A structure made up of several bars (or members) riveted or welded together is known as truss is composed of such members which are just sufficient to keep the truss in equilibrium, when the truss is supporting an external load, then the frame is known as perfect truss. Though in actual practice the members are welded or riveted together at their joints, yet for calculation purposes the joints are assumed to be hinged or pinjoined. A plane truss is a truss which carries applied loads in a plane. The basic plane truss consists of three straight members which are connected by pins at their ends to form a triangle.
2.12.1. Types of Frames or Trusses. The different types of frames or trusses are:
i. Perfect frame, and
ii. Imperfect frame.
Imperfect frame may be a deficient frame or a redundant frame.
1. Perfect Frame. The frame which is composed of such members, which are just sufficient to keep the frame in equilibrium, when the frame is supporting an external load, is known as perfect frame, The simplest perfect frame is a triangle as shown in Fig. 2.55 which consists three members and three joints. The three members are: AB, BC and AC whereas the three joints are A, B and C. This frame can be easily analysed by the condition of equilibrium.
Let the two members CD and BD and a joint D are added to the triangular frame ABC. Now, we get a frame ABCD as shown in Fig.2.56 (a). This frame can also be analysed by the conditions of equilibrium. This frame is also known as perfect frame.
* The couple at C is 4 kN m clockwise. Hence its sense is positive.
** The couple at D is 8 kN m anticlockwise. Hence its sense is negative. The couple is also moment.
Suppose we add a set of two members and a joint again, we get a perfect frame as shown in Fig.2.56 (b). Hence for a perfect frame, the number of joints and number of members are given by,
n = 2j – 3
where n = Number of members, and
j = Number of joits.
Fig. 2.56 (a)
Fig. 2.56 (b)
2. Imperfect Frame. A frame in which number of members and number of joints are not given by
n = 2j – 3
is known, an imperfect frame. This means that number of members in an imperfect frame will be either more or less than (2j – 3).
(i) If the number of members in a frame are less than (2j – 3), then the frame is known deficient frame.
(ii) If the number of members in a frame are more than (2j – 3), then the frame is known as redundant frame.
2.12.2. Assumptions Made in Finding Out the Forces in a Frame. The assumptions made in finding out the forces in a frame are:
i. The frame is a perfect frame
ii. The frame carries load at the joints
iii. All the members are pinjoined.
2.12.3. Reactions of Supports of a Frame. The frames are generally supported
i.on a roller support or
ii.on a hinged support.
If the frame is supported on a roller support, then the line of action of the reaction will be at right angles to the roller base as shown in Figs. 2.57 and 2.58.
If the frame is supported on a hinged support, then the line of action of the reaction will depend upon the load system on the frame.
The reaction at the support of a frame are determined by the conditions of equilibrium frame and the reactions at the supports must form a system of equilibrium.
1.12.4. Analysis of a Frame or Truss. Analysis of a frame consists of:
(i) Determinations of the reactions at the supports and
(ii) Determinations of the reactions at the supports and
The reactions are determined by the condition that the applied load system and the
induced reactions at the supports form a system in equilibrium.
The forces in the members of the frame are determined by the condition that every joint
should be in equilibrium and so, the forces acting at every joint should form a system in
equilibrium
A frame is analysed by the following methods:
(i) Method of joints,
(ii Method of sections, and
(iii) Graphical method.
1.12.5. Method of Joints. In this method, after determining the reactions at the supports, the equilibrium of every joint is considered. This means the sum of all the vertical forces as well as the horizontal forces acting on a joint is equated to zero. The joint should be selected in such a way that at any time there are only two members, in which the forces are unknown. The force in the member will be compressive if the member pushes the joint to which it is connected whereas the force in the member will be tensile if the member pulls the joint to which it is connected.
Problem 2.31. Find the forces in the members AB, AC and BC of the truss shown in Fig. 2.59
Sol. First determine the reactions R_{B} and R_{C}. The line of action of load of 20 kN acting at A is vertical. This load is at a distance of AB × cos 60° from the point B. Now let us find the distance AB.
The triangle ABC is a rightangled triangle with angle BAC = 90°. Hence AB will be equal to BC × cos 60°.
AB = 5 × cos 60° = 5 × 1/2 = 2.5 m
Now the distance of line of action of 20 kN from B is
AB × cos 60° or 2.5 × 1/2 = 1.25 m.
Resolving the forces acting on the joint B, vertically
F_{1} sin 60° = 15
Taking the moments about B, we get
R_{C} × 5 = 20 × 1.25 = 25
As F_{1} is pushing the joint B, hence this force will be compressive. Now resolving the forces horizontally, we get
F_{2} = F_{1} cos 60° = 1.7.32 × 1/2 = 8.66 kN (tensile)
As F_{2} is pulling the joint B, hence this force will be tensile.
Joint C
Let F_{3} = Force in the member AC
F_{2} = Force in the member BC
The force F_{2} has already been calculated in magnitude and direction. We have seen that force F_{2} is tensile and hence it will pull the joint C. Hence it must act away from the joint C as shown in Fig. 2.59 (b).
and R_{C} = Total load R_{C} = 20 – 5 = 15 kN
Now let us consider the equilibrium of the various joints.
Joint B
Let F_{1} = Force in member AB
F_{2} = Force in member BC
Let the force F_{1} is acting towards the joint B and the force F_{2} is acting away* from the joint B as shown in Fig. 2.59 (a). (The reaction R_{B} is acting vertically up. The force F_{2} is horizontal. The reaction R_{B} will be balanced by the vertical component of F_{1}. The vertical component of F_{1} must act downwards to balance R_{B}. Hence F_{1} must act towards the joint B so that its vertical component is downward. Now the horizontal component of F_{1 }towards the joint B. Hence force F_{2} must act away from the joint to balance the horizontal component of F_{1}):
Resolving forces vertically, we get
F_{3} sin 30° = 5 kN
As the force F_{3} is pushing the joint C, hence it will be compressive. Ans.
*The direction of F_{2} can also be taken towards the joint B. Actually when we consider the equilibrium of the joint B, if the magnitude of F_{1} andF_{2} comes out to be positive then the assumed direction of F_{1} and F_{2} are correct. But if anyone of them is having a negative magnitude then the assumed direction of that force is wrong. Correct direction then will be the reverse of the assumed direction.
Problem 2.32. A truss of span 7.5 m carries a point load of 1 kN at joint D as shown in Fig.2.60. Find the reactions and forces in the members of the truss.
Sol. Let us first determine the reactions R_{A} and R_{B}.
Taking moments about A, we get R_{B} × 7.5 = 5 × 1
∴ R_{A} = Total load – R_{B}
∴ = 1 – 0.667 = 0.333 kN
Now consider the equilibrium of the various joints.
Joint A
Let F_{1} = Force in member AC
F_{2} = Force in member AD.
Let the force F_{1} is acting towards the joint A and F_{2} is acting away from the joint A as shown in Fig. 2.60 (a).
Resolving the forces vertically, we get
F_{1} sin 30° = R_{A}
= 0.666 kN (Compressive)
Resolving the forces horizontally, we get
F_{2} = F_{1} × cos 30°
= 0.666 x 0.866 = 0.5767 kN (Tensile)
Joint B
Let F_{4} = Force in member BC
F_{5} = Force in member BD
Let the direction of F_{4} and F_{5} are assumed as shown in Fig. 2.60 (b).
Resolving the forces vertically, we get
F_{4} sin 30° = R_{B} = 0.667
Resolving the forces horizontally, we get
F_{5} = F_{4} cos 30° = 1.334 × 0.866 = 1.155 kN (Tensile)
Joint D
Let F_{3} = Force in member CD. The forces F_{2} and F_{5} have been already calculated in magnitude and direction. The forces F_{2} and F_{5} are tensile and hence they will be pulling the joint D as shown in Fig. 2.60 (c). Let the direction* of F_{3} is assumed as shown in Fig. 2.60 (c).
