Question

Now the friends try a homework problem. Two forces of the same magnitude 3.00 x 10' N but with opposite direction are applied to parallel faces of a metal block as shown in the Active Figure. The shear modulus of the block is 5 = 5.00 x 1010 N/m2. The two faces that sustain the shearing force are a distance h = 0.10 m apart, and have cross sectional area A = 0.020 m2. Find the value of the displacement Ax between the edges of the steel block. Submit | Skip (you cannot come back)
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Answer 1

1.

Shear Modulus, T = ?? Shear Stress Shear Strain shear eshrar

Oshear ==
  
Eshear

$\therefore \tau = \frac{\frac{F}{A}}{\frac{\Delta x}{h}} = \frac{Fh}{A\Delta x}$   

Fh 4.1 =

Given,

Shear Modulus,  
T = 5 * 1010 N/m
Force Applied
F = 3 * 107N

Area,
A= 0,020m2

h = 0.10m

===> Answer

Fh 3* 10' * 0.1 Ar = À = 0.02 * 5 * 1010 o = 0.003m = 3mm

2.

Young Modulus, Y = - Linear Stress Linear Strain

$\sigma = \frac{F}{A}$$e = \frac{\Delta L}{L}$

Y =
$\Rightarrow A = \frac{FL}{Y\Delta L}$

Given,

Youngs Modulus,
Y = 20 * 100 N/m
Applied Force,
F = 2 * 106N

Elongation ,
62000 = 7V
Length,
L = 2m

FL 2* 100 * 2 YAL - 20 * 1010 000 = 25.31 * 10-4 m2