A beam is of T section, as shown in fig. The beam is simply supported over a span of 4m and carries a UDL of (1.7+Your R...

Given,

UDL w = (1.7 + Roll number ) kN/m

L = 4 m

Support = Simply supported

Centroid (NA) ,

$$\begin{gathered} \overline{x}=0 \\ \overline{y}=\frac{y_1{A}_1+y_2{A}_2}{A_1+A_2} \\ \overline{y}=\frac{175\left(7500\right)+75\left(7500\right)}{7500+7500} \\ \overline{y}=125 mm \end{gathered}$$

Moment of Inertia,

$$\begin{gathered} I=I_1+I_2 \\ I=\frac{150{\left(50\right)}^3}{12}+\left(150\right)\left(50\right){(175-75)}^2]+[\frac{50{\left(150\right)}^3}{12}+\left(150\right)\left(50\right){(125-75)}^2] \\ I=1562500+7.5\times {10}^7+1.40625\times {10}^7+1.875\times {10}^7 \\ I=1.09375\times {10}^8 m{m}^4 \end{gathered}$$

Maximum Moment,

Suppose your roll number is 1 .

$$\begin{gathered} w=1.7+1 \\ w=2.7 kN/m \\ {M}_{max}=\frac{\left(2700\right){\left(4\right)}^2}{8} \\ {M}_{max}=5400 Nm \\ {M}_{max}=5400\times {10}^3 Nmm \end{gathered}$$

Maximum Compressive Stress and Tensile stress is at Top fiber and bottom fiber respectively.

$$\begin{gathered} {\sigma }_C=\frac{M_{max}(200-125)}{I} \\ {\sigma }_C=\frac{5400\times {10}^3(200-125)}{1.09375\times {10}^8} \\ {\sigma }_C=0.049(200-125) \\ {\mathit{\sigma }}_{\mathbf{C}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{675}\mathbf{ }\mathit{M}\mathit{P}\mathit{a}\mathbf{ }\mathbf{ }\mathbf{\left(}\mathit{M}\mathit{a}\mathit{n}\mathit{g}\mathit{i}\mathit{t}\mathit{u}\mathit{d}\mathit{e}\mathbf{\right)} \\ {\sigma }_T=\frac{5400000\left(125\right)}{1.09375\times {10}^8} \\ {\mathit{\sigma }}_{\mathbf{T}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{171}\mathbf{ }\mathit{M}\mathit{P}\mathit{a} \end{gathered}$$