if force of 2000N acts on a rectangular slab of 10cm *15cm find stress acting on it
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Answer:
The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure. Area of the face parallel to which this force is applied is
A=50cm×10cm=0.5m×0.1m=0.05m
2
If ΔL is the displacement of the upper edge of the slab due to tangential force, F, then
η=
ΔL/L
F/A
orΔL=
ηA
FL
Substituting the given values, we get
ΔL=
5.6×10
9
Nm
−2
×0.05m
2
(9×10
4
N)(0.5m)
=1.6×10
−4
m=0.16mm
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Answer:133.33×10³ N/m²
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