Suspension 120 m: BMD and BM at 25, 40, 80 m

Single three-hinged stiffening girder: $L$ = 120 m, $d$ = 12 m, loads 240 kN at 25 m and 300 kN at 80 m from the left end A.

Cable pull $H$

$$\begin{aligned} R_A &= \frac{240(95)+300(40)}{120} = 290.00\ \text{kN} \\ M_{0C} &= 290.00(60) - 240(35) = 9000.0\ \text{kN m} \\ H &= \frac{M_{0C}}{d} = \frac{9000}{12} = 750.00\ \text{kN} \end{aligned}$$

Bending moment in the girder

$M = M_0 - H\,y$ with $y = \dfrac{4d\,x(L-x)}{L^2} = \dfrac{4(12)x(120-x)}{14400}$.

$M = 0$ at both supports and at the central hinge C (60 m).

In each region, $M$ is a quadratic in $x$:

• $0 \le x \le 25$: $M = 2.5x^2 - 10x$
• $25 \le x \le 80$: $M = 2.5x^2 - 250x + 6000$ (zero at 40 m and 60 m, minimum $-250$ kN m at 50 m)
• $80 \le x \le 120$: $M = 2.5x^2 - 550x + 30000$ (zero at 100 m and 120 m, minimum $-250$ kN m at 110 m)

BMD (kN m): sagging above the line, hogging below
                                *

                               * *
          *                   *   *
         * *                 *     *
        *
       *    *               *       *
      *      *             *         *
    **        **         **           **
****------------**-----**---------------**-----**
                  *****                   *****
(top = 2000.0, bottom = -250.0; '-' is zero line)

The BMD is sagging from A to 40 m, with a peak of 1312.5 kN m under the 240 kN load (apart from a negligible hogging of about -10 kN m within the first 4 m). It is hogging from 40 m to 60 m (minimum -250 kN m at 50 m, just left of hinge C), sagging from 60 m to 100 m with a peak of 2000 kN m under the 300 kN load, and hogging again from 100 m to B (minimum -250 kN m at 110 m).

Answer: $H = 750.00$ kN; $M_{25} = 1312.50$ kN m (sagging), $M_{40} = 0.00$ kN m, $M_{80} = 2000.00$ kN m (sagging, positive).