Resolving the forces vertically, we get
F_{3} sin 60° = 1
= 1.1547 kN (Tensile)
Hence the forces in the members are:
F_{1} = 0.666 kN (Compressive)
F_{2} = 0.5767 kN (Tensile)
F_{3} = 1.1547 kN (Tensile)
F_{4} = 1.334 kN (Compressive)
F_{5} = 1.155 kN,(Tensile). Ans
Problem 2.33. A truss of span 5 m is loaded as shown in Fig. 2.61. Find the reactions and forces in the members of the truss.
Sol. Let us first determine the reactions R_{A} and R_{B}.
Triangle ABD is a rightangled triangle having angle ADB = 90°.
∴AD = AB cos 60° = 5 × 0.5 = 2.5 m
The distance of the line of action of the vertical load 10 kN from point A will be AD
cos 60°
or 2.5 × 0.5 = 1.25 m.
From triangle ACD, we have AC =AD = 2.5 m
∴ BC = 5 – 2.5 = 2.5 m
In rightangled triangle CEB, we have
*The horizontal force F_{5} is more than F_{2} Hence the horizontal component of F_{3} must be in the direction of F_{2.} This is only possible if F_{3} is acting away from D.
The distance of the line of action of the load of 12 kN from point A will be (BE × cos )
The distance of the line of action of the load of 12 kN from point A will be (51.875)=3.125 m.
Now taking the moments about A, we get
R_{B} × 5 = 10 × 1.25 + 12 × 3.125 = 50
R_{A} = Total load RB = (10 + 12) 10 = 12 kN
Now consider the equilibrium of the various joints.
Joint A
F_{1} = Force in member AD, and
F_{2} = Force in member AC
Let the directions of F_{1} and F_{2} are assumed as shown in Fig. 2.61 (a)
Resolving the forces vertically,
F_{1} sin 60° = 12
= 13.856 kN (Compressive)
Resolving the forces horizontally,
F_{2} = F_{1} cos 60° = 13.856 × 0.5
= 6.928 kN (Tensile)
Now consider the joint B .
Joint B
Let F_{3} = Force in member BE, and
F_{4} = Force in member BC
Let the directions of F_{3} and F_{4} are assumed as shown in Fig. 2.61 (b).
Resolving the forces vertically, we get
F_{3} sin 30° = 10
Now resolving the forces horizontally, we get
F4 = F3 cos 30^{o} = 20 × 0.866
= 17.32 kN (Tensile)
Now consider the joint C.
Joint C
Let F_{5} = Force in member CE
F_{6} = Force in member CD
Let the directions of F_{5} and F_{6} are assumed as shown in Fig.2.61 (c).
The forces F_{2} and F_{4} are already known in magnitude and directions. They are tensile and hence will be pulling the joint C as shown in Fig. 2.61 (c).
Resolving forces vertically, we get
F_{6} sin 60° + F_{5} sin 60° = 0
or F_{6} = F_{5}
Resolving forces, horizontally, we get
F_{2}  F_{6} cos 60° = F_{4}  F_{5} cos 60°
The magnitude of F_{6} is ve, hence the assumed direction of F 6 is wrong. The correct direction F_{6} will be as shown in Fig. 2.61 (d).
F_{5} = 10.392 (Compressive)
F_{6} = 10.392 (Tensile)
Now consider the joint E.
Joint E
Let F_{7} = Force in member ED
Let F_{7} is acting as shown in Fig. 2.61 (e).
The forces F_{3} and F_{5} are known in magnitude and directions. They are compressive hence they will be pushing the joint E as shown in Fig. 2.61 (e).
Resolving the forces along BED, we get
F_{7} + 12 cos 60° = F_{3}
F_{7} = F_{3}  12 × 0.5
= 20 – 6 = 14 kN (Compressive)
As F_{7} is positive hence the assumed direction of F_{7} is correct. Ans.
Problem 2.34. A truss of span 9 m is loaded as shown in Fig. 2.62. Find the reaction and forces in the member of the truss.
Sol. Let us first calculate the reactions R_{A} and R_{B}.
