Question

A cantilever is 3 meters long. The free end is on the left. On the left is a point load of 12kN. One meter from the left a bending moment of 10 kNm anticlockwise is applied.Take E = 200 GPa and I = 37 x 10^-6m⁴

Use Macaulay' s method and take x = 0 on the left.

Answer 1

Macaulay's method:

Moment equation; M(x)=-12000(x-0)¹-10000(x-1)⁰

$EI\frac{d^2 y}{dx^2}=M(x)=-[12(x)^1+10(x-1)^0]*10^3 Nm$

integrating;

$EI\frac{dy}{dx}=-[6(x)^2+10(x-1)^1]*10^3 +A$

at x=3, $\frac{dy}{dx}=0$

$0=-[6(3)^2+10(3-1)^1]*10^3 +A$

A=74000

So,

$EI\frac{dy}{dx}=-[6(x)^2+10(x-1)-74]*10^3$

on integrating again,

$EIy=-[2(x)^3+5(x-1)^2-74x]*10^3 +B$

at x=3, y=0

$0=-[2(3)^3+5(3-1)^2-74*3]*10^3 +B$

B=-148000

So,

$EIy=-[2(x)^3+5(x-1)^2-74x+148]*10^3$

A=74000

B=-148000

Slope at free end;

$EI\frac{dy}{dx}$ at x=0 is 74000

$\frac{dy}{dx}$ at x=0 is $\frac{74000}{2*10^{11}*37*10^{-6}}$

=0.01 rad