Question 440.
The framework is supported by the member AB which rests on a smooth floor.
When loaded the pressure distribution on AB is linear as shown in the figure. Determine the length d of member AB and the intensity w for this case.
Answer 440.
Let a = bottom surface area of the beam AB.
First we find the moment of the force distribution wa about A and then equate it to the moment of 800 lb. Consider a small element dx of AB at a distance x
Force on it,
dF = wa * (x/d) dx
Moment of this force about A
dM = (wa/d) x^2 dx
Total moment found by integrating between limits x = 0 to x = d is
(wa/d) ∫ x^2 dx ... (x = 0 to x = d)
= wad^2/3
Equatting to the moment of force 800 lb
=> wad^2/3 = 800 * 4
=> wad^2 = 9600 ... ( 1 )
Equatting the force in the vertical direction,
800 = (wa/2) d
=> wad = 1600 ... ( 2 )
Solving ( 1 ) and ( 2 ),
d = 6 ft and
wa = 1600/6 lb = 267 lb.
Link to YA!
The framework is supported by the member AB which rests on a smooth floor.
When loaded the pressure distribution on AB is linear as shown in the figure. Determine the length d of member AB and the intensity w for this case.
Answer 440.
Let a = bottom surface area of the beam AB.
First we find the moment of the force distribution wa about A and then equate it to the moment of 800 lb. Consider a small element dx of AB at a distance x
Force on it,
dF = wa * (x/d) dx
Moment of this force about A
dM = (wa/d) x^2 dx
Total moment found by integrating between limits x = 0 to x = d is
(wa/d) ∫ x^2 dx ... (x = 0 to x = d)
= wad^2/3
Equatting to the moment of force 800 lb
=> wad^2/3 = 800 * 4
=> wad^2 = 9600 ... ( 1 )
Equatting the force in the vertical direction,
800 = (wa/2) d
=> wad = 1600 ... ( 2 )
Solving ( 1 ) and ( 2 ),
d = 6 ft and
wa = 1600/6 lb = 267 lb.
Link to YA!
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