Products
Rewards
from HOLOOLY

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

HOLOOLY

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

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

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

HOLOOLY
TEXTBOOKS

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

HOLOOLY
HELP DESK

Need Help? We got you covered.

## 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

$F_{ EF }=+15.5 kN$

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.