Answer:
45.3 MN
Explanation:
The forging force at the end of the stroke is given by
F = Y.π.r².[1 + (2μr/3h)]
The final height, h is given as h = 100/2
h = 50 mm
Next, we find the final radius by applying the volume constancy law
volumes before deformation = volumes after deformation
π * 75² * 2 * 100 = π * r² * 2 * 50
75² * 2 = r²
r² = 11250
r = √11250
r = 106 mm
E = In(100/50)
E = 0.69
From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula
F = Y.π.r².[1 + (2μr/3h)]
F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]
F = 35.3 * [1 + 0.2826]
F = 35.3 * 1.2826
F = 45.3 MN
Answer:
Explanation:
relating to, measuring, or measured by the quantity of something rather than its quality.Often contrasted with qualitative.
Answer:
P.E.T
Explanation:
Stud is a short form of stud wall which are typically vertical framing members in a wall of a building. Wall studs are available in standard lengths but since the height of ceiling vary, sometimes one has to cut the stud walls on site to fit. Therefore, there are pre-cut studs which are more efficient in implementation. These are usually referred as Precision-End Trimmed studs, (P.E.T). Therefore, stud walls which are cut to exact length at the mill for modern construction are designated by letters P.E.T
Answer:
86584N
132.306 mm
Explanation:
Q = 274
Modulus of elasticity = 118 gpa
1.
Area = 316mm² without plastic deformation
F = QA
= 274x10⁶x316x10^-6
= 274000000 x 0.000316
= 86584 N
This is the maximum load.
2.
Max length =
L = 132(1 + 274x10⁶/118x10⁹)
L = 132(1+274000000/118000000000)
L = 132(1+0.002322)
L = 132(1.002322)
L = 132.306
This is the maximum length to which it may be stretched without causing plastic deformation.
Answer:
A shaft is made of an aluminum alloy having an allowable shear stress of T_allow = 100 MPa. If the diameter of the shaft is 100 mm, determine the maximum torque T that can be transmitted.
Explanation:
