2. Analysis of Perfect Frames (Graphical Method)
Chapter 14
Analysis of Perfect Frames (Graphical Method)

3. Learning Objectives Introduction Construction of Space Diagram
Construction of Vector Diagram
Force Table
Magnitude of Force
Nature of Force
Cantilever Trusses
Structures with One End Hinged (or Pin-jointed) and the Other Freely Supported on Rollers and Carrying Horizontal Loads
Structures with One End Hinged (or Pin-jointed) and the Other Freely Supported on Rollers and Carrying Inclined Loads
Frames with Both Ends Fixed
Method of Substitution

4. Structures with One End Hinged (or Pin-jointed) and the Other Freely Supported on Rollers and Carrying Horizontal Loads
We have already discussed in Art that sometimes a structure is hinged or pin-jointed at one end and freely supported on rollers at the others end. If such a structure carries vertical loads only, the problem does not present any special features. Such a problem may be solved just as a simply supported structure.
But, if such a structure carries horizontal loads (with or without vertical loads) the support reaction at the roller supported end will be normal to the support; whereas the support reaction at the hinged end will consist of :
Vertical reaction, which may be found out by subtracting the vertical support reaction at the roller supported end from the total vertical load.
Horizontal reaction, which may be found out by algebraically adding all the horizontal loads. After finding out the reactions, the space and vector diagram may be drawn as usual.

5. Example
Figure shows a pin-jointed frame carrying vertical loads of 1 kN each at B and G and horizontal load of 4 kN at D. Find graphically, or otherwise, force in the various members of the truss. Also prepare a table stating the nature of forces.
Fig
Solution
Since the frame is supported on rollers at the right hand support (E), therefore the reaction at this support will be vertical (because of horizontal support). The reaction at the left hand support (A) will be the resultant of vertical and horizontal forces.
Taking moments about A and equating the same,

8. Structures with One End Hinged (or Pin-jointed) and the Other Freely Supported on Rollers and Carrying Inclined Loads.
We have already discussed in Art 14.8 that if a structure is hinged at one end, freely supported on rollers at the other and carries inclined loads (with or without vertical loads), the support reaction at the roller supported end will be normal to the support. The support reaction at the hinged end will be the resultant of :
Vertical reaction, which may be found out by subtracting the vertical component of the support reaction at the roller supported end from the total vertical load.
Horizontal reaction, which may be found out by algebraically adding all the horizontal loads.

9. Example
A truss hinged at A and supported on rollers at D is loaded as shown in Fig Find by any method the forces in all the members of the truss and mention the nature of forces.
Solution
Since the truss is supported on rollers at the right end D, therefore reaction at this support will be inclined at 45°, with the vertical (because the support is inclined at 45° with the horizontal). Now find out the reactions as done in example We know that horizontal component of reaction at D.
  RDH = RDV = 5.4 kN
 and RAH = 5.4 kN and RAV = 1.6 kN
First of all, draw the space diagram and name the members and forces according to Bow’s notations as shown in Fig (a).

10. Now draw the vector diagram as shown in Fig. 14. 50 (b)
Now draw the vector diagram as shown in Fig (b). Measuring the various sides of the vector diagram, the results are tabulated here :
S.No.
Member
Magnitude of force in kN
Nature of force 1
AB (2-7)
6.6
Compression 2
BC (2-8) 3
CD (2-10)
6.75 4
DE (4-10)
1.35 5
EF (5-9) 6
FA (6-7)
2.0
Tension 7
BF (7-8) — 8
CF (8-9)
4.25 9
CE (9-10)
* We
have already solved this example analytically in the last chapter.

11. Frames with Both Ends Fixed
Sometimes, a frame or a truss is fixed or built-in at its both ends. In such a case, the reactions at both the supports can not be determined, unless some assumption is made. The assumptions usually made are :
The reactions are parallel to the direction of the loads and
In case of inclined loads, the horizontal thrust is equally shared by the two reactions. Generally, the first assumption is made and the reactions are determined as usual by taking moments about one of the supports.
Example
Figure shows as roof truss with both ends fixed. The truss is subjected to wind loads normal to the main rafter. Find the force in various members of the truss.
Fig

12. Solution
The reactions may be obtained by any one assumption as mentioned. With the help of first assumption the reactions have been found out as shown in Fig (a).
 Equating the anticlockwise moments and the clockwise moments about A,

13. Method of Substitution
Sometimes work of drawing the vector diagram is held up, at a joint which contains more than two unknown force members and it is no longer possible to proceed any further for the construction of vector diagram. In such a situation, the forces are determined by some other method. Here we shall discuss such cases and shall solve such problem by the method of substitution.

14. Example
A french roof truss is loaded as shown in Fig Find the forces in all the members of the truss, indicating whether the member is in tension or compression.
Fig
Solution
Since the truss and loading is symmetrical, therefore both the reactions will be equal.
First of all, draw the space diagram and name all the members according to Bow’s notations and also name the joints as shown in Fig (a).
While drawing the vector diagram, it will be seen that the vector diagram can be drawn for joint Nos. 1, 2 and 3 as usual. Now when we come to joint No. 4, we find that at this joint there are three members (namely DP, PO and ON) in which the forces are unknown.

15. we cannot draw the vector diagram for this joint
we cannot draw the vector diagram for this joint. Now, as an alternative attempt, we look to joint No. 5. We again find that there are also three members (namely NO, OR and RK) in which the forces are unknown. So we can not draw the vector diagram for this joint also. Thus we find that the work of drawing vector diagram is held up beyond joint No. 3. In such cases, we can proceed by the substitution of an imaginary member. Now, consider (for the time being only) the members OP and PQ as removed and substitute an imaginary member joining the joints 5 and 6 (as shown by the dotted line) as shown in Fig (a). Now we find that this substitution reduces the unknown force members at joint 4, from three to two (i.e., members DI and IN; assuming the letter I in place of P and O) and thus we can draw the vector diagram for this joint (i.e., No. 4).
Fig (a).

16. Fig (b).
Now after drawing the vector diagram for joint 4, proceed to joint 6 at which there are only two members (i.e., EQ and QI) in which the forces are unknown. The vector diagram, at this joint will give the forces in EQ by the side eq of the vector diagram. After drawing vector diagram at joint 6 and determining the forces in EQ (i.e., eq) replace the imaginary member by the original members PQ and PO and again draw vector diagram for the joint No. 6 as shown in Fig (b). This will give the force in the member PO. Now proceed to joint No. 5 as usual and complete the whole vector diagram as shown in Fig (b). Meausring the various sides of the vector diagram, the results are tabulated here :

17. S. No.
Member
Magnitude of force in kN
Nature of force 1
BL, IX
15,720
Compression 2
LM, WX
1,750 3
CM, HW
14,750 4
MN, VW
2,000
Tension 5
DP, GT
13,780 6
NO, UV
3,500 7
OP, TU
1,875 8
PQ, ST
1,685 9
EQ, FS
12,810 10
KL, KX
14,050 11
NK, VK
12,060 12
OR, RU
4,000 13
QR, RS
5,815 14 RK
8,080