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 3.3, Problem 55P

Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. For the loading shown, determine (a) the reaction at each support, (a) the maximum shearing stress in shaft AB, (c) the maximum shearing stress in shaft BC.

Chapter 3.3, Problem 55P, Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A

Fig. p3.55

(a)

Expert Solution
Check Mark
To determine

The reaction at the supports.

Answer to Problem 55P

The reaction at the supports are 1,105Nm_ and 295Nm_.

Explanation of Solution

Given information:

The modulus of rigidity of solid shafts is 77.2GPa.

Calculation:

The radius of the shaft AB is c=25mm.

The polar moment of inertia of shaft AB of radius c is,

JAB=π2c4=π2(25mm)4=6.1359×105mm4=6.1359×107m4

The torque carried by the shaft AB (TAB) is expressed as shown below:

TAB=GJABLABϕB (1)

Here, G is the modulus of rigidity, ϕB is angle of twist at B, and LAB is the length of the shaft AB.

Substitute 77.2GPa for G, 6.1359×107m4 for JAB, and 200 mm for LAB in Equation (1).

TAB=(77.2GPa)(6.1359×107m4)200mmϕB=(77.2GPa×109Pa1GPa)(6.1359×107m4)200mm×103m1mmϕB=236.8457×103ϕB (2)

The radius of the shaft BC is c=19mm.

The polar moment of inertia of shaft BC of radius c is,

JBC=π2c4=π2(19mm)4=2.0471×105mm4=2.0471×107m4

The torque carried by the shaft BC (TBC) is expressed as shown below:

TBC=GJBCLBCϕB (3)

Here, LBC is the length of the shaft BC.

Substitute 77.2GPa for G, 2.0471×107m4 for JBC, and 250 mm for LBC in Equation (3).

TBC=(77.2GPa)(2.0471×107m4)250mmϕB=(77.2GPa×109Pa1GPa)(2.0471×107m4)250mm×103m1mmϕB=63.2144×103ϕB (4)

The value of total torque in the shaft is 1.4kNm.

The total torque (T) is the sum of the torques TAB and TBC. It is expressed as follows:

T=TAB+TBC1.4kNm=236.8457×103ϕB+63.2144×103ϕB300.0601×103ϕB=1.4kNm×103N1kNϕB=4.6657×103rad

Substitute 4.6657×103rad for ϕB in Equation (2).

TAB=236.8457×103(4.6657×103rad)=1,105.051Nm1,105Nm

Substitute 4.6657×103rad for ϕB in Equation (4).

TBC=63.2144×103(4.6657×103rad)=294.94Nm295Nm

Therefore, the reaction at the supports are 1,105Nm_ and 295Nm_.

(b)

Expert Solution
Check Mark
To determine

The maximum shearing stress in the shaft AB.

Answer to Problem 55P

The maximum shearing stress in the shaft AB is 45MPa_.

Explanation of Solution

Given information:

The modulus of rigidity of solid shafts is 77.2GPa.

Calculation:

Refer (a).

The value of torque in the shaft AB is 1,105.051Nm.

The polar moment of inertia of shaft AB of radius c is JAB=6.1359×107m4.

The maximum shearing stress in the shaft AB (τAB) is expressed as shown below:

τAB=TABcJAB (5)

Substitute 1,105.051Nm for TAB, 25 mm for c, and 6.1359×107m4 for JAB in Equation (5).

τAB=(1,105.051Nm)(25mm)6.1359×107m4=(1,105.051Nm)(25mm×103m1mm)6.1359×107m4=45.024×106Pa45MPa

Therefore, the maximum shearing stress in the shaft AB is 45MPa_.

(c)

Expert Solution
Check Mark
To determine

The maximum shearing stress in the shaft BC.

Answer to Problem 55P

The maximum shearing stress in the shaft BC is 27.4MPa_.

Explanation of Solution

Given information:

The modulus of rigidity of solid shafts is 77.2GPa.

Calculation:

Refer (a).

The value of torque in the shaft BC is 294.94Nm.

The polar moment of inertia of shaft BC of radius c is JBC=2.0471×107m4.

The maximum shearing stress in the shaft BC (τBC) is expressed as shown below:

τBC=TBCcJBC (6)

Substitute 294.94Nm for TBC, 19 mm for c, and 2.0471×107m4 for JBC in Equation (6).

τBC=(294.94Nm)(19mm)2.0471×107m4=(294.94Nm)(19mm×103m1mm)2.0471×107m4=27.375×106Pa27.4MPa

Therefore, the maximum shearing stress in the shaft BC is 27.4MPa_.

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Chapter 3 Solutions

Mechanics of Materials, 7th Edition

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