Question

The beam shown (Figure 1) is supported by a pin at A

A and a cable at B

B. Two loads P

PP = 12 kN

kN are applied straight down from the centerline of the bottom face. Determine the state of stress at the point shown (Figure 2) in a section 2 m from the wall. The dimensions are w

ww = 6 cm

cm , h

hh = 10.5 cm

cm , L

LL = 0.8 m

m , a

aa = 1.5 cm

cm , and b

bb = 4 cm

cm .

What is the normal stress at the point due to the axial load in the beam? Let a compressive stress be negative.

Answer 1

$\sum F_h=Fcos30=R_H$

$\sum F_v=Fsin30+R_v=12+12$

$\sum M_B=0 \implies 12*0.8+12*1.6=R_v*2.4 \implies R_v=12$

$F=\frac{24-12}{sin30}=24kN$

$R_H=Fcos30=24cos30=20.78kN$

A. Axial load$=R_H=Fcos30=-20.78kN$

$\sigma_a=\frac{R_H}{A}=\frac{-20.78 *1000}{0.105*0.06}=-3.298*10^6 N/m^2 =-3.298*10^6MPa$