A simply supported beam with overhang is loaded as shown in fig. Calculate maximum tensile and compressive stress due to bending.

1. Calculation of maximum bending moment:

Let $V_B$ and $V_D$ be the support reaction at B and D.

$\sum F_y=0$

$V_B$ + $V_D$=$20*1+25=45$

$\sum M_D=0$

$V_B*(1.8+1.8)-20*1*(\frac{1}{2}+1.8+1.8)-25*1.8=0$

$V_B=35.278KN$

$V_D=45-35.278=9.72KN$

S.F analysis:

1. SF at A=0

2. SF at B but just on LHS of B=-20*1=-20KN

3. SF at B but just on RHS of B=-20*1+35.278=15.278KN

4. SF at C from A but just on RHS of $C=-20*1+35.278-25=-9.722KN$

5. SFat D=9.722KN

BM analysis:

1. BM at A=0, BM at D=0

2. BM at B$=-20*1*\frac{1}{2}$=-10KNm

3. BM at C =9.72*1.8=17.49KNm

Maximum bending moment=$17.5KNm=17.5*10^6 Nmm$ ( Sagging )

$\bar y_t=\frac{\sum ay}{A}=\frac{350*10^3}{800}=43.75mm$

Moment of inertia of section about axis (1)-(1)

$I_{1-1}=\sum I_{self}+\sum_{ay^2}$

=$7116.67*10^3+35.15*10^6=42.267*10^6 mm^4$

$I_{x-x}=I_{1-1}-A y_{-2}=42.267*10^6-8000*43.75^2$

$I_{x-x}=26.95*10^6 mm^4$

Let, $F_c and F_t$ be the maximum compressive and tensile stress due to bending.

$(\frac{M}{I}=\frac{f}{y}$

$f=\frac{M}{I}*y$

Maximum tensile stress=$\frac{17.5*10^6}{26.95*10^6}*43.75=28.40 N/mm^2$

Maximum compressive stress=$\frac{17.5*10^6}{26.95*10^6}*(230-\bar y_t)$

$\frac{17.5*10^6}{26.95*10^6}*(230-43.75)$

$=120.94N/mm^2$

Note :

• If Maximum bending moment is sagging then part of the section above the neutral axis i.e x-x will be in compression and below the neutral axis it will be in tension.

• If Maximum bending moment is hogging then part of the section above the neutral axis i.e x-x will be in tension and below the neutral axis it will be in compression.