Answer:
thrust washer
can be used to eliminate rubbing friction of wheel touching frame
Answer: 0.95 inches
Explanation:
A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.
The length= 64 inches
Ends are fixed Le= 64/2 = 32 inches
Factor Of Safety (FOS) = 3. 0
E= 10.6× 10^6 ps
σy= 4000ps
The square cross-section= ia^4/12
PE= π^2EI/Le^2
6500= 3.142^2 × 10^6 × a^4/12×32^2
a^4= 0.81 => a=0.81 inches => a=0.95 inches
Given σy= 4000ps
σallowable= σy/3= 40000/3= 13333. 33psi
Load acting= 6500
Area= a^2= 0.95 ×0.95= 0.9025
σactual=6500/0.9025
σ actual < σallowable
The dimension a= 0.95 inches
Answer:
y ≈ 2.5
Explanation:
Given data:
bottom width is 3 m
side slope is 1:2
discharge is 10 m^3/s
slope is 0.004
manning roughness coefficient is 0.015
manning equation is written as

where R is hydraulic radius
S = bed slope



P is perimeter 

![Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}](https://tex.z-dn.net/?f=Q%20%3D%20%282%2B2y%29%20y%29%20%5Ctimes%201%2F0.015%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%200.004%5E%7B1%2F2%7D)
solving for y![100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}](https://tex.z-dn.net/?f=100%20%3D%282%2B2y%29%20y%29%20%5Ctimes%20%281%2F0.015%29%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%20%5Ctimes%200.004%5E%7B1%2F2%7D)
solving for y value by using iteration method ,we get
y ≈ 2.5
Answer:
Tso = 28.15°C
Explanation:
given data
t2 = 21 mm
ki = 0.026 W/m K
t1 = 9 mm
kp = 180 W/m K
length of the roof is L = 13 m
net solar radiation into the roof = 107 W/m²
temperature of the inner surface Ts,i = -4°C
air temperature is T[infinity] = 29°C
convective heat transfer coefficient h = 47 W/m² K
solution
As when energy on the outer surface at roof of a refrigerated truck that is balance as
Q =
.....................1
Q =
.....................2
now we compare both equation 1 and 2 and put here value
solve it and we get
Tso = 28.153113
so Tso = 28.15°C