The three hinged stiffening girder of a suspension bridge of span 120 m is subjected to two point loads of 250 kN and 300 kN at a distance of 25 m and 80 m from the left end. Find the shear force and bending moment for the girder at a distance of 40 m from the left end. The supporting cables has a central dip of 12 m. Find also the maximum tension in the cable and draw the B.M.D. for the girder.

(1)To find $V_A$ & $V_B$:

Considering the stiffening girder asasis beam supporting the given external loud system.

$\sum M_A=0$

$250\times 25+300\times 80-V_B\times 120=0$

$V_B=252kN(\uparrow)$

$\sum F_y=0(\uparrow +ve)$

$V_A-250-300+252=0$

$V_A=298kN(\rightarrow)$

(2) To find: H

B.M at centre of girder

$B.M_C=298\times 60-250\times 35=9130kN.m$

$B.M_D=298\times 25=7450kN.m$

$B.M_E=252\times 40=10080kN.m$

Horizontal sere-action at each end of the cable

$H=\frac{M_C}{h}=\frac{9130}{12}=760.83kN$

(3)Find equivalent Udl(we):

$H=\frac{Wel^2}{8h}$

$760.83=\frac{We\times (120)^2}{8\times 12}=5.07kN/m$

$\therefore V_{A_1}=V_{B_1}=\frac{WeL}{2}=\frac{5.07\times 120}{2}=304.22kN$

(4)$T_{max}:$

$T_max=\sqrt{V_{A_1}^2+H^2}$

$=\sqrt{(304.33)^2+(760.83)^2}$

$T_max=819.44kN$

(5)S.F calculation:

For the girder,

S.F. at any section x,

$S.F_x=\text{Beam shear} - H\tan \theta$

where,

$\tan \theta=\frac{dy}{dx}=\frac{4h}{l^2}(l-2x)$

At 40m from left end:

At x=40 m,

$\tan \theta=\frac{4\times 12}{120^2}(120-2\times 40)=\frac{2}{15}$

$\therefore Beam shear=298-250=48kN$

$\therefore S.F_x=48-760.83\times \frac{2}{15}=-53.44kN$

(6) Bending Moment Calculation:

B.M at any section is given by

$B.M._x=\text{Beam moment}-H_y$

At x=40m,

$y=\frac{4h}{l^2}(lx-x^2)=\frac{32}{3}m$

Beam moment at 40m from left end

$B.M=298\times 40-250\times 15=8170kNm$

$B.M=8170-760.83\times \frac{32}{3}$

$B.M=54.48kNm$