Answer:
a)Wt =25.68 lbf
b)Wt = 150 lbf
F= 899.59 N
Explanation:
Given that
m= 150 lbm
a)
Weight on the spring scale(Wt) = m g
We know that
Wt = 150 x 5.48/32 lbf
Wt =25.68 lbf
b)
On the beam scale
This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.
Wt = 150 lbf
If the plane is moving upward with acceleration 6 g's then the for F
F = m a
We know that
a=6 g's
So
F = 90 x 9.99 N
F= 899.59 N
Explanation:
Step1
Factor of safety is the number that is taken for the safe design of any component. It is the ratio of failure stress to the maximum allowable stress for the material.
Step2
It is an important parameter for design of any component. This factor of safety is taken according to the environment condition, type of material, strength, type of component etc.
Step3
Different material has different failure stress. So, ductile material fails under shear force. Ductile material’s FOS is based on yield stress as failure stress as after yield point ductile material tends to yield. Brittle material’s FOS is based on ultimate stress as failure stress.
The expression for factor of safety for ductile material is given as follows:
Here, is yield stress and
is allowable stress.
The expression for factor of safety for brittle material is given as follows:
Here, is ultimate stress and
is allowable stress.
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
Read more about shear force & bending moment diagram at; brainly.com/question/14834487
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