# Determine the reactions on supports A and C and write the functions of the internal forces for the beam sections AB

The figure shows a metallic structure composed of two beams $$AB$$ and $$BC$$ connected to each other by a pin in $$B$$, the structure being supported in A and C through fixed supports. In the AB beam, a constant distributed load $$w= 300 N /m$$ and a torque counterclockwise $$M_D = 600 Nm$$ are applied. Disregard the proper weight of the structure

Determine the reactions on supports $$A$$ and $$C$$ and write the functions of the internal forces for the beam sections $$AB$$

I've done this free body diagramm

Then i got $$R(Ax)=0$$ and $$R(Ay)=0$$and $$R(Cy)=800N$$, $$R(Ay)=1000N$$,$$F=-800N$$

But in the solutions they said that $$R(Ax)=-1920N$$and that$$R(Cx)=1920 N$$ How to i get to this results?

For the writing the internal forces for the beam i've calculated correctly for $$0 but for $$3 i am not getting the correct result (the correct result $$V(x) = -300 x+ 1000$$ ; $$M(x) = -150 x^2 +1000 x -600$$)

However i got

and i know that this is wrong because $$d/dx M(x)$$ is not igual to $$V(x)$$

Could someone help me?

Edit: i notice that i forget to draw a Force in X for B

UPDATE: i already understand why my body diagramm for 3<x<6 and why my expression for V(x) and M(x) are wrong (i forgot to draw the moment and i should't have drawn the 800 force and the (x-3) is also incorrect)

• Homework.......? Jan 3 at 18:58

This is a statically determinate structure (four reactions minus one release equals three static equilibrium equations).

So we should be able to do this using simple statics:

\begin{align} \sum F_x &= A_x + C_x = 0 \\ \therefore A_x &= -C_x \\ \sum F_y &= A_y + C_y - 300\cdot6 = 0 \\ \therefore C_y &= 300\cdot6 - A_y \\ \sum M_A &= -300\cdot6\cdot\dfrac{6}{2} + 600 + 2.5C_x = 0 \\ \therefore C_x &= \dfrac{300\cdot6\cdot\dfrac{6}{2} - 600}{2.5} = 1920\text{ N} \\ \therefore A_x &= -1920\text{ N} \end{align}

But how can we determine $$A_y$$ and $$C_y$$? Well, thankfully we have the hinge at B, which we can use to ensure that the bending moment to the left of B is zero:

\begin{align} \sum M_B^- &= -6A_y + 300\cdot6\cdot\dfrac{6}{2} + 600 = 0 \\ \therefore A_y &= \dfrac{300\cdot6\cdot\dfrac{6}{2} + 600}{6} = 1000\text{ N} \\ \therefore C_y &= 300\cdot6 - 1000 = 800\text{ N} \end{align}

As for the internal force equations, it's worth noting that AB has no concentrated forces (just a bending moment). We therefore know that the shear diagram will be continuous and therefore we can have one shear equation for all of $$x\in[0, 6]$$.

Since shear is the integral of the distributed load, we can easily get that

\begin{align} q(x) &= -300 \\ V(x) &= \int q(x) = -300x + C \\ V(0) &= C = 1000 \\ \therefore V(x) &= -300x + 1000 \end{align}

And we also know that the bending moment is the integral of the shear diagram, so we get:

\begin{align} V(x) &= -300x + 1000 \\ M(x) &= \int V(x) = -150x^2 + 1000x + C \end{align}

But here we must remember that the bending moment diagram will have a discontinuity due to the concentrated moment at D. So:

\begin{align} M(x) &= -150x^2 + 1000x + C \\ M(0) &= C_{[0,3)} = 0 \\ M(6) &= -5400 + 6000 + C_{(3,6]} = 0 \\ \therefore C_{(3,6]} &= -600 \\ \therefore M(x) &= \begin{cases} -150x^2 + 1000x &x\in[0, 3) \\ -150x^2 + 1000x - 600 &x\in(3, 6] \\ \end{cases} \end{align}