VECTOR GEOMETRY – TRUSS ANALYSIS

Consider a member (bar) with a pin joint at each end as shown. A pin joint cannot rotate (torque) from one o another so each can only push or pull on the joint along the direction of its length.  Remember, the force in the other end of each member also pushes or pulls and so acts in the opposite direction with equal force.

A member in tension is called a TIE and is shown with arrows pointing inwards at each end.  A member in compression is called a STRUT and is shown with arrows pointing outwards at each end.

BOW’S NOTATION

When several members are pinned together and the joint is in total equilibrium (not moving), the resultant force must be zero. This means that if we add up all the forces as vectors, they must form a closed polygon. If one or even two of these forces is unknown, then it must be the vector, which closes the polygon. Consider three members joined by a pin as shown. Only one of these forces is known. Bow’s notation helps us to identify and label each member and draw the Polygon or (in this case) the triangle of forces. The process is as follows.

SPACE DIAGRAM

1. Label the spaces between each member. This is why the diagram is called a

2. Starting at any space, say A, identify each member by moving clockwise around the joint so the first becomes a-b, the next b-c and the last c-a (in this case only).

3. Draw the known vector a-b. We know that the next vector b-c starts at b but we do not know its length. Draw a ‘c’ line from ‘b’ in the direction of member b-c. We know that when all the vectors are added, they must form a closed triangle so c-a must end at ‘a’. Draw a ‘c’ line through ‘a’ in the direction of member c-a. Where the two ‘c’ lines cross must be point ‘c’.

4. Starting at any space, say A, identify each member by moving clockwise around the joint so the first becomes a-b, the next b-c and the last c-a (in this case only).

Draw the known vector a-b. We know that the next vector b-c starts at b but we do not know its length. Draw a ‘c’ line from ‘b’ in the direction of member b-c. We know that when all the vectors are added, they must form a closed triangle so c-a must end at ‘a’. Draw a ‘c’ line through ‘a’ in the direction of member c-a. Where the two ‘c’ lines cross must be point ‘c’

Finally, transfer the arrows back to the space diagram in the same direction as on the triangle of forces (figure 6). If they push onto the pin joint, the member must be in compression and so is a strut. If the arrow pulls on the joint, the member must be in tension and so is a tie

WORKED EXAMPLE No.1

A strut is held vertical as shown by two guy ropes. The maximum allowable compressive force in the strut is 20 kN. Calculate the forces in each rope. Note that ropes can only be in tension and exert a pull. They cannot push

Solution

SOLVING THE FORCES IN PIN JOINTED FRAMES

Let’s now apply our knowledge to unknown forces in latticework frames. Here are some examples of lattice work frames.

• Roof
• Electric
• Bridges

Many of these structures are riveted and not entirely free to rotate at the joint but the theory of pin jointed frames seems to work quite well for them. We will apply Bow’s notation to each joint in turn and so solve the forces in each member. By transferring the direction back from the polygon to the framework diagram, it can be deduced which are struts and which are ties. Knowing this, the force direction is determined at the other end of the member and this is needed to solve the other pin joints.

WORKED EXAMPLE No.2

Solve the forces and the reactions for the frame shown.

SOLUTION

 First draw the space diagram and label the spaces

Next solve the joint with the known force

By scaling or use of trigonometry b-c = 173 N (strut)

c-a = 100 N (strut)

Next solve the other joint

By scaling or trigonometry b-d =R1 = 150 N. c-d = 86.5 N (Tie)

R2 may easily be deduced since the total upwards force is 200N then R2 must be 200-150 = 50N. The solution for the other joint is not really needed but here it is

From this a-d = R2 = 50 N as expected.

When you are proficient at this work, you may find it convenient to draw all the solutions together as one diagram. For this example, we would have