Graphical method

In this case, the method described in Chapter 3 will be used to analyse statically determinate plane pin-jointed trusses. The method will be described with the aid of worked examples.

Problem 5. Determine the internal forces that occur in the plane pin-jointed truss of Figure 4.8, due to the externally applied vertical load of 3 kN. A B C R1 R2 D 3 kN 30° 60° Figure 4.8

Firstly, we will fill the spaces between the forces with upper case letters of the alphabet, as shown in Figure 4.8. It should be noted that the only reactions are the vertical reactionsR1 andR2; this is because the only externally applied load is the vertical load of 3 kN, and there is no external horizontal load. The capital lettersA, B, C and D can be used to represent the forces between them, providing they are taken in a clockwise direction about each joint. Thus the letters AB represent the vertical load of 3 kN. Now as this load acts vertically downwards, it can be represented by a vector ab, where the magnitude of ab is 3 kN and it points in the direction from a to b. As ab is a vector, it will have a direction as well as a magnitude. Thus ab will point downwards froma tobas the 3 kN load acts downwards.

This method of representing forces is known as Bow’s notation.

To analyse the truss, we must first consider the joint ABD; this is because this joint has only two unknown forces, namely the internal forces in the two members that meet at the jointABD. Neither joints BCD and CAD can be considered first, because each of these joints has more than two unknown forces.

Now the 3 kN load is between the spacesA and B, so that it can be represented by the lower case lettersab, point fromatoband of magnitude 3 kN, as shown in Figure 4.9. a b d 30° 3 kN 60° Figure 4.9

Similarly, the force in the truss between the spaces B and D, namely the vector bd, lies at 60° to the horizontal and the force in the truss between the spaces D and A, namely the vector da, lies at 30° to the horizontal. Thus, in Figure 4.9, if the vectors bdandadare drawn, they will cross at the pointd, where the point d will obviously lie to the left of the vector ab, as shown. Hence, if the vector ab is drawn to scale, the magnitudes of the vectorsbdand da can be measured from the scaled drawing. The direction of the force in the member between the spaces B and D at the joint ABD point upwards because the vector from b to d points upwards. Similarly, the direction of the force in the member between the spaces D and A at the joint ABD is also upwards because the vector fromd toapoints upwards. These directions at the joint ABD are shown in Figure 4.10. Now as the framework is in equilibrium, the internal forces in the membersBD and DA at the joints (2) and (1) respectively, will be equal and opposite to the internal forces at the jointABD; these are shown in Figure 4.10.

A _B C R1 R2 Joint D Joint 1 2 3 kN −2.6 kN −1.55 kN Figure 4.10

Comparing the directions of the arrows in Figure 4.10 with those of Figure 4.3, it can be seen

that the members BD and DA are in compression and are defined as struts. It should also be noted from Figure 4.10, that when a member of the framework, say,BD, is so defined, we are referring to the top joint, because we must always work around a joint in a clockwise manner; thus the arrow at the top of BD points upwards, because in Figure 4.9, the vectorbd points upwards fromb to d. Similarly, if the same member is referred to as DB, then we are referring to the bottom of this member at the joint (2), because we must always work clockwise around a joint. Hence, at joint (2), the arrow points downwards, because the vectordb points downwards fromd tobin Figure 4.9.

To determine the unknown forces in the horizontal member between joints (1) and (2), either of these joints can be considered, as both joints now only have two unknown forces. Let us consider joint (1), i.e. jointADC. Now the vectoradcan be measured from Figure 4.9 and drawn to scale in Figure 4.11.

30° a c d

Figure 4.11

Now the unknown force between the spaces D and C, namely the vector dc is horizontal and the unknown force between the spacesCandA, namely the vectorca is vertical, hence, by drawing to scale and direction, the point c can be found. This is because the point c in Figure 4.11 lies below the pointa and to the right ofd.

In Figure 4.11, the vectorcarepresents the mag- nitude and direction of the unknown reaction R1 and the vector dc represents the magnitude and direction of the force in the horizontal member at joint (1); these forces are shown in Figure 4.12, whereR1=0.82 kN anddc=1.25 kN. A _B C R1=ca= 0.82 kN R2 = ? D 3 kN −2.6 kN −1.55 kN +1.25 kN Figure 4.12

Comparing the directions of the internal forces in the bottom of the horizontal member with Figure 4.3,

it can be seen that this member is in tension and therefore, it is a tie.

