The shear modulus of aluminum is 2.4 × 10 10 N/m 2 . An aluminum nail of radius

Question

The shear modulus of aluminum is 2.4 × 1010 N/m2. An aluminum nail of radius 7.5 × 10–4 m projects 0.035 m horizontally outward from a wall. A man hangs a wet raincoat of weight 25.5 N from the end of the nail.
Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?
[Calculate your answer in meters, but do NOT include units in your answer. Provide a numerical value only. This answer SHOULD be entered in scientific notation. Use only 3 significant digits, use "E" to indicate the power of 10, and use a two digit number for the exponent, for example 1.23E+04 ]

What is the stress for the situation described in question #1?

[Do NOT include units in your answer. Provide a numerical answer only for this question. Do NOT use scientific notation for this answer.]

The correct units for the answer to the previous question are?

What is the strain for the situation described in question #1?[Do NOT include units in your answer. Provide a numerical answer only for this question. This answer SHOULD be entered in scientific notation. Use only 3 significant digits, use "E" to indicate the power of 10, and use a two digit number for the exponent, for example 1.23E+04]

Explanation / Answer

Given:-

G = 2.4×10^10 N/m^2
Y= 7.0 x 10^10 N/m^2 (youngs modulus of aluminum)

A= 1.77 x 10 ^-6 N/m^2

F = 25.5 N, L = 0.035 m

1) Vertical deflection due to shear only :

deltaY= (F/A)(L/G) (25.5 N/(1.77 x 10 ^-6 N/m^2)) / (0.035m/(2.4×10^10 N/m^2))

= 0.210E-04 m

= 2.10E-05 m

2) Shear stress= F/A = 25.5 N/(1.77 x 10 ^-6 N/m^2)= 1.44 x 10^7 N/m^2

3) Y= stress/strain ---> strain = stress/Y

strain = (1.44 x 10^7 N/m^2)/(7.0 x 10^10 N/m^2)

strain = 2.00E-04

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