Answer:
Modulus of resilience will be 
Explanation:
We have given yield strength 
Elastic modulus E = 104 GPa
We have to find the modulus
Modulus of resilience is given by
Modulus of resilience
, here
is yield strength and E is elastic modulus
Modulus of resilience
Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
Answer:
lubricating all moving parts in the engine
Explanation:
like the pistons, pushrods, and the crank
Answer:
0.024 m = 24.07 mm
Explanation:
1) Notation
= tensile stress = 200 Mpa
= plane strain fracture toughness= 55 Mpa
= length of a surface crack (Variable of interest)
2) Definition and Formulas
The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.
By definition we have the following formula for the tensile stress:
(1)
We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for 
Multiplying both sides of equation (1) by 
(2)
Sequaring both sides of equation (2):
(3)
Dividing both sides by
we got:
(4)
Replacing the values into equation (4) we got:
![\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5B%5Cfrac%7B55%20Mpa%5Csqrt%7Bm%7D%7D%7B1.0%28200Mpa%29%7D%5D%5E2%20%3D0.02407m)
3) Final solution
So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.