Taking moments about A, we get
RA × 9 = 9 × 3 + 12 × 6 = 27 + 72 = 99
and R_{A} = Total load  R_{B} = (9 + 12) 11 = 10 kN
In this problem, there are some members in which force is zero.
These members are obtained directly as given below:
“If three forces act at a joint and two of them are along the same straight line, then for the equilibrium the joint, the third force should be equal to zero.”
1. Three forces are acting at the point A (i.e., R_{A}, F_{AC} and F_{AG}), two of which (i.e., R_{A}, R_{AC}) the same straight line. Hence the third force (i.e., R_{AG}) is zero.
2. Similarly, three forces are acting at the joint B (i.e., R_{B}, F_{BF} and F_{BH}), two of which (i.e., R_{B} and F_{BH}) are along the same straight line. Hence the third force F_{EH} should be zero.
3. At the joint E also, three forces (i.e., F_{ED}, F_{EF} and F_{EH}) are acting, two of which (i.e., F_{ED} and F_{EF}) are along the same straight line. Hence the third force F_{EH} must be zero.
Now the equilibrium of various joints can be considered.
F_{AG} = Force in member AG = 0
F_{AG} = Force in member AC
= R_{A} = 10 kN (Compressive)
Now consider the equilibrium of joint C.
Joint C [See Fig.2.62 (b)]
Let F_{CD} = Force in member CD
F_{CG} = Force in member CG
F_{AC} = 10 kN (Compressive)
Let the directions of F_{CG} and F_{CD} are assumed as shown in Fig. 2.62 (b).
Resolving forces horizontally, we get
F_{CG} cos θ = 10
Now consider the equilibrium of joint G.
Joint G
The force in member CG is 12.5 kN (Tensile).
Hence at the joint G, this force will be pulling the joint G as shown in Fig. 2.62 (c).
Resolving the forces vertically, we get
12.5 cos θ +F_{GD} = 9
F_{GD} = 9 – 12.5 cos θ
As the magnitude of F_{GD} is negative, hence its assumed direction is wrong. The correct direction will be as shown in Fig. 2.62 (d).
Then, F_{GD} = 1 kN (Compressive)
Resolving the forces horizontally, we get
12.5 sin θ = F_{GH}
= 7.5 kN (Tensile)
Now consider the equilibrium of joint D.
Joint D
The forces in the members CD and GD have been already calculated. They are 7.5 kN and 1 kN respectively. Both are compressive.
Let F_{DH} = Force in member DH, and
F_{DE} = Force in member DE
Resolving the forces vertically, we get
F_{DH} θ = 1 kN
Resolving the forces horizontally, we get
7.5 + F_{DH} sin θ = F_{DE}
= 7.5 + 0.75 = 8.25 kN (Compressive)
Now consider the equilibrium of joint E.
Joint E
As shown in Fig. 2.62 (f), a joint E three forces are acting. The forces i.e., F_{DE} and F_{EF} are in the same straight line.
Hence force F_{EH} must be zero.
Force in EF, i.e., F_{EF} = F_{DE}
= 8.25 kN (Compressive)
Now consider the joint H.
Joint H
It is already shown that forces in the members EH and BH are zero.
Also the forces in the member GH is 7.5 kN tensile and in the member DH is 1.25 kN tensile.
Let F_{HF} is the force in the member HF.
Resolving forces vertically, we get
1.25 cos θ + F_{HF} cos θ = 12
Now consider the joint B.
Joint B
See Fig. 2.62 (h)
The force in member BF = 11 kN (Compressive)
Now the forces in each member are known.
They are shown in Fig. 2.63. Also these forces are shown below.
Member  Force in member 
AC  10 kN (Comp.) 
AG  0 
CG  12.5 kN (Tens.) 
CD  7.5 kN (Comp.) 
DG  10 kN (Comp.) 
DE  8.25 kN (Comp.) 
DH  1.25 kN (Tens.) 
GH  7.5 kN (Tens.) 
EH  0 
EF  8.25 kN (Comp.) 
HB  0 
HF  13.75 kN (Tens.) 
BF  11 kN (Comp.) 
Problem 2.35. A plane truss is loaded and supported as shown in Fig. 2.64. Determine the nature and magnitude of the forces in the members 1, 2 and 3.
