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 x104 N. The lower edge is riveted to the floor. 1 1 1 1 50 cm How much will the upper edge be displaced?
Answers
Answered by
4
Answer:
The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure. A=50cm×10xm=0.5m×0.1m=0.05m2.
Explanation:
I hope this is helpful for you.
please dear mark me as brainliest.
Similar questions