I was solving problems from the Meriam and Kraige Engineering Mechanics Statics vol-1 and I couldn't get the correct way through this problem.

The question is to find out the force in member $DK$ of the loaded overhead sign truss.

My attempt

I tried eliminating member NM,FG,GT,HI.i tried using the method of section but any section passing through DK is of more than 3 members.

Please suggest how to approach this problem.Thanks.

This is a K-Truss. I will try to provide a solution.

Corrections: There should be a cyan member in the truss, otherwise, it is unstable.

As shown above, the system will be cut by curved line (as highlighted by the red line). This is a special case to the method of sections wherein the rule allows cutting of more than 3 members provided that all member except one are collinear at a point. Separating the left side of the system gives:

The reactions have been solved by static equations. In the above image, we only need the horizontal unknowns $LK$ and $CD$, thus:

Summing up moments at C yields:

$$15\ (8)=LK\ (5)$$ $$LK\ =\ 24\ kN\ (Tension)$$

I will leave the sign as is so as to simplify the analysis of sections.

Next is summing up moments at L yields:

$$15\ (8)\ -\ CD\ (5) = 0$$ $$CD\ =\ -24\ kN$$

Note that we do not need to solve the values of member BP and MP as they are not needed in the solution.

Next, we cut another section from the system as shown below:

and separating the left section:

Note that members $CD$ and $LK$ has been solved as denoted in yellow color. Two equations and two unknowns will be needed, thus, summing up forces along x-axis and y-axis yields the following equations respectively:

$$QD\ [4/sqrt(22.25)]\ +\ QK\ [4/sqrt(22.25)]\ +\ CD\ +\ LK\ =\ 0$$ $$QD\ [2.5/sqrt(22.25_]\ -\ QK\ [2.5/sqrt(22.25)]\ -\ 5\ +\ 15\ =\ 0$$ $$QD\ =\ -9.434\ kN$$ $$QK\ =\ +9.434\ kN$$

Note that we cannot still use method of joints at joint $D$, therefore, we will continue to solve for more members of the system as:

Separating the right sections:

Summing up moments at J yields:

$$20\ (8)\ +\ ED\ (5)\ =\ 0$$ $$ED\ = \ -32\ kN$$

Now we can use method of joints at joint $D$ because there are only two unknowns left as:

Summing up forces along x-axis yields:

$$-DC\ -\ DQ\ [4/sqrt(22.25)]\ +\ DE\ +\ DR\ [(4/sqrt(22.25)]\ = \ 0$$ $$DR\ = \ 0$$

Summing up forces in the y-axis yields:

$$-DQ\ [2.5/sqrt(22.25)]\ +\ -DK\ =\ 0$$ $$DK\ =\ 5\ kN$$

• Thank you for such a nice explanation.it cleared my concepts a lot. Sep 22 '17 at 2:54

At first glance it’s a fairly daunting problem, but it’s actually easily solved using K-truss analysis.

1. Solve for the external reactions by summing the moments about Point U, and summing vertical forces.

2. Use K-truss analysis to determine the net vertical force in diagonals QD/QJ and RD/RK. This is done by making vertical section cuts and then summing the vertical forces. We need only consider the vertical component of the force in the diagonals.

3. Distribute that equivalent vertical force we calculated in step 2 between the diagonals. (Basically we look at another free body diagram and consider vertical equilibrium again.) Because the diagonals are at the same angle, the equivalent vertical force distributes equally.

4. Use method of joints at Joint K to solve for the force in element DK. (5kN in tension)

• Thanks a lot .I was unaware of K truss analysis and now am able to understand it.it simplified the problem a lot. Sep 22 '17 at 3:02