Sol. First calculate the reactions R_{A} and R_{B}
Taking moments about A, we get
R_{B} × 4 = 1 × 1000
Consider the equilibrium of joint A.
Joint A [See Fig. 2.64 (a)]
Resolving the forces vertically,
F_{AD} sin θ = 750
Resolving the forces horizontally, we get
F_{AE} = F_{AD} cos θ = 1250 × 0.8
= 1000 N (Tensile)
Now consider joint E.
Joint E
Three forces, i.e., F_{AE}, F_{EF} and F_{ED} are acting at the joint E. Two of the forces, i.e., F_{AE} and F_{EF} are in the same straight line. Hence the third force, i.e., F_{ED} should be zero.
and F_{EF} = F_{AE} = 1000 N (Tensile)
Now consider the equilibrium of joint D.
Joint D
Let F_{1} = Force in member DG
F_{DF} = Force in member DF
Let us assume their directions as shown in Fig. 2.64 (b).
The forces in the member AD and DE are 1250 N (Compressive) and 0 respectively.
Resolving forces vertically, we get
1250 sin θ + F_{1} sin θ + F_{DF} sin θ = 1000
or 1250 × 0.6 + F_{1} × 0.6 + F_{DF} × 0.6 = 1000 ∵ sin θ = 0.6)
or F_{1} + F_{DF} = 1666.66 – 1250 = 416.66 … (i)
Resolving the forces horizontally, we get
1250 cos θ + F_{1} cos θ = F_{DF} × cos θ
or 1250 + F_{1} = F_{DF} or F_{1} – F_{DF} = 1250 … (ii)
Adding (i) and (ii), we get
2F_{1} = 416.66 – 1250 = – 833.34
Substituting the value of F_{1} in equation (i), we get
 416.67 + F_{DF} = 416.66
or F_{DF} = 416.66 + 416.67
= 833.33 N (Comp.)
The magnitude of F_{1} is negative. Hence its assumed direction is wrong. The correct direction of F_{1} is shown in Fig. 2.64 (c).
∴ F_{1} = 416.67 N (Compressive). Ans.
To find the forces F_{2} and F_{3} consider the joint F.
Joint F
The forces in the members DF and EF are already known. They are:
F_{DF} = 833.33 N (Compressive)
F_{EF} = 1000 N (Tensile).
These forces are acting at the joint F as shown in
F_{2} = Force in member FG, and
F_{3} = Force in member FH
Resolving forces vertically, we get
833.33 sin θ = F_{2}
F_{2} = 833.33 × 0.6
= 499.998 N = 500 N (Tensile). Ans.
Resolving forces horizontally, we get
F_{3} + 833.33 cos θ = 1000
F_{3} = 1000 – 833.33 × 0.8
= 333.336 N (Tensile). Ans.
Method of Joints Applied to Cantilever Trusses. In case of cantilever trusses, it is not necessary to determine the support reactions. The forces in the members of cantilever truss can be obtained by starting the calculations from the free end of the cantilever.
Problem 2.36. Determine the forces in all the members of a cantilever truss shown in Fig. 2.65.
Sol. Here the calculations can be started from end C. Hence consider the equilibrium of the joint C.
Joint C
Let F_{CD} = Force in member CD, and
F_{CA} = Force in member CA.
Their assumed directions are shown in Fig. 2.65,
Resolving the forces vertically, we get
FCD × sin 60^{o} = 1000
Resolving the forces horizontally, we get
F_{CA} = F_{CD} × cos 60^{o}
= 1154.7 × 0.5
=577.35 N (Compressive)
Now consider the equilibrium of the point D.
Joint D
[See Fig. 2.65 (a)]The force F_{CD} = 1154.7 N (tensile) is already calculated.
