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nexus9112 [7]
3 years ago
14

A cyclic tensile load ranging from 0 kN to 55 kN force is applied along the length of a 100 mm long bar with a 15 mm x 15 mm squ

are cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.
Required:
a. What is the nominal maximum tensile stress on the bar?
b. If there were an initial 1.2 mm deep surface crack on the right surface of the bar, what would the critical stress needed to cause instantaneous fast fracture of the bar be?
Engineering
1 answer:
Yuliya22 [10]3 years ago
7 0

Answer:

square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.

Required:

a. What is the nominal maximum tensile stress on the bar?

b. If there were an initial 1.2 mm deep surface crack on the right surface of the bar, what would the critical stress needed to cause instantaneous fast fracture of the bar be?

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The solid spindle AB is connected to the hollow sleeve CD by a rigid plate at C. The spindle is composed of steel (Gs = 11.2 x 1
dalvyx [7]

Answer:

T_max = 12.63 kip.in

Ф_a = 1.093°

Explanation:

Given:

- The modulus of rigidity of solid spindle G_ab = 11.2 * 10^6 psi

- The diameter of solid spindle d_ab = 1.75 in

- The allowable stress in solid spindle τ_ab = 12 ksi

- The modulus of rigidity of sleeve G_cd = 5.6 * 10^6 psi

- The outer diameter of sleeve d_cd = 3 in

- The thickness of sleeve t = 0.25

- The allowable stress in sleeve τ_cd = 7 ksi

Find:

- The largest torque T that can be applied to end A that does not exceed allowable stresses and sleeve angle of twist 0.375°

- The corresponding angle through which end A rotates.

Solution:

- Calculate the polar moment of inertia of both spindle AB and sleeve CD.

   Spindle AB:    c_ab = 0.5*d_ab = 0.5(1.75) = 0.875 in

                           J_ab = pi/2 c^4 = pi/2 0.875^4 = 0.92077 in^4

   Sleeve CD:  c_cd1 = 0.5*d_cd = 0.5(3) = 1.5 in , c_cd2 = c_cd1 - t = 1.25

                     J_cd = pi/2 (c_cd1^4 - c_cd2^4)= pi/2(1.5^4-1.25^4) = 4.1172 in^4

- The stress criteria for maximum allowable torque in spindle AB:

                    T_ab = J_ab*τ_ab / c_ab

                    T_ab = 0.92077*12 / 0.875

                    T_ab = 12.63 kip.in

- The stress criteria for maximum allowable torque in sleeve CD:

                    T_cd = J_cd*τ_cd / c_cd1

                    T_cd = 4.1172*7 / 1.5

                    T_cd = 19.21 kip.in

- The angle of twist criteria for point D:

                    T_d = J_cd*G_cd*Ф / L

                    T_d = 4.1172*5.6*10^6*0.006545 / 8

                    T_d = 18.86 kip.in

- The maximum allowable Torque for the structure is:

                    T_max = min ( 12.63 , 19.21 , 18.86 )

                    T_max = 12.63 kip.in

- The angle of twist of end A:

                    Ф_a = Ф_a/d = Ф_a/b + Ф_c/d:

                    T_max* ( L_ab / J_ab*G_ab + L_cd / J_cd*G_cd)

                    12.63*(12/0.92*11.2*10^6 + 8/4.117*5.6*10^6)

                    0.01908 rads = 1.093°

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Why would an aerospace engineer limit the maximum angle of deflection of the control surfaces?
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It has the same limit of each angle
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An ice hockey player is skating on an ice rink. The rink has a coefficient of kinetic friction of roughly 0.1. If the normal for
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Option C (IaaS) will be the correct answer.

Explanation:

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