A circular copper bar with diameter d 5 3 in. is subjected to torques T 5 30 kip-in. at its ends. Find the maximum shear, tensil
2 answers:
Answer:
Maximum shear stress= 5.66 ksi
Maximum tensile stress= 5.66 ksi
Maximum compressive stress=-5.66 ksi
Maximum shear strain=0.000943
Maximum tensile strain= 0.0004715
Maximum compressive strain= -0.0004715
Explanation:
For acircular bar, the maximum shear stress will be given by
where d is the diameter and T is torque.
By substituting 30 kip-in for torque and 3 in for d then
Maximum shear stress=
Also, the maximum tensile and compressive stresses will be 5.66 ksi and -5.66 ksi respectively.
The maximum shear strain will be given by stress divided by modulus of elasticity, in this case 6000 G
Maximum shear strain will be
The maximum tensile strain will be the above divided by 2 whereas the maximum compressive strain will be negative of tensile strain hence
Maximum compressive strain will be
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Answer:
Explanation:
The viscosities above and below the plate are given by
where
and
are viscosities of fluid below and above plate respectively
Force on plate due to top layer of the fluid
where
and
are the velocity of plate and gap between the plate and upper surface respectively.
Force on plate due to bottom layer of the fluid is given by
where
and
are the velocity of plate and gap between the plate and upper surface respectively
Total force per unit area is the sum of two shear forces
hence
but since
hence