A square steel plate has area `1 m^(2)` and thickness 5 cm. The lower surface is fixed. A tangential force applied to top surface displace it through

Question

A square steel plate has area `1 m^(2)` and thickness 5 cm. The lower surface is fixed. A tangential force applied to top surface displace it through

A square steel plate has area `1 m^(2)` and thickness 5 cm. The lower surface is fixed. A tangential force applied to top surface displace it through 0.005 cm. Find the modulus of rigidity of steel, if shearing stress of modulus of rigidity of steel. If shearing stress of `4.2 xx 10^(3) N//m^(2)` is being applied.
A. `5.2 xx 10^(6) Nm^(-2)`
B. `4.2 xx 10^(6) Nm^(-2)`
C. `3.6 xx 10^(4) Nm^(-2)`
D. `4.6 xx 10^(4) Nm^(-2)`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Given, area `A = 1 m^(2)`
Thickness, `d = 5 cm = 5 xx 10^(-2) m`
lateral displacement = 0.005 cm
`= 5 xx 10^(-5) m`
Shearing strain `= ("Lateral displacement")/("Distance from fixed layer")`
`= (5 xx 10^(-5))/(5 xx 10^(-2)) = 10^(-3)`
Modulus of ragidity `= ("Shearing stress")/("Shearing strain")`
`= (4.2 xx 10^(2))/(10^(-3))`
Modulus of rigidity `eta = 4.2 xx 10^(6) Nm^(-2)`