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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. y 222.7 lb A 8”

1000 in-lb B 8” 350 lb

127.3 lb

451 lb

C

8”

D 6” 1000 in-lb

A

x

B 123 lb z

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

6” D C 1000 in-lb

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

Figure 1. Free body diagrams for the shaft.

1

1000 in-lb 8”

328 lb

x

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. 127.3 lb

123 lb

A C

B

D

A

-222.7 lb

C

B

D -328 lb

Shear Force

1968 in-lb

984 in-lb C

B

A

D A -763.8 in-lb

-1781.6 in-lb

C

B

Bending Moment x-z plane

x-y plane

1000 in-lb Torque A

C

B

D

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

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

M B = 222.7 × 8 = −1781.6 in-lb M C = 127.3 × 6 = −763.8 in-lb and the bending moments in the x-z plane on sections B and C are

M B = 123 × 8 = +984 in-lb M C = 328 × 6 = +1968 in-lb Therefore, the resultant bending moments on these two sections are M B = [( −1781.6) 2 + ( +984) 2 ]1/ 2 = 2035.3 in-lb M C = [(+1968) 2 + ( −763.8) 2 ]1/ 2 = 2111.0 in-lb

The torques at sections B and C are

TB = (300 − 50)(4) = 1000 in-lb (clockwise, positive x-axis) 2

D

TC = (392 − 59)(3) ≈ 1000 in-lb (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 M C > M B , (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:

σx =

32M C 32 × 2111.0 21502 = = psi d3 πd3 πd3

τ trans =

4VC 4 × 127.32 + 1232 300.5 = = psi d2 3A 3 × (π d 2 / 4)

τ torsion =

16TC 16 ×1000 5093 = = 3 psi d πd3 πd3

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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y

MC My f Mz P1 P2

f

TC x

z

Neutral axis 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 M y = +1968 in-lb

and

M z = −763.8 in-lb

and the resultant bending moment is M C = [(+1968) 2 + ( −763.8) 2 ]1/ 2 = 2111.0 in-lb

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

φ = tan −1 (M z / M y ) = tan −1 (−763.8 / +1968) = −21.21o 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 P1 and P2 . The question is: Is it possible to state which of these two points is the more critical? Note that the critical point P1 is in tension and the critical point P2 is in compression. Since materials are usually weaker in tension than compression then the logical choice is that the critical point is P1. Therefore, the following discussion will assume that P1 is the critical point. Note, also, that the location of the critical point P1 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. 4

The bending moment M C produces a tensile normal stress σ x at P1 and a compressive normal stress −σ x at P2. The clockwise torque TC (into the paper) produces a torsional shear stress τ tors at P1 and at P2 . These stress patterns are shown in the following Figure 4. y

τtors

σx is out of the page at P1 (the point is in tension)

τtors P1

τtrans

z

P2

τtors

τtors

σx is into the page at P2 (the point is in compression)

Neutral axis 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 P1 and P2 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 P1 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.

τtors

τtors

σx 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 y = 67 kpsi and the elongation is listed as 12%. Recall that 5

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

σ′ =

Sy

(1)

ny

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

σ ′ = σ x2 − σ xσ y + σ y2 + 3τ xy2

(2a)

For the problem here, the normal stress σ y = 0, therefore Eq, (2a) reduces to 2 σ ′ = σ x2 + 3τ tors

Substituting σ x =

(2b)

21502 5093 psi and τ torsion = 3 psi into Eq. (2b), the von-Mises stress is 3 d d  21502 

2

 5093 

2

σ′ =   + 3 3  3  d   d 

(3)

Note that here we have neglected the transverse shear stress at point P1 . Simplifying Eq. (3) gives

σ′ =

23, 241 d3

(4)

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

23, 241 67, 000 = d3 2.0

(5)

Rearranging this equation, the diameter of the shaft is d = 0.8852 inches Based on this diameter, the normal stress, torsional shear stress, and transverse shear stress are

σx =

21,502 21,502 = = 30,999 psi d3 0.88523

τ torsion = τ trans =

5093 5093 = = 7342 psi 3 d 0.88523

300.5 300.5 = =383.5 psi 2 d 0.88522 6

(6)

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

τ max =

Sy

(7)

2n y

The maximum shear stress is given by Equation (3.14), see page 82, that is σ  2 =  x  + τ tors 2   2

τ max Substituting σ x =

(8)

21502 5093 psi and τ torsion = 3 psi into this equation gives: 3 d d 2

τ max

 21, 502   5093  =   + 3  3  2d   d 

2

(9)

Then simplifying this equation gives

τ max =

11,896 d3

(10)

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

11,896 67, 000 = d3 2(2)

(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

σ ′ = Sy which gives

23, 241 = 67, 000 psi d3 Rearranging this equation, the diameter is 7

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.

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