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Question: Two strips of metal are riveted together at their ends by four rivets, each of diameter 6 mm. Assume...

Two strips of metal are riveted together at their ends by four rivets, each of diameter 6 mm. Assume that each rivet is to carry one quarter of the load. If the shearing stress on the rivet is not to exceed 6.9×1076.9 \times 10 ^ { 7 }Pa, the maximum tension that can be exerted by the riveted strip is

A

2×103 N2 \times 10 ^ { 3 } \mathrm {~N}

B

3.9×103 N3.9 \times 10 ^ { 3 } \mathrm {~N}

C

7.8×103 N7.8 \times 10 ^ { 3 } \mathrm {~N}

D

15.6×103 N15.6 \times 10 ^ { 3 } \mathrm {~N}

Answer

7.8×103 N7.8 \times 10 ^ { 3 } \mathrm {~N}

Explanation

Solution

: Radius of a rivet,

Maximum stress = 6.9×107 Pa6.9 \times 10 ^ { 7 } \mathrm {~Pa}

Maximum load on a rivet

= Maximum stress × Area of cross-section

= 6.9×107×π×(3×103)26.9 \times 10 ^ { 7 } \times \pi \times \left( 3 \times 10 ^ { - 3 } \right) ^ { 2 }

= 1950 N

Maximum tension that can be exerted by rivet strip

=4×1950 N=7800 N=7.8×103 N= 4 \times 1950 \mathrm {~N} = 7800 \mathrm {~N} = 7.8 \times 10 ^ { 3 } \mathrm {~N}