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Solution:-\rightarrow According to NSCP code for timber(a) maximum bending strees of YAKAL =24.50 \mathrm{MPa} with 80 \% selenderness carrection factor \rightarrow allowubl bending strers =24.80 \times 0.8=19.84 \mathrm{MPa}(b) maximum compressive load =15.8 \mathrm{MPa}allowable compressive load =15.8 \times 0.8=12.64 \mathrm{MPa}(C) check for compressive stress\rightarrow \text { compressive load }=4 \text { cour }\rightarrow comprerive spres = Compressive loadArea of crossection\begin{aligned}\rightarrow \text { compressive stress } & =\frac{400 \times 1000}{200 \times 350}=5.7 \pm \mathrm{N} / \mathrm{mm}^{2} \\& =5.71 \mathrm{MPa}<12.64 \mathrm{MPa}\end{aligned}\rightarrow check for Bending compressivestress\rightarrow Bending moment diagram of beam\begin{array}{l}\rightarrow \text { maximum the Bendingmoment }=\frac{\omega l^{2}}{24}=\frac{26 \times 4^{2}}{24} \\=17.33 \mathrm{kN}-\mathrm{m} \\\end{array}\rightarrow maximum - ve Bending moment =\frac{\omega l^{2}}{12}=\frac{26 \times 4^{2}}{12}=34 \cdot 67 \mathrm{kN}-\mathrm{m}\rightarrow maximum Bending moment on the beam =34.67 \mathrm{ur}-\mathrm{m}\rightarrow So maximum Bending stress \left(f_{b}\right)\begin{array}{l}f_{\text {bmax }}=\frac{B M_{\text {max }}}{I} \times y_{\text {amax }} \\f_{\text {max }}=\frac{39.67 \times 10^{6} \mathrm{~N}-\mathrm{mm}}{200 \times \frac{350^{3}}{12}} \frac{350}{2} \mathrm{~mm} \\f_{\text {bomax }}=8.99 \mathrm{mPa}\end{array}\sum_{\frac{m}{I} x y}^{200}but allowable bending strees is 19.84 \mathrm{MPa}\rightarrow Hence beam is safe in both compresion as well as in Bending: ...