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ME 452: Machine Design II Solution to Homework Set 1

Problem 5-63, page 260 Problem Statement: The figure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The forces shown acting on the pulley surfaces represent the belt tensions. The shaft is to be made of AISI 1035 CD steel. Using a conservative failure theory with a design factor of 2, determine the minimum shaft diameter to avoid yielding.

Solution:

The solution to this homework problem follows the stress analysis procedure presented in lecture and in the course text book. The six steps are: (1) draw free-body diagrams to determine the external loads acting on the shaft; (2) draw shear force, bending moment, and torque diagrams; (3) determine the locations of the critical section(s) or plane(s); (4) superimpose the stress patterns on the critical sections; (5) determine the critical elements (or points) of the shaft; and (6) apply the appropriate theories of failure to determine the factor(s) of safety at the critical element(s).

Step (1): The free body diagrams of the shaft in the x-y and the x-z planes are shown in figure 1 below.

x-y plane z-axis pointing out of the paper

x-z plane y-axis pointing out of the paper

x A D

123 lb

451 lb

328 lb B C

8” 8” 6”

z

1000 in-lb

1000 in-lb

1000 in-lb

1000 in-lb

C B

Figure 1. Free body diagrams for the shaft.

350 lb

222.7 lb 127.3 lb

A D 8” 8” 6” x

y

2

Step (2): The shear-force and bending-moment diagrams for the shaft in the x-y and x-z planes and the torque diagram are shown in Figure 2 below. Note that the sign conventions are presented on page 75 in the book.

Step (3): The bending moments in the x-y plane on sections B and C are

222.7 8 1781.6 in-lbBM = × = −

127.3 6 763.8 in-lbCM = × = −

and the bending moments in the x-z plane on sections B and C are

123 8 984 in-lbBM = × = +

328 6 1968 in-lbCM = × = +

Therefore, the resultant bending moments on these two sections are

2 2 1/2[( 1781.6) ( 984) ] 2035.3 in-lbBM = − + + =

2 2 1/2[( 1968) ( 763.8) ] 2111.0 in-lbCM = + + − =

The torques at sections B and C are

(300 50)(4) 1000 in-lbBT = − = (clockwise, positive x-axis)

x-y plane x-z plane

-222.7 lb

127.3 lb

A B C D

A B C

D

-1781.6 in-lb

123 lb

-328 lb

A B C D

A B C D

1968 in-lb

-763.8 in-lb

984 in-lb

Shear Force

Bending Moment

A B C D

1000 in-lb Torque

Figure 2. Shear force, bending moment and torque diagrams for the shaft

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(392 59)(3) 1000 in-lbCT = − ≈ (counterclockwise, negative x-axis)

In general, the location of the critical section of the shaft is determined by: (1) the magnitude of the internal loads; (2) the change in the shaft diameter along the length of the shaft; (3) the effects of stress concentration; and (4) the strengths of the shaft material.

In this problem we assume the shaft diameter and the material properties do not change along the length of the shaft, and there are no significant stress concentration effects (more on these issues in later homework problems). Therefore, the critical plane is determined by internal load variations only. For long slender shafts (that is, the length of the shaft is greater than 5 times the diameter), transverse shear stresses are usually negligible, so bending and torsion determine the critical plane. In this problem, we can eliminate most of the planes in the shaft by inspection, except for the plane just to the right of section B and the plane just to the left of section C. These are potential critical planes because they contain the maximum bending moments and the largest torsion. Since the bending moment C BM M> ,

(though only slightly) then the critical plane is just to the left of section C. Step (4): There are three stress patterns to consider on the critical plane: (i) the normal stress due to bending which peaks furthest from the neutral axis, and drops to zero on the neutral axis; (ii) the transverse shear stress (due to the bending) which peaks on the neutral axis, and drops to zero at points furthest from the neutral axis; and (iii) the torsional shear stress (due to the torque) which peaks on the circumference of the shaft, and drops to zero at the center of the shaft.

The maximum stress values for the three stress patterns (in terms of the diameter d of the shaft) are:

3 3 3

32 32 2111.0 21502 psiC

x

M

d d dσ

π π×= = =

2 2

2 2

4 4 127.3 123 300.5 psi

3 3 ( / 4)C

trans

V

A d dτ

π× += = =

×

3 3 3

16 16 1000 5093 psiC

torsion

T

d d dτ

π π×= = =

In this problem it is sufficient to know that the maximum normal stress due to bending (referred to simply as the bending stress) and the maximum shear stress due to torsion (referred to simply as the torsional stress) will occur somewhere on the surface of the shaft in the critical plane at C. The exact location on the surface of the shaft does not matter when the material is homogeneous and isotropic (page 60). However, in other problems when the stresses vary with time (fatigue) and the shaft also rotates, the exact locations of the maximum stresses will become very important because the time history determines the damage and likelihood of failure. So we will use this problem to illustrate how one would identify which element (point) on plane C is the critical element (point).

The normal stresses on sections B and C in the x-direction due to the bending moments will add. This means that the maximum bending stress on these two sections can be obtained from the resultants of the bending moments on these sections.

Consider the critical plane (that is, the plane to the left of section C) as shown in Figure 3 below where the x axis points out of the paper.

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Figure 3. The critical plane to the left of point C.