The reactionR2can be determined by considering joint (2), i.e. joint BCD, as shown in Figure 4.13, where the vectorbcrepresents the unknown reaction R2 which is measured as 2.18 kN.

b c d

Figure 4.13

The complete vector diagram for the whole frame- work is shown in Figure 4.14, where it can be seen that R₁ +R₂ = 3 kN, as required by the laws of equilibrium. It can also be seen that Figure 4.14 is a combination of the vector diagrams of Figures 4.9, 4.11 and 4.13. Experience will enable this problem to be solved more quickly by producing the vector diagram of Figure 4.14 directly.

a

b c d

Figure 4.14

The table below contains a summary of all the measured forces. Member Force (kN) bd −2.6 da −1.55 cd 1.25 R1 0.82 R2 2.18

Problem 6. Determine the internal forces in the members of the truss of Figure 4.15. due to the externally applied horizontal force of 4 kN at the jointABE.

A B E D _C R₁ R2 H₂ Joint Joint 1 2 4 kN 30° 60° Figure 4.15

In this case, the spaces between the unknown forces areA, B, C, D and E. It should be noted that the reaction at joint (1) is vertical because the joint is on rollers, and that there are two reactions at joint (2) because it is firmly anchored to the ground and there is also a horizontal force of 4 kN which must be balanced by the unknown horizontal reaction H2. If this unknown horizontal reaction did not exist, the structure would ‘float’ into space due to the 4 kN load.

Consider joint ABE, as there are only two unknown forces here, namely the forces in the mem- bers BE and EA. Working clockwise around this joint, the vector diagram for this joint is shown in Figure 4.16. By measurement,ae =3.5 kN and be=2.1 kN. 30_° 60° a e b 4 kN Figure 4.16

Joint (2) cannot be considered next, as it has three unknown forces, namelyH2, R2 and the unknown member forceDE. Hence, joint (1) must be consid- ered next; it has two unknown forces, namelyR1and the force in member ED. As the member AE and its direction can be obtained from Figure 4.16, it can be drawn to scale in Figure 4.17. By measurement, de=3 kN.

As R1 is vertical, then the vector da is vertical, hence, the position d can be found in the vector diagram of Figure 4.17, where R1 = da (pointing

downwards). By measurement,R1=1.8 kN. To determine R2 and H2, joint (2) can now be considered, as shown by the vector diagram for the joint in Figure 4.18.

The complete diagram for the whole framework is shown in Figure 4.19, where it can be seen that

3.5 kN a e d 30° Figure 4.17 b c e d 3 kN 2.1 kN Figure 4.18 b a,c d e Figure 4.19

this diagram is the sum of the vector diagrams of Figures 4.16 to 4.18.

The table below contains a summary of all the measured forces Member Force (kN) be −2.1 ae 3.5 de −3.0 R1 −1.8 R2 1.8 H₂ 4.0

Couple and moment

Prior to solving Problem 7, it will be necessary for the reader to understand the nature of acouple; this is described in Chapter 9, page 109.

The magnitude of a couple is called itsmoment; this is described in Chapter 5, page 57.

Problem 7. Determine the internal forces in the pin-jointed truss of Figure 4.20.

2 m 4 kN 3 kN 5 kN Joint 2 Joint 1 R1 R2 2 m 2 m 2 m 30° 30° 30° 30° Figure 4.20

In this case, there are more than two unknowns at every joint; hence it will first be necessary to calculate the unknown reactionsR1 andR2.

To determineR1, take moments about joint (2):

Clockwise moments about joint(2)=counter-clock- wise (or anti-clockwise) moments about joint (2) i.e. R1×8 m=4 kN×6 m+3 kN×4 m +5 kN×2 m =24+12+10=46 kN m Therefore, R1= 46 kN m 8 m =5.75 kN

Resolving forces vertically:

Upward forces=downward forces i.e. R1+R2=4+3+5=12 kN

However,R1 =5.75 kN, from above,

hence, 5.75 kN+R2=12 kN

from which, R2=12−5.75

=6.25 kN

Placing these reactions on Figure 4.21, together with the spaces between the lines of action of the forces, we can now begin to analyse the structure.

R1 = 5.75 kN R2 = 6.25 kN 5 kN 4 kN 3 kN B A C D E F G H J Figure 4.21

Starting at either jointAF Eor jointDEJ, where there are two or less unknowns, the drawing to scale of the vector diagram can commence. It must be remembered to work around each joint in turn, in a clockwise manner, and only to tackle a joint when it has two or less unknowns. The complete vector diagram for the entire structure is shown in Figure 4.22. a b c d e f g h j R₁ = 5.75 kN R₂ = 6.25 kN Figure 4.22

The table below contains a summary of all the measured forces. Member Force (kN) af −11.5 f e 10.0 j d −12.5 ej 10.9 bg −7.5 gf −4.0 ch −7.6 hg 4.6 j h −5.0 R1 5.75 R2 6.25

Now try the following exercise

Exercise 20 Further problems on a graph-