Deflection at the centre - maximum deflection:
A simply supported beam of span 6 m is subjected to Udl of 24 kN/m for a length of 2 m from left support. Discover the deflection at the centre, maximum deflection & slopes at the ends and at the centre. Take EI = 20 × 10^{6} N-m^{2}.
Solution
∑ F_{y} = 0, so that R_{A} + R_{B} = 24 × 2 = 48 kN --------- (1)
Taking moments around A,
24 × 2 × 1 = R_{B} × 6
R_{B} = 8 kN (↑) -------- (2)
R_{A} = 48 - 8 = 40 kN (↑). ------------(3)
By apply the Udl over the portion DB downwards and upwards,
Figure
M = 40 x - 24 x × (x/2) + 24 ( x - 2) ( (x - 2)/2)
Note down that the third term vanishes if x < 2 m.
= 40 x - 12 x^{2 } + 12 ( x - 2)^{2} ------- (4)
EI d ^{2} y/ dx^{2} = 40 x - 12 x^{ 2} + 12 ( x - 2)^{2} ------- (5)
EI dy / dx = 40 x^{2}/2- 12 x^{3} /3+ 12 ( x - 2)^{3}/3 + C_{1}
= 20 x^{2} - 4 x^{3} + 4 ( x - 2)^{3} + C_{1} -------- (6)
EIy = 20 x^{ 2}/3 - x^{4} + (x - 2)^{4 }+ C_{1} x + C_{2} -------- (7)
Here again note that the third term vanishes for x < 2 m.
at A, x = 0, y = 0 ∴ C_{2} = 0
at B, x = 6 m, y = 0
0 = 20 × 6^{3} /3 - 64 + (6 - 2)^{4} + C_{1} × 6
C1 =- 20 × 12 + 36 × 6 - ((16 × 16 )/6)=- 200/3
∴ EI dy/dx = 20 x^{2 } - 4 x^{3} + 4 ( x - 2)^{3} - 200/3 -------- (8)
The third term vanishes.
Slope at A, (x = 0), 27
θ_{A} = -200/3EI =- (200 × 10^{3})/ (3 × 20 ×10^{6})
= -(1/300) rad = - 3.33 × 10^{- 3} rad
Slope at B, (x = 6 m),
EI θ _{B} = 200 × 6^{2} - 4 × 6^{3 } + 4 (6 - 2)^{3} - (200/3)
θ = 136/ 3 EI = (136 × 10^{3 })/(3 × 20 ×10^{6})
= + 2.27 × 10^{- 3} radian
Slope at C, (x = 3 m), i.e. x > 2 m
EI θ _{C} = 20 × 3^{2 } - 4 × 3^{3} + 4 (3 - 2)^{3} - (200/3)
θ_{C} = 20 /3 EI = 0.47 × 10^{- 3} radians
EIy =( 20 x ^{3}/3)- x^{4} + ( x - 2)^{4} - (200/3) x -------- (9)
Deflection at centre, (x = 3 m),
EIy_{C} = (20/3) × 33 - 34 + (3 - 2)4 - (200 /3)× 3
y_{C} = - 100 / EI = - 100 × 10^{3} × 10^{3 }/ (20 × 10^{6})
= - 5 mm
For maximum deflection,
dy/ dx = 0
0 = 20 x^{2 } - 4x^{3} + 4 ( x - 2)^{3} - (200/3)
= 20 x^{2} - 4x^{3} + 4x^{3} - 32 - 24 x^{2} + 48 x - (200 /3)
=- 4x^{2} + 48 x - (296 /3)
∴ x2 - 12x + (74 /3 )= 0
x = 2.63 m , x > 2m
EIy _{max} = (20/3) × 2.633 - 2.634 + (2.63 - 2)4 - (200/3) × 2.63 = - 101.7
∴ y_{max} = - 5.087 mm; - 5.1 mm