Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a10 mm diameter, determine the maximum value of the average normal stress in the link BD if θ = 15 degrees.

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Question

The cable is subjected to a uniform loading of w = 60 kN/m. Determine the maximum and minimum tension in cable.

Solution

Step 1 Solution:-

Free Body Diagram

 

Step 2 Solution:-

$$\begin{gathered} y=\frac{1}{F_H}\int \left(\int wdx\right)dx \\ Integrating we get, \\ y=\frac{1}{F_H}\int wdxdx+C_1 \\ Again by integrating we get, \\ y=\frac{1}{F_H}\left(\frac{w{x}^2}{2}+C_{1}x+C_2\right)---equation 1 \\ U\sin g the boundary conditions, \\ x=0 and \frac{dy}{dx}=0 \\ {C}_1=0 \\ Now, \\ Substitutig the value of C_{1}and C_2 into the 1 st equation we get, \\ y=\frac{1}{F_H}\left(\frac{w{x}^2}{2}\right)---equation 2 \end{gathered}$$

 

Step 3 Solution:-

$$\begin{gathered} The above equation is parabolic. \\ Now, \\ x=50 m , y = 12m \\ Now, \\ from equation 2 \\ 12=\frac{1}{F_H}\left(\frac{60*{50}^2}{2}\right) \\ {F}_H=6250kN \\ now, \\ x=50m and \theta =\theta max \\ Hence, \\ {\overline{)\frac{dy}{dx}}}_{x=30m}=\tan \theta max---equation 3 \\ Now, \\ from equation 2 \\ \frac{dy}{dx}=\frac{wx}{F_H} \\ {\overline{)\frac{dy}{dx}}}_{x=30m}=\frac{60*50}{6250} \\ {\overline{)\frac{dy}{dx}}}_{x=30m}=0.48 \\ Now, from equation 3, \\ 0.48=\tan {\theta }_{max} \\ {\theta }_{max}={\tan }^{-1}0.48 \\ {\theta }_{max}=25.641 degree \end{gathered}$$

 

 

Step 4 Solution:-

$$\begin{gathered} F_H=T_{max}\cos {\theta }_{max} \\ 6250=T_{max}*\cos 25.641 \\ {T}_{max}=6932.71kN \\ T=6.93MN \\ Maximum tension in the cable is T_{max}=6.93MN \\ Minimum tension in cable exists when \theta =0^o \\ Now, \\ {F}_H=T_{min}\cos 0 \\ {T}_{min}=F_H \\ {T}_{min}=6250kN \\ {T}_{min}=6.250MN \end{gathered}$$

Step 5 Answer:-

T_max=6932.71kN

T_min=6.250MN