,180mm3(1m103mm)3=24.18×106m3

Find the first moment area (Qy) at point b as follows:

Qy=Ax¯ (6)

Substitute 780mm2 for A and 45mm for y¯ in Equation (6).

Qy=780×(45)=35,100mm3(1m103mm)3=35.1×106m3

Find the first moment area (Qx) and (Qy) at point e as follows:

The point e is located at edge. Since (Qx) is zero.

The point e is located at edge. Since (Qy) is zero.

Determine the normal stress at point d using the relation:

σ=MxyIxMyxIy (7)

Here, Mx and My is couple moments, Iy is moment of inertia about y axis, Iz is moment of inertia about z axis, and x is the distance.

Substitute 5,200Nm for Mx, 37.5mm for y, 7.3215×106m4 for Iy, 3.9303×106m4 for Ix, 3,600Nm for My, and 15mm for x in Equation (7).

σ=5,200×37.5mm(1m103mm)3.9303×1063,600×15mm(1m103mm)7.3215×106=5,200×0.03753.9303×1063,600×(0.015)7.3215×106=49,614,533.247,375,537.8=42,238,995Pa(1MPa106Pa)

σ=42.2MPa

Thus, the normal stress at point d is 42.2MPa_.

Determine the shear stress at point a due to Vx using the relation:

τ=|Vx||Qy|Iyt (8)

Substitute 9kN for Vx, 35.1×106m3 for Qy, and 7.3215×106m4 for Iy, and 13mm for t in Equation (8).

τ=9kN(103N1kN)35.1×1067.3215×106(13mm(1m103mm))=9000×35.1×1067.3215×106×0.01=33,318,992.01Pa(1MPa106Pa)=3.32MPa

Sketch the horizontal direction of shear stress as shown in figure 3.

Mechanics of Materials, 7th Edition, Chapter 8.3, Problem 54P , additional homework tip  3

Determine the shear stress at point a due to Vy using the relation:

τ=|Vy||Qx|Ixt (8)

Substitute 13kN for Vx, 24.18×106m3 for Qx, and 3.9303×106m4 for Ix, and 13mm for t in Equation (8).

τ=13kN(103N1kN)24.18×1063.9303×106(13mm(1m103mm))=13000×24.18×1063.9303×106(0.013)=6,152,202.12Pa(1MPa106Pa)=6.15MPa

Sketch the vertical direction of shear stress as shown in figure 4.

Mechanics of Materials, 7th Edition, Chapter 8.3, Problem 54P , additional homework tip  4

Find the shear stress at point d using superposition method.

τ=6.153.32=2.83MPa

Thus, the shear stress at point d is 2.83MPa_.

Determine the normal stress at point e using the relation:

σ=MxyIxMyxIy (8)

Substitute 5,200Nm for Mx, 37.5mm for y, 7.3215×106m4 for Iy, 3.9303×106m4 for Ix, 3,600Nm for My, and 75mm for x in Equation (8).

σ=5,200×37.5mm(1m103mm)3.9303×1063,600×75mm(1m103mm)7.3215×106=5,200×0.03753.9303×1063,600×(0.075)7.3215×106=49,614,533.2436,877,689=12,736,844Pa(1MPa106Pa)

σ=12.74MPa

Thus, the normal stress at point b is 12.74MPa_.

Determine the shear stress at point e using the relation:

The point e is located on edge.

The shear stress at point e is zero.

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Chapter 8 Solutions

Mechanics of Materials, 7th Edition

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