We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program


Advertise your business, and reach millions of students around the world.


All the data tables that you may search for.


For Arabic Users, find a teacher/tutor in your City or country in the Middle East.


Find the Source, Textbook, Solution Manual that you are looking for in 1 click.


Need Help? We got you covered.

Chapter 4

Q. 4.5

Calculate the forces in the members CD, CF and EF in the Pratt truss shown in Fig. 4.18.


Verified Solution

Initially the support reactions are calculated and are readily shown to be

R_{ A , V }=4.5 kN \quad R_{ A , H }=2 kN \quad R_{ B }=5.5 kN

We now ‘cut’ the members CD, CF and EF by a section mm, thereby dividing the truss into two separate parts. Consider the left-hand part shown in Fig. 4.19 (equally we could consider the right-hand part). Clearly, if we actually cut the members CD, CF and EF, both the left- and right-hand parts would collapse. However, the equilibrium of the left-hand part, say, could be maintained by applying the forces F_{ CD }, F_{ CF } \text { and } F_{ EF } to the cut ends of the members. Therefore, in Fig. 4.19, the left-hand part of the truss is in equilibrium under the action of the externally applied loads, the support reactions and the forces F_{ CD }, F_{ CF } \text { and } F_{ EF } which are, as in the method of joints, initially assumed to be tensile; Eq. (2.10) are then used to calculate the three unknown forces.
Resolving vertically gives

\sum F_{x}=0 \quad \sum F_{y}=0 \quad \sum M_{z}=0                                          (2.10)


F_{ CF } \cos 45^{\circ}+4-4.5=0                                      (i)

so that

F_{ CF }=+0.71 kN

and is tensile.

Now taking moments about the point of intersection of F_{ CF } \text { and } F_{ EF } we have

F_{ CD } \times 1+2 \times 1+4.5 \times 4-4 \times 1=0                                              (ii)

so that

F_{ CD }=-16 kN

and is compressive.
Finally F_{EF} is obtained by taking moments about C, thereby eliminating F_{ CF } \text { and } F_{ CD } from the equation. Alternatively, we could resolve forces horizontally since F_{ CF } \text { and } F_{ CD } are now known; however, this approach would involve a slightly lengthier calculation. Hence

F_{ EF } \times 1-4.5 \times 3-2 \times 1=0                                      (iii)

which gives

F_{ EF }=+15.5 kN

he positive sign indicating tension.
Note that Eqs (i)–(iii) each include just one of the unknown member forces so that it is immaterial which is calculated first. In some problems, however, a preliminary examination is worthwhile to determine the optimum order of solution.
In Ex. 4.5 we see that there are just three possible equations of equilibrium so that we cannot solve for more than three unknown forces. It follows that a section such as mm which must divide the frame into two separate parts must also not cut through more than three members in which the forces are unknown. For example, if we wished to determine the forces in CD, DF, FG and FH we would first calculate F_{ CD } using the section mm as above and then determine F_{ DF }, F_{ FG } \text { and } F_{ FH } using the section nn. Actually, in this particular example F_{ DF } may be seen to be zero by inspection (see Section 4.5) but the principle holds.