1.A metal bar of
10mm dia when subjected to a pull of 23.55KN gave and elongation of 0.3mm on a
gauge length of 200mm. In a torsion test maximum shear stress of 40.71N/mm2
was measured on a bar of 50mm dia. The angle of twist measured over a length of
300mm being 0°21’.
diad = 10mm
pull P = 23.55KN
elongation SL = 0.3mm
Gauge length = 200mm
(or) = 40.71
Dia = 500mm
Twist angle = 0o21’
length = 300 mm
A hollow shaft dia
ratio 3/5 is required to transmit 450Kw at 1200pm, the shearing stress in the
shaft must not exceed 60N/mm2 and the twist in a length of 2.5m is
not to exceed 1o. Calculate the minimum external of the shaft. Take,
What must be the length
of a 5mm dia aluminium wine so that it can be twisted through 1 complete
revolution without exceeding a shear of 42N/mm2. Take, G=27 GPO.
A solid steel shaft has
to transmit 75Kw power at 200 pm. Taking allowable shear stress 70Mpo. Find
suitable dia of shaft with the maximum torque transmitted on each revolutions
exceeds by mean by 30% 1.3 times mean.