The signboard truss is designed to support a horizontal wind load of 800 lb. A separate analysis shows that 5 8 of this force is
transmitted to the center connection at C and the rest is equally divided between D and B. Calculate the forces in members BE and BC.
1 answer:
Answer:
BE = 559lb T
BC = 300lb T
Explanation:
Base on the figure attached, let us first find the load on joint B, C and D
FC = 5/8 * 800 = 500lb
FD = 13/28 * 800 = 150lb
FB is also equal to 150lb
The angle alpha α at joint D will be:
Tan α = 6/12
α = tan ^-1 (0.5)
α = 26.6
by using resolution of forces, resolve the force at that point to horizontal and vertical component.
Horizontal component
Fd + DE sin α = 0
DE sinα = -Fd
DE = -Fd/sinα
DE = -150/26.6
DE = - 335.4lb
Vertical component
-CD - DEcosα = 0
CD = -DEcosα
CD = -(-335.4)cos26.6
CD = 300lb
So let us do the same at joint C and joint E
At joint C
Horizontal component
Fc + CE = 0
CE = -Fc
CE = -500lb
Vertical component
CD - BC = 0
BC = CD
BC = 300lb T
At joint E
The horizontal component will be:
-CE - DEsinα - BEsinα + EFsinα = 0
-(-500) - (-335.4)sin26.6 - BEsin26.6 + EFsin26.6 + EFsin 26.6
650 - BEsin26.6 + EFsin26.6 = 0
Make BE the subject of the formular
BEsin26.6 = 650 + EFsin26.6
BE = (650 - EFsin26.6)/ sin26.6
BE = 1453.2 + EF ................ (1)
The vertical component
DEcosα - BEcosα - EFcosα = 0
substitute all the parameters
- 335.4cos26.6 - BEcos26.6 - EFcos26.6 = 0
Substitute BE in the equation 1 into the equation above
-300 - (1453.2 + EF)cos26.6 - EFcos26.6 = 0
-1599.7 - 1.789EF = 0
EF = -1599.7 / 1.789
EF = -894.2lb
Substitute EF in equation 1
BE = 1453.2 - 894.2
BE = 559lb T
Therefore the forces in members BE and BC
BE = 559lb T
BC = 300lb T
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(a) Work out put=692.83
(b) Change in specific entropy=0.0044
Explanation:
Properties of steam at 1 MPa and 200°C
We know that if we know only one property in side the dome then we will find the other property by using steam property table.
Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.
Properties of saturated steam at 40°C
So the enthalpy of steam at the exit of turbine
(a)
Work out put =
=2827.4-2134.57
Work out put =692.83
(b) Change in specific entropy
Change in specific entropy =0.0044
Answer:
Temperature at center of apples = 11.2⁰C
Temperature at surface of apples = 2.7⁰C
Amount of Heat transferred = 17.2kJ
Explanation:
The properties of apple are given as:
k = 0.418 W/m.°C
ρ = 840 kg/m³
Cр = 3.81 kJ/kg.°C
α = 1.3*10 ⁻⁷ m²/s
h = 8 W/m².°C
d = 0.09m
r = 0.045m
t = 1 hour = 3600s
<h2>Solution</h2>Biot number is given as:
The constants λ₁ and A₁ corresponding to Biot number (from the table) are:
λ₁ = 1.476
A₁ = 1.239
Fourier Number is:
As Fourier Number > 0.2 , one term approximates solutions are applicable
The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.
