Please don't post this question on google D W L Problem 2 264.0mm 27.8kN/m 7.70m Determine the Maximum Shear and Bending...

Given data:

Circular cross section:

$$\begin{gathered} D=264 mm \\ L=7.7 m \\ w=27.8 kN/m \end{gathered}$$

wt = w+ weight of beam.

unit weight of beam = 5.6 kN/m³.

Determining propped reaction Using consistence deformation method:

assuming reaction at B is redundant.

Deflection at B.

${\left({\delta }_B\right)}_p=\frac{w_t\times L^4}{8EI}$

Redundant beam:

Deflection at B.

${\left({\delta }_B\right)}_R=\frac{R_B\times L^3}{3EI}$

Compatibility equations:

$$\begin{gathered} {\left({\delta }_B\right)}_p={\left({\delta }_B\right)}_R \\ \frac{w_t\times L^4}{8EI}=\frac{R_B\times L^3}{3EI} \\ {\mathit{R}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{3}{\mathbf{w}}_{\mathbf{t}}\mathbf{L}}{\mathbf{8}} \end{gathered}$$

Determining total load on the beam:

$$\begin{gathered} w_t=w+w_b \\ A=\frac{\pi }{4}\times {\left(0.264\right)}^2=0.0547 m^2 \\ {w}_b=5.6\times A kN/m \\ {w}_b=5.6\times 0.547=0.306 kN/m \\ {w}_t=27.8+0.306=28.106 kN/m \end{gathered}$$

Beam with loading:

$$\begin{gathered} R_{\mathrm{B}}=\frac{3{\mathrm{w}}_{\mathrm{t}}\mathrm{L}}{8} \\ {R}_{\mathrm{B}}=\frac{3\times 28.106\times 7.7}{8}=81.156 kN \end{gathered}$$

Maximum shear and bending moment:

$$\begin{gathered} V_X=-81.156+28.106X \\ {V}_A=135.26 kN, V_B=-81.156 kN \\ {M}_X=81.156X-28.106\frac{X^2}{2} \\ {M}_B=0, M_A=-208.3 kN/m \end{gathered}$$

Zero shear location:

$$\begin{gathered} 0=-81.156+28.106X \\ X=2.887 m \left(from B\right) \end{gathered}$$

Maximum positive moment:

$$\begin{gathered} M_{max}=81.156\times 2.887-28.106\frac{2.{887}^2}{2} \\ {M}_{max}=117.17 kN.m \end{gathered}$$

Overall maximum shear force and bending moment:

$$\begin{gathered} V_{max}=135.26 kN \\ {M}_{max}=208.3 kN.m \end{gathered}$$

occur at support.

Moment of inertia for circular section:

$$\begin{gathered} I=\frac{\pi d^4}{64} \\ I=\frac{\pi \times {264}^4}{64}=238443564.9 m{m}^4 \end{gathered}$$

Maximum bending stress:

Bending equation:

$$\begin{gathered} \frac{M}{I}=\frac{f}{y} \\ {y}_{max}=\frac{d}{2}=\frac{264}{2}=132 mm \\ {f}_{max}=\frac{M_{max}}{I}{y}_{max} \\ {f}_{max}=\frac{208.3\times {10}^6}{238443564.9}\times 132 N/m{m}^2 \\ {f}_{max}=115.313 N/m{m}^2 \\ {\mathit{f}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{115}\mathbf{.}\mathbf{313}\mathbf{ }\mathit{M}\mathit{P}\mathit{a} \end{gathered}$$

Maximum shear stress:

$$\begin{gathered} {\tau }_{max}=\frac{4}{3}{\tau }_{avg} \\ {\tau }_{avg}=\frac{V_{max}}{A} \\ {\tau }_{avg}=\frac{135.26\times {10}^3}{{\displaystyle \frac{\pi }{4}}\times {264}^2}=2.47 N/m{m}^2 \\ {\tau }_{max}=\frac{4}{3}\times 2.47=3.29 N/m{m}^2 \\ {\mathit{\tau }}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{29}\mathbf{ }\mathit{M}\mathit{P}\mathit{a} \end{gathered}$$