How do I work out the Bending Moment here? I've managed to solve the reactions in each $2$ supports, and SF and BM from the $1$st support $($left side$).$ But I can't seem to get the BM$=-74\mathrm{.}4$kNm $($or anywhere near it$),$ I$\text{'}$m not sure why...

Activity $10\mathrm{.}3$c: Calculate the bending moment

Use the example of the rail vehicle from the tutorial clip in Activity $10\mathrm{.}3b$ but with the

loading arrangement given in the figure below.

The rail vehicle is modelled as being $20m$ long with vertical point loads downward,

$P_1$ to $P_{10},$ and a UDL of $20kN{m}^{-1}.$

The simple supports are at $x=3\mathrm{.}2m$ and $x=17\mathrm{.}4m$ from the left$-$hand end.

The loads $P_1$ to $P_{10}$ act downwards with $P_1$ at $x=1\mathrm{.}0m$ and $P_2$ to $P_{10}$ at $2\mathrm{.}0m$

intervals so that $P_{10}$ is at $x=19\mathrm{.}0m.$

The UDL $$ is $20kN{m}^{-1}$

$P_1$ and $P_2$ are $5kN$

$P_3$ and $P_4$ are $10kN$

$P_5$ and $P_6$ are $10kN$

$P_7$ and $P_8$ are $20kN$

$P_9$ and $P_{10}$ are $5kN$

$P_1$ and $P_{10}$ act at $1\mathrm{.}0m$ from either end of the vehicle and the other point

loads act at intervals of $2\mathrm{.}0m$ between $P_1$ and $P_{10}.$

Show that the bending moment at the centre of the vehicle is $564kNm$ and that the

maximum bending moment is $570kNm$ to the nearest $10kNm.$

The shear force will increase linearly with the distributed load and vertically up with

each point load and then decrease at each support.

The bending moment will change as a parabola before changing direction at the

support, reaching a maximum near the centre and finally changing direction again

at the next support.

Step $3$: Draw the SF and BM diagrams

Take a piece of squared graph paper and draw a scale version of the rail vehicle.

Add the axes for both the SF and BM diagrams. Calculate the shear forces and

bending moments at the key points of the supports.

Starting with the vehicle up to the left$-$hand support, the only forces acting on this

part of the beam are two point loads and $\{$:$P_2)$ and the UDL of that portion of the

vehicle as shown below.

$S=5+5+L=10+3\mathrm{.}220=74kN$

$M=-L\frac{L}{2}-5(L-1)-5(L-3)$

$=-20\frac{{3\mathrm{.}2}^2}{2}-52\mathrm{.}2-50\mathrm{.}2=-114\mathrm{.}4kNm$

Now, repeating the step directly above to calculate $S$ and $M$ at the right$-$hand

support, the only forces acting on this part of the beam are ten point loads to

P$1$ to $P_9$ and the UDL of that portion of the vehicle gives

$S=188kN$

and

$M=-74\mathrm{.}4$kNmActivity $10\mathrm{.}3$c: Calculate the bending moment

Use the example of the rail vehicle from the tutorial clip in Activity $10\mathrm{.}3b$ but with the

loading arrangement given in the figure below.

The rail vehicle is modelled as being $20m$ long with vertical point loads downward,

$P_1$ to $P_{10},$ and a UDL of $20kN{m}^{-1}.$

The simple supports are at $x=3\mathrm{.}2m$ and $x=17\mathrm{.}4m$ from the left$-$hand end.

The loads $P_1$ to $P_{10}$ act downwards with $P_1$ at $x=1\mathrm{.}0m$ and $P_2$ to $P_{10}$ at $2\mathrm{.}0m$

intervals so that $P_{10}$ is at $x=19\mathrm{.}0m.$

The UDL $$ is $20kN{m}^{-1}$