Answer:
hello your question is incomplete attached below is the missing diagram to the question and the detailed solution
Answer : principal stresses : 0.82 MPa, -33.492 MPa
shear stress = 17.157 MPa
∅ = 9.09 ≈ 10°
Explanation:
The principal stress ( б1 ) = 0.82 MPa
( б2 ) = -33.492 MPa
The shear stress = 17.157 MPa
∅ = 9.09 ≈ 10°
attached below is the detailed solution and the Mohr's circle
Answer:
shear plane angle Ф = 26.28°
shear strain 2.20
Explanation:
given data
angle = 16°
chip thickness t1 = 0.32 mm
cut yields chip thickness t2 = 0.72 mm
solution
we get here first chip thickness ratio that is
chip thickness ratio = ................. 1
put here value
chip thickness ratio =
chip thickness ratio r = 0.45
so here shear angle will be Ф
tan Ф = ............2
tan Ф =
tan Ф = 0.4938
Ф = 26.28°
and
now we get shear strain that is
shear strain r = cot Ф + tan (Ф - α ) ................3
shear strain r = cot(26.28) + tan (26.28 - 16 )
shear strain r = 2.20
Answer:
Explanation:
Given that
P = 4 KPa
Contact angle = 6°
Surface tension = 1 N/m
Lets assume that atmospheric pressure = 100 KPa
Lets take that density of water =
So the capillarity rise h
h= 9.61 m
We know that for capillarity rise h
