Answer: required tensile stress is 0.889 MPa
Explanation:
Given that;
tensile load is oriented along the [1 1 1] direction
shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane
first we determine
λ which is Angle between [1 1 1] and [1 0 1]
so
cosλ = [ 1(1) + 1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]
= 2 / √3√2 = 2/√6
Next, we determine ∅ which is angle between [1 1 1] and [1 1 -1]
so,
cos∅ = [ 1(1) + 1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]
cos∅ = [ 2-1] / [√3√3 ]
cos∅ = 1/3
Now, we know that;
σ = T_stress/cosλcosθ
so we substitute
σ = 0.242 / ( 2/√6 × 1/3 )
σ = 0.242 / 0.2721
σ = 0.889 MPa
Therefore, required tensile stress is 0.889 MPa
Answer:
••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.
Answer:
Explanation: Clutch Plate.
Clutch Cover.
Clutch Bearing (Release bearing)
Release Fork (clutch fork)
Answer:
The shear strain is 0.05797 rad.
Explanation:
Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.
Step1
Given:
Height of the pad is 1.38 in.
Deformation at the top of the pad is 0.08 in.
Calculation:
Step2
Shear strain is calculated as follows:
For small angle of 
, 
can take as.
Thus, the shear strain is 0.05797 rad.
