A rectangle block of aluminium is 45 cm by 45 cm by 15 cm when unstressed. A force of 1.273 N is applied
tangentially to the upper surface causing a 2.7 cm displacement relative to the lower surface. The block is
placed such that 45x45 comes on the lower and upper surface.
Calculate the shear modulus
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Shearing stress=
area parallel to the force
force
=
500mm×20mm
200N
=
(0.5×0.02)
200
N/m
2
=20000Pa
Shearing strain ϕ=
300mm
15mm
=0.05
Shearing modulus=
Shearing strain
Shearing stress
=
0.05
20000
=400000Pa
=4×10
5
Pa
⇒n=4
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