Rolling and Shearing are the types of a)-Bulk Deformation Process b)- Sheet Metal Process c)- Machining Process d)- Both a &
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The image is missing, so i have attached it.
Answer:
A) P = 65.11 KN
B) Q = 30 KN
Explanation:
We are given;
The end reaction of the beam; F = 100KN
Coefficient of static friction between two steel surfaces;μ_ss = 0.3
Coefficient of static friction between steel and concrete;μ_sc = 0.6
So, F1 = μ_ss•F =0.3 x 100 = 30 KN
F2 = μ_ss•N_EF = 0.3N_EF
From the screen shot, we see that the angle is 12°
Sum of forces in the Y-direction gives;
F2•sin12 - N_EF•cos12 + 100 = 0
Rearranging gives;
N_EF•cos12 - F2•sin12 = 100
Let's put 0.3N_EF for F2 to give;
N_EF•cos12 - 0.3N_EF•sin12 = 100
Thus;
N_EF(0.9158) - 0.1247 = 100
N_EF(0.9781) = 100 + 0.1247
N_EF = 100.1247/0.9158
N_EF = 109.33 KN
Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN
Wedge will move if;
P = (F1 + F2cos12 + N_EFsin12)
Thus;
P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)
P ≥ 65.11 KN
B) For static equilibrium, Q = F1
Thus, Q = 30 KN
Answer:
Explanation:
The Young's module is:
Let assume that both specimens have the same geometry and load rate. Then:
The displacement rate for steel is:
Answer:
Explanation:
Given that
L= 50 m
Pressure drop = 130 KPa
For Copper tube is 3/4 standard type K drawn tube
Outside diameter=22.22 mm
Inside diameter=18.92 mm
Dynamic viscosity for kerosene
Pressure difference given as
Where
L is length of tube
μ is dynamic viscosity
Q is volume flow rate
d is inner diameter of tube
ΔP is pressure drop
Now by putting the values
So flow rate is