# Calculating the reactions and using method of joints

Determine the vertical displacement of joint C of the steel truss. The cross-sectional area of each member is A = 400 mm2 and Est = 200 GPa.

I think I know how to calculate the displacement using the method of virtual work, but I am just having trouble calculating the reactions at the supports and then using the method of joints. Can someone explain why $$DC$$ is $$1$$KN?

$$\sum y = 1 + V_D + V_A = 0$$

$$\sum x = h_D + h_A = 0$$

If I take moment at $$C$$, can I add $$V_D$$ and $$V_A$$ together because they share the same line of action?

As there are 4 unknowns but only 3 equilibrium equations available, this truss is structurally indeterminate to the first degree, thus you need to find another equation to solve the problem.

1. Release one restraint at the supports to make it a simply supported truss.

2. Analyze the truss with the real load, and get the reactions and joint displacements.

3. Apply virtual load on the released restraint, and get the reactions and joint displacements.

4. By closing the gap at the released restraint, proportion the reactions from the vertual load analysis and superimpose the results on the reactions from the real load analysis to get the final reactions.

• Hello r13 Thanks for the help, what does "Release one restraint at the supports to make it a simply supported truss." exactly mean? I don't think I know this method. There is only one real load of $100$KN downwards on $B$. So that is still statically indeterminate. Apr 20 at 4:34
• Also another question that I have is, does the reactions on the virtual system the same as the reactions on the real system? I suspect no? Because different loads are applied? (real vs virtual) Apr 20 at 4:47
• Here is a good example of the method I mentioned. structville.com/2018/01/…. Though your problem may not need to go that route, but I think it is good to know.
– r13
Apr 20 at 16:19

Let's call the length of $$DC=L,$$ then $$AC=(\sqrt{2}/2)*L$$ The deflection on an bar under axial load:

$$\delta = \frac{PL}{EA}$$

assuming small deflections (which is what we have here)

$$\delta_{DC}= \frac{PL}{EA}=\frac{1kNL}{EA}$$

$$\delta AC =-\frac{1.414kN*1.414L}{EA}=-2\delta DC$$

The joint c will move $$1kNL/EA$$ to the right and $$2kNL/EA$$ down 45degrees along AC.

It looks from your picture like the values for $$F_{DC}$$ and $$F_{AC}$$ are given and not derived.

This tells you that the angle $$\bar{AC}$$ makes with the horizontal is $$45^{\circ}$$. This, in turn, allows you to calculate the displacements. Since $$\bar{DC}$$ is in tension, joint C will displace to the right due the tensile load. Since $$\bar{AC}$$ is in compression, joint C will displace down to the left at a $$45^{\circ}$$ angle. Thus the net vertical displacement of the joint will be the vertical component of the displacement along $$\bar{AC}$$.

The tension in the bar $$DC$$ is given: $$1$$ kN.
The tension in the bar $$AC$$ is also given: $$-1.414$$ kN.
Neglecting gravitational forces, there is only one external force.

Free-body diagram of the $$DC$$ bar: Let $$\theta$$ be the angle of the bar $$AC$$ to the horizontal to get \begin{align}D_x&+T\cos(\theta)=0\\ D_y&+T\sin(\theta)-1\text{ kN}=0\end{align} It is given that $$T=1.414\text{ kN}$$ and that $$D_x=-1\text{ kN}$$ (the bar $$DC$$ is in positive tension) so we can find the angle $$\theta$$ from the first equation: $$T\cos(\theta)=-D_x\hspace{0.5cm}\Longrightarrow\hspace{0.5cm}1.414 \cos(\theta)=1.0\hspace{0.5cm}\Longrightarrow\hspace{0.5cm}\theta=45^\circ$$ A free-body diagram of the bar $$AC$$ would show the force $$T$$ at point $$C$$, but directed oppositely of what's shown in this figure. Also, an oppositely directed reaction force would occur at the point $$A$$, so we know that the point $$C$$ moves in the direction of the bar-axis. This means (due to the $$45^\circ$$ angle), that the point $$C$$ moves an equal amount in both the vertical and horizontal direction.

Let $$L$$ be the length $$\overline{AC}$$. Then the absolute displacement of joint $$C$$ is $$\delta_L=\frac{TL}{EA}$$ and the vertical displacement is $$1/\sqrt{2}$$ times that amount.