Part A Determine the state of stress at point A on the cross

Part A

Determine the state of stress at point A on the cross section of the beam at section aa. Take P = 27 kN .

Part B

Find A. (Tau_A)

0.5 m 0.5 m 100 mm 10 mm 50 mm 180 mm 10 mm 10 mm Section a-a

Taking Moment about the right support (hinged)

R1x3 – Px2=0

Given P= 27 KN

R1 = 54/3=18 KN

R1+R2 = P= 27KN

R2 = 9KN

Shear Force at section aa = 18 KN

For I section shear stress at distance y from neutral axis is

T= (F/Ib)[B(D^2-d^2)/8+b/2(d^2/4-y^2)]

Where Y is the distance of point A from Neutral line

y=180/2-50=40mm

B=100mm

D=200

d=180

b=10mm

Calculating I of the Beam

I = I1+A1d1^2+ I2d2^2+I3d3^2

1->top segment

2->Web

3->Bottom Segment

I 1=100x10^3/12=8333.33 mm4

I2= 10x180^3/12=4860000mm4

I3=100x10^3/12=8333.33 mm4

I = 8333.33 +100x10x10^2+ 4860000+ 180x10x180^2+8333.33 +100x10x10^2

I = 63658333.3 mm4

Ixb =63658333.3 mm5

B(D^2-d^2)/8+b/2(d^2/4-y^2) = 127500mm3

putting all values

T=18x127500/63658333.3 KN/mm2 = 0.03605184 KN/mm2

T= 36.05184 Mpa is the shear stress at the point A