The neutral axis of the cross-section of the shaft at C is obtained as follows. Recall that the bending moments at section C, about the y-axis and the z-axis, respectively, are

1968 in-lbyM = + and 763.8 in-lbzM = −

and the resultant bending moment is

2 2 1/2[( 1968) ( 763.8) ] 2111.0 in-lbCM = + + − =

Since the resultant moment acts about the neutral axis then the orientation of the neutral axis, with respect to the y-axis, is

1 1 otan (M / M ) tan ( 763.8 / 1968) 21.21z yφ − −= = − + = −

According to the above discussion, the critical points are at the circumference of the shaft on the line perpendicular to the neutral axis. The elements are denoted on the figure as 1P and 2P . The question is:

Is it possible to state which of these two points is the more critical? Note that the critical point 1P is in tension and the critical point 2P is in compression. Since

materials are usually weaker in tension than compression then the logical choice is that the critical point is 1P . Therefore, the following discussion will assume that 1P is the critical point.

Note, also, that the location of the critical point 1P in the positive y-z quadrant agrees with our

intuition, that is, the net loads on the shaft from the pulleys at B and C are in the positive y and z directions, therefore the top-right fibers of the shaft must be in tension.

x

y

z

My

Mz

MC

Neutral axis

f

f

P1

P2 TC

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The bending moment CM produces a tensile normal stress xσ at 1P and a compressive normal

stress xσ− at P2. The clockwise torque CT (into the paper) produces a torsional shear stress torsτ at 1P

and at 2P . These stress patterns are shown in the following Figure 4.

Figure 4. The critical plane to the left of point C.

(The x axis points out of the paper)

The transverse shear stress (due to bending) is also shown slightly shaded on the figure. Note that the maximum transverse shear stresses for this problem do not necessarily occur at points P1 and P2. Since the length/diameter ratio for this shaft is probably much greater than 5, we will initially neglect the effects of the transverse shear stresses at points 1P and 2P assuming they are negligibly small. After we

have estimated the appropriate diameter for the shaft based on this assumption, we will calculate the magnitude of the peak transverse shear stress on the section to confirm the validity of this assumption.

The above discussion leads to the state-of-stress element for the critical point 1P as shown in the

following Figure 5. Since the torsional shear stress is not in the y-direction or in the z-direction then we will omit the direction subscripts.

Figure 5. The biaxial state of stress of element P1. The x-axis points out of the paper. (The face on the top of the element is the free surface).

Step (5): The material specification calls for AISI 1035 CD. The “10” indicates that this is a plain carbon steel without other alloying elements (see page 45), and the “CD” indicates it is cold drawn. The drawing process work hardens the material and increases its strength. From Table A-20, see page 1040, the yield strength for this material is S 67 kpsiy = and the elongation is listed as 12%. Recall that

z

τtors

τtors

τtors

τtors

y

τtrans

Neutral axis

P1

P2 σx is into the page at P2

(the point is in compression)

σx

τtors τtors

σx is out of the page at P1

(the point is in tension)

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materials with elongations greater than 5% are considered ductile (see page 238), so the best failure theory here is the distortion energy (von Mises) theory. This theory is more accurate than the maximum shear stress failure theory (and, therefore, generally preferred), however, it is slightly less conservative. Calculations based on both theories will be included here.

According to Von-Mises theory, see Equation (5-19), page 224, the design equation is

y

y

S

nσ ′ = (1)

The von-Mises stress for a biaxial state of stress, see Equation (5-15), page 223, can be written as

2 2 23x x y y xyσ σ σ σ σ τ′ = − + + (2a)

For the problem here, the normal stress 0,yσ = therefore Eq, (2a) reduces to

2 23x torsσ σ τ′ = + (2b)

Substituting 3

21502 psix d

σ = and 3

5093 psitorsion d

τ = into Eq. (2b), the von-Mises stress is

2 2

3 3

21502 50933

d dσ ′ = +

(3)

Note that here we have neglected the transverse shear stress at point 1P . Simplifying Eq. (3) gives

3

23,241

dσ ′ = (4)

Then substituting the yield strength and Eq. (4) into Eq. (1) gives

3

23,241 67,000

2.0d= (5)

Rearranging this equation, the diameter of the shaft is

d = 0.8852 inches (6)

Based on this diameter, the normal stress, torsional shear stress, and transverse shear stress are

3 3

21,502 21,502 = = 30,999 psi

0.8852x dσ =

3 3

5093 5093 = = 7342 psi

0.8852torsion dτ =

2 2

300.5 300.5 = =383.5 psi

0.8852trans dτ =

7

Note that the transverse shear stress is small, only about 5.2% of the torsional shear stress. Note also that the material fails at 67,000 psi normal stress, and with our factor of safety of 2.0, the safe limit (often called an allowable stress) is 67,000/2 or 33,500 psi. So the normal stress, at about 31,000 psi, is doing most of the damage here.

The slightly less accurate (but more conservative) maximum shear stress theory. The design equation for maximum shear stress theory is given by Equation (5.3), see page 220, that is

max 2y

y

S

nτ = (7)

The maximum shear stress is given by Equation (3.14), see page 82, that is

22

max 2x

tors

στ τ = +

(8)

Substituting 3

21502 psix d

σ = and 3

5093 psitorsion d

τ = into this equation gives:

2 2

max 3 3

21,502 5093

2d dτ = +

(9)

Then simplifying this equation gives

max 3

11,896

dτ = (10)

Substituting Eq. (10) into Eq. (7) gives

3

11,896 67,000

2(2)d= (11)

Rearranging this equation, the diameter of the shaft gives

d = 0.8922 inches

Note that this solution is less than 1% greater than (more conservative) the distortion energy solution.

Because the distortion energy theory is more accurate than the maximum shear stress theory, and about

as easy to calculate, it is the generally preferred theory. Step (6): For a yield factor of safety, ny = 1, the distortion-energy failure-theory design equation reduces to

ySσ ′ =

which gives

3

23,24167,000psi

d=

Rearranging this equation, the diameter is

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d = 0.7026 inches

Note the power of the d3 terms. A 26% change in the diameter (from 0.7026 to 0.8852) doubles the factor of safety from 1 to 2.