Homework 5 Solutions - HW 5 Spring 2016 PDF

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HW 5 Spring 2016...

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530.215 MECHANICS-BASED DESIGN, SPRING 2016

HW 5 solutions 1 This problem is from Shigley’s 5-38. A 1020 CD steel shaft is to transmit 20 hp while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a minimum factor of safety of 3 based on the maximum-shear- stress theory. Assume Sy = 390 MPa. (Hint: Use 1 hp = .7457 kW). Solution: The power is the shaft is .7457 ∗ 20 = 14.914. Since power is given by (Shigley’s 3-43): P = 2πT

N 60

(1)

we can substitute power P=14.914 kW and the rotational speed N=1750 rpm. Solving for the torque, we get T = 81.38 N m . Looking up the yield strength of AISI 1020 CD Steel in table A-20 of Shigley’s, we see that Sy = 390 MPa. Therefore, the maximum allowable shear stress (considering a factor of safety n=3) is: Sy n 390 = 3 = 130 MPa

τ=

Now we can solve for the diamter using the relationship τ = 16T πd 3 16(81.38) 6 130 × 10 = πd 3 → d = 14.72mm = .58in τ=

(2) (3) (4) Tr J

: (5) (6) (7)

2 This problem is from Shigley’s 5-58. A spherical pressure vessel is formed of 16-gauge (0.0625-in) cold-drawn AISI 1020 sheet steel. If the vessel has a diameter of 15 in, use the distortion-energy theory to estimate the pressure necessary to initiate yielding. What is the estimated bursting pressure? Assume Sy = 57 kpsi, and Sut = 68 kpsi. Solution: Looking again at table A-20 in Shigley’s, the yield strength Sy = 57 kpsi and the tensile strength Sut = 68 kpsi for cold drawn AISI 1020 steel. For spherical pressure vessels: σθ =

Homework 5

Pd 4t

(8)

p. 1

530.215 MECHANICS-BASED DESIGN, SPRING 2016

Substituting our known values of d = 15 in., and t = .0625 in, we find that the hoop stress σ θ = 60P . For the principal stresses we have σ 1 = σ 2 = 60P and σ r = σ 3 = 0. To use the distortion-energy theory, we first compute the von Mises stress as: r ′ 1 [(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 ] (9) σ = 2 r 1 [(60P − 60P )2 + (60P − 0)2 + (0 − (−60P ))2 ] (10) = 2 = 60P (11) Equating the von Mises stresses and the yield condition: ′

σ = Sy 60P = 57 × 10

(12) 3

(13)

Therefore at yield, P=950 psi. For the rupture condition the von Mises would exceed the ultimate tensile strength: ′

σ = Sut

(14)

60P = 68

(15)

→ P = 1.13 kpsi

(16)

3 A hollow cylinder of length L = 200 mm is made of an Al alloy and subjected to torsion at one end, as shown in Figure 2. The outer diameter D0 = 60 mm, and the inner diameter Di = 40 mm. (a) Determine the maximum allowable torque for a safety factor of 2 and a shear yield strength τy = 190 MPa. (b)Write out the stress matrix in cylindrical coordinates for a point at the cylinder’s surface. What are the principal stresses and their direction from the z-axis (longitudinal axis)?

Figure 1: Hollow cylinder subjected to torsion for problem 3 Solution: (a) Since n = Homework 5

τy , τ

thus τy =

T

Do 2

J

n

, and the maximum allowable torque is p. 2

530.215 MECHANICS-BASED DESIGN, SPRING 2016

simply: T = = 4

Plugging in J = π2 [( D2o −

Di 4 ] 2

(b)

  0  σ =  0  0

0 0 τmax

0 τmax 0

τy J Do n 2

(17)

τy J Do

(18)

= 1.021 × 10−6 m4 , we get T=3.233 kNm.      

  0  =  0  0

 0 0  0 95  MPa 95 0 

(19)

The principal stresses are σ 1 = τmax and σ 2 = −τmax . They occur at ±45◦ from the longitudinal z-axis. 4 Consider a solid cylinder with a diameter of 50mm that is fixed to the wall on one end. The other end is under a combined loading of torsion and tension. The applied torque is (500π)N · m and the tensile force is P . The yield strength of this material is Sy = 300M P a and the factor of safety is n = 1.2. Determine the maximum tensile force P , using (a) Maximum Shear Stress (Tresca) criterion; (b) Distortion-Energy (Von Mises) criterion. Solution: The normal stress caused by tension is σ z = P/A =

4P πD2

(20)

And the shear stress at the edge of the cylinder caused by torsion is τzθ =

Tr 16T = J πD3

(21)

Since σ r = τrz = τrθ = 0, this is a plane stress problem. In addition, σθ = 0 (a): Maximum shear stress theory The two principal stresses in plane are r σθ + σz 2 σθ + σz 2 = + ( σA = ) + τzθ 2 2 r σθ + σz 2 σθ + σz 2 = − ( σB = ) + τzθ 2 2

Homework 5

r

σz 2 2 >0 ) + τzθ 2 r σz σz 2...