Let F_{AD} = Force in member AD, and
F_{BD} = Force in member BD, and
Their assumed directions are shown in Fig.2.65 (a)
Resolving the forces vertically, we get
F_{AD} cos 30^{o} = 1154.7 cos 30^{o}
Resolving the forces horizontally, we get
F_{BD} = F_{AD} sin 30° + F_{DC} sin 30°
= 1154.7 × 0.5 + 1154.7 × 0.5 = 1154.7 N (Tensile)
Now the forces are shown in a tabular form below :
Member 
Force in the member 
Nature of force 
AC 
577.35 N 
Compressive 
CD 
1154.7 N 
Tensile 
AD 
1154.7 N 
Compressive 
BD 
1154.7 N 
Tensile 
Problem 2.37. Determine the forces in all the members of a cantilever truss shown in Fig. 2.66.
Sol. Start the calculations from joint C.
From triangle ACE, we have
Joint C
The direction of forces at the joint C are shown in Fig. 2.66.
Resolving the forces vertically, we get
F_{CD} sin θ = 1000
Resolving the forces horizontally, we get
F_{CB} = F_{CD} × cos θ = 1666.66 × 0.8 = 1333.33 N (Tensile)
Now consider the equilibrium of joint B.
Joint B
Resolving vertically, we get
F_{BD} = 1000 N (Compressive)
F_{BA} = F_{CB} = 1333.33 (Tensile)
Now consider the joint D.
Joint D
The forces in member CD and BD have already been calculated. They are 1666.66 N and 1000 N respectively as shown in Fig. 2.66 (a).
Let F_{DA} = Force in member DA, and
F_{DE} = Force in member DE
Resolving forces vertically, we get
1000 + 1666.66 sin θ = F_{AD} sin θ + F_{ED} sin θ
or 1000 + 1666.66 × 0.6 = F_{AD} × 0.6 + F_{ED} × 0.6
Resolving forces horizontally, we get
1666.66 cos θ + F_{AD} cos θ = F_{ED} cos θ
1666.66 + F_{AD} = F_{ED} or F_{ED}  F_{AD} = 1666.66
Adding equations (i) and (ii), we get
2F_{ED} = 3333.32 + 1666.66 = 4999.98
Substituting this value in equation (i), we get
F_{AD} + 2500 = 3333.32
∴ F_{AD} = 3333.32 – 2500 = 833.32 N (Tensile)
Now the forces are shown in a tabular form below:
Member 
Force in the member 
Nature of force 
AB BC CD DE AD BD 
1333.33 N 1333.33 N 1666.66 N 2500 N 833.32 N 1000 N 
Tensile Tensile Compressive Compressive Tensile Compressive 
Method of Joints Applied to Trusses Carrying Horizontal Loads. If a truss carries horizontal loads (with or without vertical loads), hinged at one end and supported on rollers at the other end, then the support reaction at the roller supported end will be normal, whereas the support reactions at the hinged end will consists of: (i) horizontal reaction and (ii) vertical reaction.
The horizontal reaction will be obtained by adding algebraically all the horizontal loads; whereas the vertical reaction will be obtained by subtracting the roller support reaction from the total vertical loads. Now the forces in the members of the truss can be determined.
Problem 2.38. Determine the forces in the truss shown in Fig. 2.67 which carries horizontal load of 12 kN and a vertical load of 18 kN.
Sol. The truss is supported on rollers at B and hence the reaction at B must be normal to the roller base i.e., the reaction at B, in this case, should be vertical,
At the end A, the truss is hinged and hence the support reactions at the hinged end A will consists of a horizontal reaction H_{A} and a vertical reaction R_{A}.
Taking moments of all forces at A, we get
R_{B} × 4 = 18 × 2 + 12 × 1.5 = 36 + 18 = 54
Let us first consider the equilibrium of joint A.
Joint A
The reactions R_{A} and H_{A} are known in magnitude and direction. Let the directions of the forces in the members AC and AD are as shown in Fig. 2.67(a).
Resolving the forces vertically, we get
F_{AD} sin θ =R_{A}
Resolving the forces horizontally, we get
F_{AC} =H_{A} +F_{AD} cos θ
= 12 + 7.5 × 0.8 = 18 kN (Tensile)
Now consider the joint C.
Joint C
At the joint C, the force 0 in member CA and vertical load 18 kN are known in magnitude and directions. For equilibrium of the joint C.
