Chapter-5
Deflection of Beam
S K Mondal’s
2
dy
³ EI dx or EI
2
dy dx
P ³ x dx P.
x2 A 2
.............(i)
Again integrating both side we get § x2 · EI³ dy = ³ ¨ P A ¸ dx © 2 ¹ 3 Px or EIy = ..............(ii) Ax +B 6 Where A and B is integration constants. Now apply boundary condition at fixed end which is at a distance x = L from free end and we also know that at fixed end at x = L,
y=0
at x = L,
dy dx
0
from equation (ii) EIL = -
PL3 + AL +B 6
from equation (i) EI.(0) = -
PL2 +A 2
Solving (iii) & (iv) we get A =
Therefore,
y=-
..........(iii) …..(iv)
PL2 PL3 and B = 2 3
Px 3 PL2 x PL3 6EI 2EI 3EI
The slope as well as the deflection would be maximum at free end hence putting x = 0 we get ymax = -
PL3 (Negative sign indicates the deflection is downward) 3EI
(Slope)max = T max =
PL2 2EI
Remember for a cantilever beam with a point load at free end.
3
Downward deflection at free end,
And slope at free end,
T
G
PL 3EI
PL2 2EI Page 201 of 429