Question The big end of a connecting rod is subjected to a load of 40 KN. The diameter of the circular part adjacent to the strap is 50 mm. Design the joint assuming the permissible tensile stress in the strap as 30 Mpa and the permissible shear stress in the cotter and jib as 20 Mpa.

Ans:- Given that, P=40 \mathrm{KN}=40 \times 10^{3} \mathrm{~N}Diameter of circularpart, d=50 \mathrm{~mm}Tensile stress, \sigma_{t}=30 \mathrm{MPa}Shear & tress, \tau=20 \mathrm{MPa}Width of strayLet B_{1}= width of the strupThe vidth of strap is generally made equal to the diameter of the adjacent end of the round part of the rod.B_{1}=d=\operatorname{som} mA=12.5 \mathrm{~mm}Thickness of gib = Thickners of cotter =12.5 \mathrm{~mm}2) Thickners of the Strap at the thinnest part t_{1}= Thickners of the strap at the thinnest partConsidering ferilun of stap in tension,\begin{aligned}\operatorname{load} P & =2 B_{1} \times t_{1} \times \sigma_{t} \\t_{1} & =\frac{P}{2 B_{1} \sigma_{t}}=\frac{40 \times 10^{3}}{2 \times 50 \times 10^{-3} \times 30 \times 10} \\t_{1} & =13.33 \mathrm{~mm}\end{aligned}Thickness of the Strap at the cotterLat t_{3}= thickney of the strap of the catterthe thickness of the strep cet the cotter is increased such that the arpa of cross section of the spap at the cotter hole is not less them the area of the strap at the thinnest part.\begin{array}{l}2 t_{3}\left(B_{1}-t\right)=2 t_{1} \times B_{1} \\2 t_{3}(50-12.5)=2 \times 13.3 \times 50 \\t_{3}=17.73 \mathrm{~mm}\end{array}Total width of giband cotterLet B= Total width of gibconsidesing failure of g ib and cotter in donblesheqr\text { lorad; } \begin{aligned}P & =2 B \times 5 \times \tau \Rightarrow B=\frac{P}{2+\tau} \\B & =\frac{40 \times 10^{3}}{2 \times 12.5 \times 10^{-3} \times 20 \times 10^{6}}=80 \mathrm{~mm} \\B & =80 \mathrm{~mm}\end{aligned} ...