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:
its a view point for auto cad
Explanation:
from my knowlege in IED we learned about it as a way of sing how an object would look in inventor or auto CAD
Answer:
i) SF: 
ii) BM : 
Explanation:
Let's take,
Making y the subject of formula, we have :
For shear force (SF), we have:
This is the area of the diagram.
The shear force equation =
For bending moment (BM):
The bending moment equation =
