Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 12M

Year: Dec 2013

The first step will be to find out the reaction at the supports.

Support Reactions

We know that the Moment about the point A is 0.

Using $ΣM_A = 0$, we have, (AC| +)

$-(8 × 5) × 2.5 - (-20 sin 60) × 7 + 10 - 4 - (12 sin 45) × 13 + R_D × 10 = 0$

$R_D = 32.55kN↑$

Using $ΣF_Y = 0 \ \ \ (↑ | ↓ +ve)$

$V_A - (8 × 5) - 20 sin 60 - 12 sin 45 + 32.55 = 0$

$V_A = 33.25kN↑$

Using $ΣF_X = 0 \ \ \ (→ +ve)$

$H_A - 20 cos 60 + 12 cos 45 = 0$

$H_A = 1.15 kN→$

Shear Force Calculations: (↑ | ↓ +ve )

$SF_E = 12 sin45 = +8.485kN$

$SF_D (Just \ \ Right \ \ Of \ \ D) = +8.485kN$

$SF_D (Just \ \ Left \ \ Of \ \ D) = +8.485kN - 32.55kN = -24.065kN$

$SF_C (Just \ \ Right \ \ Of \ \ C) = -24.065kN$

$SF_C (Just \ \ Left \ \ Of \ \ C) = -24.065kN + 20 sin 60 = -6.744kN$

$SF_B = -6.744kN$

$SF_A = -6.744 + 40 = +33.25kN$

There is a sudden sign change in the value of the shear force between the points A and B. This implies that there is a point of Zero shear between these two points. Let this Point be called P. Let P be located at a distance of ‘l’ from the point A.

By the law of similar triangles, we have,

$\frac{33.25}{l} = \frac{6.744}{5 - l}$

$l = 4.157m$

Bending Moment Calculations: (AC+)

$BM_E = 0$

Since there are couples present at D and C, there is going to be a sudden change in the bending moments at these points. Hence to know these values, we find the values of the bending moments just left and right of these two points.

$BM_D (Just \ \ Right \ \ Of \ \ D) = (-12 sin 45 × 3) = -25.456kNm$

$BM_D (Just \ \ Left \ \ Of \ \ D) = -25.456 - 4 = -29.456kNm$

$BM_C (Just \ \ Right \ \ Of \ \ C) = (-12 sin 45 × 6) - 4 + (32.55 × 3 ) = 42.74kNm$

$BM_C (Just \ \ Left \ \ Of \ \ C) = 42.74 + 10 = 52.74kNm$

$BM_B = (-12 sin 45 × 8) - 4 + (32.55 × 5) + 10- (20 sin 60 × 2) = 66.22kNm$

$BM_P = (32.25 × 4.157) - (8 × 4.157) × [\frac{4.157}{2}] = 69.1kNm$

$BM_A = 0$

In the B.M.D, we find that the point of Inflexion lies between the points CD. Let it be denoted by point O. Let it be at a distance ‘z’ from the point D.

By the law of similar triangles, we have:

$\frac{29.45}{z} = \frac{42.74}{3 - z}$

$z = 1.22 m$

The point of Inflexion O lies at a distance of 1.225m from the support ‘D’

Axial Force Calculations: (→ + )

$AF_E = 12cos45 = +8.484kN$

$AF_C (Just \ \ Right \ \ Of \ \ C) = +8.484kN$

$AF_C (Just \ \ Left \ \ Of \ \ C) = +8.484kN - 20cos60 = -1.515kN$

$AF_A = -1.515kN$