Question

Example 9.4 A square lead slab of side 50
cm and thickness 10 cm is subject to a
shearing force (on its narrow face) of 9.0 x
104 N. The lower edge is riveted to the floor.
How much will the upper edge be displaced?



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Answer 1

Answer:

hey mate ....

Explanation:

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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