Find out the maximum deflection in beam:
A beam of span 8m is loaded with UDL of 10 kN/m over the middle half portion. Discover the maximum deflection. EI is constant.
Solution
By symmetry,
R_{A} = R_{B}
= (10 × 4 )/2
= 20 kN
Figure
M = 20 x -(10 ( x - 2)^{2}) /2-(10 ( x - 6)^{2})/2
= 20 x - 5 ( x - 2)^{2} - 5 ( x - 6)^{2} ---------(1)
EI ( d ^{2} y/ dx^{2}) = M= 20 x - 5 ( x - 2)^{2} - 5 ( x - 6)^{2} -------- (2)
EI dy/ dx = 20 x^{2 }/2- 5 ( x - 2)^{3 }/3-( 5/3) ( x - 6)^{3} + C_{1} . . . (3)
EIy = 10 x^{3} /3- (5/12) ( x - 2)^{4} - (5/12) ( x - 6)^{4 } + C_{1} x + C _{2 } --------- (4)
at A, x = 0, y = 0, C2 = 0
at B, x = 8 m, y = 0
0 =10 × 8^{3}/3 - (5 /12 ) (8 - 2)^{4} - (5/12 ) (8 - 6)^{4} + C_{1} × 8
C_{1 }= - 145
EI (dy/ dx) = 10 x2/3 - (5/3) ( x - 2)3 - (5 /3)( x - 6)3 - 145
EIy = 10 x ^{3 } /3 - (5 /12) ( x - 2)^{4} - (5/12) ( x - 6)^{4} - 145 x
The maximum deflection at centre, since the beam is symmetrical and symmetrically loaded.
∴ EIy_{max } =((10 × 4^{3} )/3) - (5/12) (4 - 2)^{4} - (5/12) ( 4 - 6)^{4} - 145 × 4
y_{max}= - 1120/3EI