Even though the content of many alcohol blends doesn’t affect engine drive ability using gasoline with alcohol in warm weather m
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Maximum absolute values of the shear = 28 KN
Maximum absolute values of bending moment = 5.7 KN.m
<h3>How to draw Shear Force and Bending Moment Diagram?</h3>A) We can see the beam loaded in the first image attached.
For the shear diagram, let us calculate the shear from point load to point load.
From A to C, summing vertical to zero gives; ∑fy = 0: -20 - V = 0
V = -20 KN
From C to D, summing vertical to zero gives; ∑fy = 0: -20 + 48 - V = 0
V = 28 KN
From D to E, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - V = 0
V = 8 KN
From E to B, summing vertical to zero gives; ∑fy = 0: -20 + 48 - 20 - 20 - V = 0
V = -12 KN
For the bending moment diagram, let us calculate the bending moment from point load to point load.
At point A, the bending moment would be zero. Thus, M_A = 0 KN.m
At point C, taking moment about point C and equating to zero gives;
M_C = 0. Thus; 20(0.225) + M = 0
M = -4.5 KN.m
At point D, taking moment about point D and equating to zero gives;
M_D = 0. Thus; 20(0.525) - 48(0.3) + M = 0
M = 3.9 KN.m
At point E, taking moment about point E and equating to zero gives;
M_D = 0. Thus; 20(0.75) - 48(0.525) + 20(0.225) + M = 0
M = 5.7 KN.m
At point B, taking moment about point E and equating to zero gives;
M_E = 0. Thus; 20(1.05) - 48(0.825) + 20(0.525) + (20 * 0.3) + M = 0
M = 2.1 KN.m
2) From the attached diagrams, we can deduce that;
Maximum absolute values of the shear = 28 KN
Maximum absolute values of bending moment = 5.7 KN.m
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Answer:
a). 8.67 x m
b).0.3011 m
c).0.0719 m
d).0.2137 N
e).1.792 N
Explanation:
Given :
Temperature of air, T = 293 K
Air Velocity, U = 5 m/s
Length of the plate is L = 6 m
Width of the plate is b = 5 m
Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X Pa-s
We know density of air is ρ = 1.21 kg /
Now we can find the Reyonld no at x = 1 m from the leading edge
Re =
Re =
Re = 332052.6
Therefore the flow is laminar.
Hence boundary layer thickness is
δ =
=
= 8.67 x m
a). Boundary layer thickness at x = 1 is δ = 8.67 X m
b). Given Re = 100000
Therefore the critical distance from the leading edge can be found by,
Re =
100000 =
x = 0.3011 m
c). Given x = 3 m from the leading edge
The Reyonld no at x = 3 m from the leading edge
Re =
Re =
Re = 996158.06
Therefore the flow is turbulent.
Therefore for a turbulent flow, boundary layer thickness is
δ =
=
= 0.0719 m
d). Distance from the leading edge upto which the flow will be laminar,
Re =
5 X =
x = 1.505 m
We know that the force acting on the plate is
=
and at x= 1.505 for a laminar flow is =
=
= 1.878 x
Therefore, =
=
= 0.2137 N
e). The flow is turbulent at the end of the plate.
Re =
=
= 1992316
Therefore =
=
= 3.95 x
Therefore =
=
= 1.792 N