F_{BC} = F_{CA} = 18 kN (Tensile)
F_{CD} = 18 kN (Tensile)
Now consider the joint B.
Joint B
At the joint B, R_{B} and force F_{BC} are known in magnitude and direction.
Let F_{BD} is the force in member BD.
Resolving the forces vertically, we get
F_{BD} × sin θ = R_{B}
Now the forces are shown in a tabular form below :
Member 
Force in the member 
Nature of force 
AC AD CD CB BD 
18 kN 7.5 kN 18 kN 18 kN 2.5 kN 
Tensile Compressive Tensile Tensile Compressive 
Problem 2.39. Determine the forces in the truss shown in. Fig. 2.68 is subjected to horizontal and vertical loads, Mention the nature of forces in each case.
Sol. The truss is supported on rollers at B and hence RB will be vertical. The truss is
hinged at A and hence the support reactions at A will consists of a horizontal reaction H_{A} and a vertical reaction R_{A}.
Taking moment about A, we get
R_{B} × 12 = 8 × 1.5 + 3 × 4 + 6 × 8
= 72
and R_{A} = Total vertical loads – R_{B}
= (3 + 6) – 6
= 3 kN (↑)
and H_{A} = Sum of all horizontal loads
= 8 kN(←)
Now the forces in the members can be calculated. Consider the joint A.
Joint A
The reactions R_{A} and H_{A} are known in magnitude and direction. Let the directions of the forces F_{CA} and F_{FA} are as shown in Fig.2.68 (b).
Resolving the forces vertically, we get
F_{CA} × sin θ = 3 kN
= 5 × 0.8 + 8 = 12 kN (Tensile)
Now consider joint C
Joint C
The force F_{CA} is known in magnitude and direction. The assumed direction of the forces F_{CD} and F_{CF} are shown in Fig. 2.68 (c).
Resolving forces vertically, we get
F_{CA} sin θ = F_{CF} sin θ
Resolving forces horizontally, we get
F_{CD} = F_{CA} cos θ + FCF cos θ
= 5 × 0.8 + 5 × 0.8 = 8 kN (Compressive)
Now consider the joint F.
Joint F
The forces F_{FA} and F_{FC} are known in magnitude and directions. The assumed directions of the forces F_{DF} and F_{GF} are shown in Fig. 2.68 (d).
Resolving the forces vertically, we get
5 × sin a + F_{DF} sin θ = 3
Resolving the forces horizontally, we get
12 + 5 cos θ = F_{GF} + F_{DF} cos θ
or 12 + 5 × 0.8 = F_{GF} + 0 or 12 + 4 = F_{GF}
F_{GF} = 12 + 4 = 16 kN (Tensile)
Now consider the joint D.
Joint D
The forces F_{DC} and F_{FD} are known in magnitude and direction. The assumed directions of F_{DG} and F_{DE} are shown in Fig. 2 68 (e).
Resolving vertically, we get
F_{DG} sin θ = F_{DF} × sin θ = 0
∴ F_{DG} = 0
Resolving forces horizontally, we get
F_{DE} = F_{CD} = 8 kN
∴ F_{DE} = 8 kN (Compressive)
Now consider the joint G.
Joint G
The forces F_{DG} and F_{FG} are known in magnitude and direction. The assumed directions of F_{GE} and F_{GB} are shown in Fig. 2.68 (f).
Resolving the forces vertically, we get
F_{GE} sin θ = F_{DG} sin θ + 6 = 6
Resolving forces horizontally, we get
F_{GB} = 16 – F_{GE} cos θ
= 16 – 10 × 0.8 = 8 kN (Tensile)
Now consider the joint E.
Joint E
The forces F_{GE} and F_{DE} are known in magnitude and directions. Let F_{BE} is acting in a direction as shown in Fig. 2.68 (g).
Resolving forces vertically, we get
FGE sin θ = F_{BE} sin θ
F_{BE} = F_{GE} = 10
F_{BE} = 10 kN (Compressive)
If we have calculated the forces in member BE and BG, considering joint B, we would have got the same results.
