Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 10M

Year: June 2014

The supports need to be found out first.

Support Reactions:

$ΣM_A = 0$

$15(1.5) - 10 - 5(8) + R_B × 10 = 0$

$R_B = 2.75kN↑$

$ΣF_Y = 0$

$R_A - 15 - 5 + R_B = 0$

$R_A - 20 + 2.75 = 0$

$R_A = 17.25kN↑$

Shear Force Calculations:(↑ | ↓ +ve)

$SF_B = -2.75kN$

$SF_E (Just \ \ right \ \ to \ \ E) = -2.75kN$

$SF_E (Just \ \ left \ \ to \ \ E) = -2.75kN + 5kN$

$SF_D = 2.25kN$

$SF_C = 2.25kN$

$SF_A = 2.25 + 151 = 17.25kN$

Bending Moment Calculations: (AC+)

$BM_B = 0$

$BM_E = (2.75)(2) = 5.5kNm$

At the point D, there is a moment acting and therefore there will be a sudden change in the value of the moment at this point. To determine this change, we find out the bending moments to the right and to the left of the point D.

$BM_D (Just \ \ Right \ \ of \ \ D) = (2.75)(5) - (5)(3) = -1.25kNm$

$BM_D (Just \ \ Left \ \ of \ \ D) = (-1.25) + (-10) = -11.25kNm$

$BM_C = (2.75)(7) - (5)(5) - 10 = -15.75kNm$

$BM_A = 0$