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 
BUSINESS MANAGER

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.

Chapter 4

Q. 4.4

A motorway warning display is supported symmetrically by two identical pin-jointed frames as shown in Fig. P.4.4. The frames must be designed to withstand a wind load of 4 kN / m ^{2} . Calculate the forces in the members of a frame and check your solution using a graphical method. Neglect the weight of the display.

Step-by-Step

Verified Solution

In Fig. S.4.4(a), the length of BC is obtained by similar triangles.

i.e.

\frac{ BC }{2.5}=\frac{3}{5.25}

which gives

BC =1.43 m

Further

\alpha=\tan ^{-1} \frac{3}{5.25}=29.7^{\circ}

and

\beta=\tan ^{-1} \frac{1.43}{2.75}=27.5^{\circ}

The total wind load on the display is 4 × 5 × 4 = 80 kN. Therefore, the horizontal forces at A and B are 20 kN.
Resolving forces horizontally at B

F_{ BC }+20=0

so that

F_{ BC }=-20 kN \text { (compression) }

Resolving forces horizontally at A

F_{ AC } \sin 29.7^{\circ}+20=0

from which

F_{ AC }=-40.4 kN \text { (compression) }

Resolving forces vertically at A

F_{ AB }+F_{ AC } \cos 29.7^{\circ}=0

which gives

F_{ AB }=35.1 kN \text { (tension) }

Resolving forces perpendicular to AC at C

F_{ CD } \cos 32.8^{\circ}+F_{ AC } \cos 29.7^{\circ}=0

so that

F_{ CD }=20.7 kN \text { (tension) }

Resolving forces parallel to CE at C

F_{ CE }-F_{ CA }-F_{ CB } \cos 60.3^{\circ}+F_{ CD } \cos 57.2^{\circ}=0

which gives

F_{ CE }=-61.5 kN \text { (compression) }

Finally, resolving forces horizontally at E

F_{ ED }+F_{ CE } \cos 60.3^{\circ}=0

giving

F_{ ED }=30.5 kN \text { (tension) }

The graphical solution is obtained by first numbering the spaces between the forces as shown in Fig. S.4.4(a) and then completed as shown in Fig. S.4.4(b).