a uniformly thick beam of length L with square cross-section with width a has a distributed force,...

Question

a uniformly thick beam of length L with square cross-section with width a has a distributed force, q(x)=Qcos pie*x/2L N/m applied x=0 to x=L. A point force P is applied at x=L. Choose the beam dimension a to reduce the deflection to below 1cm whilst minimising the mass of the beam.

Robert · Accepted Answer

Consider a cantilever beam with non uniform loading as shown in figure below.
To find the point at which the total force will act, consider a vertical differential element of area 
cos
2
x
dA ydx Q dx
L
 
   
 
as shown in figure above. We know that cAx xdA  . So using this, we 
find the equation of distance where the resultant force will act. 
Equation 1 
2
2
cos cos
2 2
2 2 2
sin sin cos
2 2 2
2 2
sin cos
2 2
2
sin
2
2
cot
2
c
c
c
c
x x
x Q dx xQ dx
L L
LQ x xL x L x
x Q
L L L
xL x L x
Q
L L
x
LQ x
L
L x
x x
L
 
  
  
 
 




   
   
   
        
         
         
      
      
       

 
 
 
 
   
 
 
 
Using Macaulay’s method, consider a section X-X at a distance x from the free end.

a uniformly thick beam of length L with square cross-section with width a has a distributed force, q(x)=Qcos pie*x/2L N/m applied x=0 to x=L. A point force P is applied at x=L. Choose the beam...

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