Answer:
Change in length = 0.1257 mm
Change in diameter= -0.03771mm
Explanation:
Given
Diameter, d = 15 mm
Length of rod, L = 200mm
F = Force= 300N
d = 0.015m
Ep=2.70 GPa, np=0.4.
First, we have to calculate the normal stress using
σ = F/A where F = Force acting on the Cross-sectional area
A = Area
Area is calculated as πd²/4 where d = 0.015m
A = 22/7 * 0.015²/4
A = 0.000176785714285m²
A = 1.768E-4m²
So, stress. σ = 300N/1.768E-4m²
σ = 1696832.579185520Pa
σ = 1.697MPa
Calculating E(long)
E(long) = σ /Ep
E(long) = 1.697E-3/2.70
E(long) = 0.0006285
At this point, we fan now calculate the change in length of the element;
∆L = E(long) * L
∆L = 0.0006285 * 200mm
∆L = 0.1257mm
Calculating E(lat)
E(lat) = -np * E(long)
E(lat) = -4 * 0.0006285
E(lat) = -0.002514
At this point, we can now calculate the change in diameter of the element;
∆D = E(lat) * D
∆L = -0.002514 * 15mm
∆L = -0.03771mm
Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²
x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K
K= 58.66 lb/ft
The damper coefficient for critically damped system
Cc= 12.7 lb.sec/ft
Answer:
I think 5 is passion and 6 is true I think and 7 is time management I'm pretty sure
Answer:
Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.
When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.
