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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Textbook Question
Chapter 7.1, Problem 14P

7.13 through 7.16 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.

Chapter 7.1, Problem 14P, 7.13 through 7.16 For the given state of stress, determine the normal and shearing stresses after

Fig. P7.14

(a)

Expert Solution
Check Mark
To determine

The normal and shearing stresses after the element has been rotated through 25° clockwise.

Answer to Problem 14P

The normal stresses are σx=56.19MPa_ and σy=86.19MPa_.

The shear stress is τxy=38.17MPa_.

Explanation of Solution

Given information:

The stress component along x direction as σx=60MPa.

The stress component along y direction as σy=90MPa.

The shear stress component as τxy=30MPa.

The orientation of the principal plane as θ=25°.

Calculation:

Calculate the normal stress along x direction (σx) as shown below.

σx=σx+σy2+σxσy2cos2θ+τxysin2θ (1)

Substitute 60MPa for σx, 90MPa for σy, 30MPa for τxy, and 25° for θ in Equation (1).

σx=60+902+60902cos(2×25°)+30sin(2×25°)=1575cos(50°)+30sin(50°)=1548.2122.98=56.19MPa

Hence, the normal stress as σx=56.19MPa_.

Calculate the normal stress along y direction (σy) as shown below.

σy=σx+σy2σxσy2cos2θτxysin2θ (2)

Substitute 60MPa for σx, 90MPa for σy, 30MPa for τxy, and 25° for θ in Equation (2).

σy=60+90260902cos(2×25°)30sin(2×25°)=15+75cos(50°)30sin(50°)=15+48.21+22.98=86.19MPa

Hence, the normal stress as σy=86.19MPa_.

Calculate the shear stress (τxy) as shown below.

τxy=σxσy2sin2θ+τxycos2θ (3)

Substitute 60MPa for σx, 90MPa for σy, 30MPa for τxy, and 25° for θ in Equation (3).

τxy=60902sin(2×25°)+30cos(2×25°)=75sin(50°)+30cos(50°)=57.45+19.28=38.17MPa

Therefore, the shear stress as τxy=38.17MPa_.

(b)

Expert Solution
Check Mark
To determine

The normal and shearing stresses after the element has been rotated through 10° counter clockwise.

Answer to Problem 14P

The normal stresses are σx=45.22MPa_ and σy=75.22MPa_.

The shear stress is τxy=53.84MPa_.

Explanation of Solution

Given information:

The stress component along x direction as σx=60MPa.

The stress component along y direction as σy=90MPa.

The shear stress component as τxy=30MPa.

The orientation of the principal plane as θ=10°.

Calculation:

Calculate the normal stress along x direction (σx) as shown below.

Substitute 60MPa for σx, 90MPa for σy, 30MPa for τxy, and 10° for θ in Equation (1).

σx=60+902+60902cos(2×10°)+30sin(2×10°)=1575cos20°+30sin20°=1570.48+10.26=45.22MPa

Hence, the normal stress σx=45.22MPa_.

Calculate the normal stress along y direction (σy) as shown below.

Substitute 60MPa for σx, 90MPa for σy, 30MPa for τxy, and 10° for θ in Equation (2).

σy=60+90260902cos(2×10°)30sin(2×10°)=15+75cos20°30sin20°=15+70.4810.26=75.22MPa

Hence, the normal stress as σy=75.22MPa_.

Calculate the shear stress (τxy) as shown below.

Substitute 60MPa for σx, 90MPa for σy, 30MPa for τxy, and 10° for θ in Equation (3).

τxy=60902sin(2×10°)+30cos(2×10°)=75sin20°+30cos20°=25.65+28.19=53.84MPa

Therefore, the shear stress as τxy=53.84MPa_.

