Question

Strength of materials I (CIV213) 1. For the simply supported beam and load shown below A. Draw shear and bending moment diagram B. Determine the absolute maximum bending stress. (10pts) I P2 = 16 kN P,= 12 kN 6 cm D 4 m BL 1.5 m. 5 m.° 3.5m 100mm -9 m

Answer 1

P2 = 16 kN P,= 12 kN D B IC 4 m 1.5 m 3.5m ng -9 m VA р

For drawing the SFD and BMD, Firstly we need the end support reaction V_A and V_D . For this, we need to take $\sum M_{A}= 0$

Sign Convention : Assume Clockwise Direction - positive and anticlockwise - negative.

$P_{1}\times 4 + P_{2}\times 5.5 - V_{D}\times 9 = 0$

$P_{1}\times 4 + P_{2}\times 5.5 = V_{D}\times 9$

12 x 4+ 16 x 5.5 = Vp X 9

$\frac{12 \times 4 + 16\times 5.5}{9} = V_{D}$

V_D = 15.11 KN.

We also Know that $\sum V_{A} = 0$   

Sign Convention : upward force - positive and downward force - negative

V_A -12 -16 + V_D = 0

V_A + V_D = 28

but V_D= 15.11 KN

So V_A= 28 - 15.11 KN = 12.89 KN

V_D = 15.11 KN.and V_A= 12.89 KN

For Bending Moment.

BM at ends of hinge and roller support is zero. So BM_A = BM_D = 0

For BM_B = V_A * 4 = 12.89 * 4 = 51.56 KNm

For BM_C = V_A * 5.5 - P₁ * 1.5 = 12.89 * 5.5 - 12*1.5 = 52.89 KNm

So SFD and BMD are as shown below:

SFD +12.89KN +0.89 KN D A B -15.11 KN 52.89 Km BMD 51.56 Km D A B

B. Absolute Bending Stress  

Bending Stress formula is given by:

$\frac{M}{I}= \frac{\sigma }{y}$

where M = Max Bending Moment = 52.89 KNm ( as calculated above)

I = Moment of Interia = for rectangular section

bd 12

b= 6 cm and d=100 mm = 10 cm

$I = \frac{6\times 10^{3}}{12} = 500 cm^{4} = 5.0 \times 10^{-6} m^{4}$

$Y = \frac{D}{2} = \frac{10}{2}= 5cm = 0.05m$

so absolute bending stress = $\sigma = \frac{My}{I} =\frac{52.89\times 0.05}{5.0\times 10^{-6}} = 528900 KN/m^{2}$