A cast iron beam is of T section as shown in fig. Sketch the bending distribution diagram at point C.

1. Calculation of maximum bending moment :

Let the support reaction at A and B be $V_a$ and $V_B$

$\sum f_y=0$ (+ve)

$V_A+V_B=1800*8=14400$

$\sum M_B=0$ (+ve)

$V_A*8-1800*8*\frac{4}{2}=0$

$V_A=7200N$

$V_B=14000-7200=7200N$

SF analysis:

SF at A=7200N

AF at any section in AB distant x from A

SF=7200-1800*x

At x=8m, SF at B=7200-1800*8=-7200N

Section f zero shear:7200-1800*x=0

x=4m

Maximum bending moment=$7200*4-1800*4*\frac{4}{2}=14400Nm (sagging)=14400*10^3 Nmm$

2. Calculation of $\bar y_t$ , $\bar y_b$ and $I_{NA}$ : -

Now, $\bar y_t=\frac{\sum_ay}{A}=\frac{212000}{4400}=48.18 mm$

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

$I_{1-1}=\sum I_{self}+\sum_{ay^2}=2.946*10^6+15.56*10^6$

$I_{1-1}=18.506*10^6 mm^4$

$I_{N-A}=I_{1-1}- Ay_t^2=18.506*10^6-4400*48.18^2$

$I_NA=8.29*10^6 mm^4$

$\bar y_b=(20+120)-\bar y_t=140-48.18=91.82mm$

3. Calculation of bending stress:

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

Let $f_t$ and $f_c$ be the bending stress in tensile and compression zone i.e

$f_t=\frac{M}{I_{NA}}*\bar y_t$

$f_t=\frac{14400*10^3}{8.29*10^6}*48.18$

$f_t=83.69 n/mm^2$

$f_c=\frac{M}{I_{NA}}*\bar y_b$

$f_t=\frac{14400*10^3}{8.29*10^6}*91.82=159.49N/mm^2$