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Chapter 7 Solutions

Mechanics of Materials, 7th Edition

Ch. 7.1 - 7.9 through 7.12 For the given state of stress,...Ch. 7.1 - 7.9 through 7.12 For the given state of stress,...Ch. 7.1 - 7.13 through 7.16 For the given state of stress,...Ch. 7.1 - 7.13 through 7.16 For the given state of stress,...Ch. 7.1 - 7.13 through 7.16 For the given state of stress,...Ch. 7.1 - 7.13 through 7.16 For the given state of stress,...Ch. 7.1 - 7.17 and 7.18 The grain of a wooden member forms...Ch. 7.1 - 7.17 and 7.18 The grain of a wooden member forms...Ch. 7.1 - Two wooden members of 80 120-mm uniform...Ch. 7.1 - Two wooden members of 80 120-mm uniform...Ch. 7.1 - The centric force P is applied to a short post as...Ch. 7.1 - Two members of uniform cross section 50 80 mm are...Ch. 7.1 - The axle of an automobile is acted upon by the...Ch. 7.1 - A 400-lb vertical force is applied at D to a gear...Ch. 7.1 - A mechanic uses a crowfoot wrench to loosen a bolt...Ch. 7.1 - The steel pipe AB has a 102-mm outer diameter and...Ch. 7.1 - For the state of plane stress shown, determine the...Ch. 7.1 - For the state of plane stress shown, determine (a)...Ch. 7.1 - For the state of plane stress shown, determine (a)...Ch. 7.1 - Determine the range of values of x for which the...Ch. 7.2 - Solve Probs. 7.5 and 7.9, using Mohr's circle. 7.5...Ch. 7.2 - Solve Probs. 7.7 and 7.11, using Mohrs circle. 7.5...Ch. 7.2 - Solve Prob. 7.10, using Mohrs circle. 7.9 through...Ch. 7.2 - Solve Prob. 7.12, using Mohr's circle. 7.9 through...Ch. 7.2 - Solve Prob. 7.13, using Mohr's circle. 7.13...Ch. 7.2 - Solve Prob. 7.14, using Mohr's circle. 7.13...Ch. 7.2 - Solve Prob. 7.15, using Mohr's circle. 7.13...Ch. 7.2 - Solve Prob. 7.16, using Mohr's circle. 7.13...Ch. 7.2 - Solve Prob. 7.17, using Mohr's circle. 7.17 and...Ch. 7.2 - Solve Prob. 7.18, using Mohr's circle. 7.17 and...Ch. 7.2 - Solve Prob. 7.19, using Mohr's circle. 7.19 Two...Ch. 7.2 - Solve Prob. 7.20, using Mohr's circle. 7.20 Two...Ch. 7.2 - Solve Prob. 7.21, using Mohrs circle. 7.21 The...Ch. 7.2 - Solve Prob. 7.22, using Mohrs circle. 7.22 Two...Ch. 7.2 - Solve Prob. 7.23, using Mohr's circle. 7.23 The...Ch. 7.2 - Solve Prob. 7.24, using Mohr's circle 7.24 A...Ch. 7.2 - Solve Prob. 7.25, using Mohrs circle. 7.25 A...Ch. 7.2 - Solve Prob. 7.26, using Mohrs circle. 7.26 The...Ch. 7.2 - Solve Prob. 7.27, using Mohr's circle. 7.27 For...Ch. 7.2 - Solve Prob. 7.28, using Mohrs circle. 7.28 For the...Ch. 7.2 - Solve Prob. 7.29, using Mohr's circle. 7.29 For...Ch. 7.2 - Solve Prob. 7.30, using Mohrs circle. 7.30...Ch. 7.2 - Solve Prob. 7.29, using Mohr's circle and assuming...Ch. 7.2 - 7.54 and 7.55 Determine the principal planes and...Ch. 7.2 - 7.54 and 7.55 Determine the principal planes and...Ch. 7.2 - 7.56 and 7.57 Determine the principal planes and...Ch. 7.2 - 7.56 and 7.57 Determine the principal planes and...Ch. 7.2 - For the element shown, determine the range of...Ch. 7.2 - For the element shown, determine the range of...Ch. 7.2 - For the state of stress shown, determine the range...Ch. 7.2 - For the state of stress shown, determine the range...Ch. 7.2 - 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The state of plane stress shown occurs in a...Ch. 7.5 - Prob. 82PCh. 7.5 - The state of plane stress shown occurs in a...Ch. 7.5 - Solve Prob. 7.83, using the...Ch. 7.5 - The 38-mm-diameter shaft AB is made of a grade of...Ch. 7.5 - Solve Prob. 7.85, using the...Ch. 7.5 - The 1.5-in.-diameter shaft AB is made of a grade...Ch. 7.5 - Prob. 88PCh. 7.5 - Prob. 89PCh. 7.5 - Prob. 90PCh. 7.5 - Prob. 91PCh. 7.5 - Prob. 92PCh. 7.5 - Prob. 93PCh. 7.5 - Prob. 94PCh. 7.5 - Prob. 95PCh. 7.5 - Prob. 96PCh. 7.5 - Prob. 97PCh. 7.6 - A spherical pressure vessel has an outer diameter...Ch. 7.6 - A spherical gas container having an inner diameter...Ch. 7.6 - The maximum gage pressure is known to be 1150 psi...Ch. 7.6 - Prob. 101PCh. 7.6 - Prob. 102PCh. 7.6 - A basketball has a 300-mm outer diameter and a...Ch. 7.6 - The unpressurized cylindrical storage tank shown...Ch. 7.6 - Prob. 105PCh. 7.6 - Prob. 106PCh. 7.6 - Prob. 107PCh. 7.6 - Prob. 108PCh. 7.6 - Prob. 109PCh. 7.6 - Prob. 110PCh. 7.6 - Prob. 111PCh. 7.6 - The cylindrical portion of the compressed-air tank...Ch. 7.6 - Prob. 113PCh. 7.6 - Prob. 114PCh. 7.6 - Prob. 115PCh. 7.6 - Square plates, each of 0.5-in. thickness, can be...Ch. 7.6 - The pressure tank shown has a 0.375-in. wall...Ch. 7.6 - Prob. 118PCh. 7.6 - Prob. 119PCh. 7.6 - A pressure vessel of 10-in. inner diameter and...Ch. 7.6 - Prob. 121PCh. 7.6 - A torque of magnitude T = 12 kN-m is applied to...Ch. 7.6 - The tank shown has a 180-mm inner diameter and a...Ch. 7.6 - The compressed-air tank AB has a 250-rnm outside...Ch. 7.6 - In Prob. 7.124, determine the maximum normal...Ch. 7.6 - Prob. 126PCh. 7.6 - Prob. 127PCh. 7.9 - 7.128 through 7.131 For the given state of plane...Ch. 7.9 - 7.128 through 7.131 For the given state of plane...Ch. 7.9 - Prob. 130PCh. 7.9 - 7.128 through 7.131 For the given state of plane...Ch. 7.9 - Prob. 132PCh. 7.9 - Prob. 133PCh. 7.9 - Prob. 134PCh. 7.9 - 7.128 through 7.131 For the given state of plane...Ch. 7.9 - 7.136 through 7.139 The following state of strain...Ch. 7.9 - 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