Shear Strain
Trending Questions
- 4.8×10−4 m
- 3.2×10−4 m
- 1.6×10−4 m
- 0.8×10−4 m
The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30o. Then angle of shear is
12°
0.12°
1.2°
0.012°
- shear
- longitudinal
- longitudinal and shear
- Volumetric
When a spiral spring is stretched by suspending a load on it, the strain produced is called
Shearing
Longitudinal
Volume
Transverse
An ISLB 350@ 486 N/m beam is laterally supported throughout. The following properties are given for the I - section
Ix=13158.3×104mm4, shape factor =1.12, fy=250MPa, γmo=1.1
Assume the section to be a compact section and a case of low shear. The design bending strength will be
- 191.39
- 300
- mgL2AY
- mgLAY
- mgL3AY
- 2mgLAY
- 0.025
- 0.05
- 0.1
- 1
A vertical metal cylinder of radius 2 cm and length 2 m is fixed at the lower end and a load of 100 kg is put on it. Find (a) the stress (b) the strain and (c) the compression of the cylinder. Young modulus of the metal = 2×1011Nm−2.
- 10−3 m
- 10−2 m
- 10−4 m
- 10−1 m
- 16.2
What is the shearing stress on this plane?
- (F/3A)sin2θ.
- (F/2A)sin2θ.
- (F/4A)sin2θ.
- (F/5A)sin2θ.
The maximum longitudinal pitch in bolted joints, subjected to tensile forces will be
- 200
For what value of θ is the shearing stress maximum?
- 45
- 55
- 65
- 775
- 3200 kg/m3
- 800 kg/m3
- 1250 kg/m3
- 5000 kg/m3
- 0.001 radians
- 0.004 radians
- 0.002 radians
- 0.04 radians
- 0.1 mm
- 0.02 mm
- 0.04 mm
- 0.004 mm
- 0.002
- 0.004
- 0.008
- 0.016
(Take g=10ms−2)
- 4km
- 8km
- 10km
- 16km
(Modulus of rigidity η=8×1010N/m2)
- 1.1×10−3
- 1.5×10−3
- 1.25×10−3
- 1.6×10−3
- 0.120
- 1.20
- 120
- 0.0120
- FlπYr1/2
- 2FlπYr1/2
- 3FlπYr1/2
- 2FIYr1r2
- tanθ
- tan2θ
- cos2θ
- cot2θ
- 2×103N
- 15.6×103N
- 7.8×103N.
- 3.9×103N
- 100 cm
- 150 cm
- 50 cm
- 200 cm
- 0.5×10−3Nm−1
- 10−3Nm−1
- 2×10−3Nm−1
- 0.1×10−3Nm−1
- 1.6×10−3
- 1.6×103
- 16×103
- 16×106