Now the forces in each member are shown in Fig. 2.68 (h).
Method of Joints Applied to Trusses Carrying Inclined Loads. If a truss carries
inclined loads, hinged at one end and supported on rollers at the other end, then the support reaction at the roller supported end will be normal, whereas the support reactions at the hinged end will consists of:
(i)Horizontal reaction and
(ii)Vertical reaction.
The inclined loads are resolved into horizontal and vertical components.
The horizontal reaction will be obtained by adding algebraically all the horizontal components of the inclined loads; whereas the vertical reaction will be obtained by subtracting the roller support reaction from the total vertical components of the inclined loads. Now the forces in the members of truss can be determined.
Problem 2.40. Determine the forces in the truss shown in Fig. 2.69 which is subjected to inclined loads.
Sol. The truss is supported on roller at B and hence R_{B} will be vertical.
The truss is hinged at A and hence the support reactions at A will consists of a horizontal reaction H_{A} and a vertical reaction R_{A}.
Now length AC = 4 × cos 30 = 4 × 0.866 = 3.464 m
and length AD = 2 × AC = 2 × 3.464 = 6.928 m
Now taking moments about A, we get
R_{B} × 12 = 2 × AC + 1 × AD + 1 × AE
= 2 × 3.464 + 1 × 6.928 + 1 × 4 = 17.856
Total vertical components of inclined loads
= (1 + 2 + 1) × sin 60° + 1.0
= 4 × 0.866 + 1.0 = 4.464 kN
Total horizontal components of inclined loads
= (1 + 2 + 1) cos 60° = 4 × 0.5 = 2 kN
Now R_{A} = Vertical components of inclined loads  R_{B }
= 4.464 – 1.49 = 2.974 kN (↑)
H_{A} = Sum of all horizontal components = 2 kN
Now the forces in the members can be calculated.
Consider the equilibrium of joint A.
Joint A
F_{AE} = Force in member AE
F_{AC} = Force in member AC
Their directions are assumed as shown in Fig. 2.69 (a).
Resolving the forces vertically, we get
F_{AC} × sin 30° + 1 × sin 60° = 2.974
F_{AC} × 0.5 + 0.866 = 2.974
= 4.216 kN (Compressive)
Resolving the forces horizontally, we get
F_{AE} = 2 + F_{AC} cos 30° – 1 × 60°
= 2 + 4.216 × 0.866 – 0.5 = 5.15 kN (Tensile)
Now consider the joint C.
Joint C
From Fig. 2.69 (b), we have
F_{CD} = F_{AC} = 4.216 (Compressive)
and F_{CE} = 2 kN (Compressive)
Now consider joint E.
Joint E
Joint E [See Fig. 2.69 (c)]
Resolving forces vertically, we get
1 + 2 × sin 60° = F_{ED} × sin 60°
Resolving forces horizontally, we get
5.15 – 2 × cos 60°  F_{ED} cos 60^{0 } F_{EF} = 0
F_{EF} = 5.15 – 1 – 1.57 = 2.58 kN (Tensile)
At the joint G, two forces, i.e., F_{BG} and F_{DG} are in the same straight line and hence the third force, i.e., F_{GF} should be zero.
F_{GF }= a
Now consider the joint F.
Joint F [See Fig. 2.69 (d)]
Resolving forces vertically, we get
F_{DF} × sin 60° = 0
F_{DF} = 0
Resolving horizontally, we get
F_{FB} = F_{EF} = 2.58 kN
F_{FB} = 2.58 kN (Compressive)
Now consider the joint B.
Joint B
Resolving vertically, we get
F_{BG} × sin 60° = 1.49
The forces are shown in a tabular form as
Member 
Force in the member 
Nature of force 
AC AE CE CD ED EF DF DG GB FB FG 
4.216 kN 5.15 kN 2 kN 4.216 kN 3.155 kN 2.58 kN 0 2.98 kN 2.98 kN 2.58 kN 0 
Compressive Tensile Compressive Compressive Tensile Tensile Nil Compressive Compressive Compressive Nil 
2.12.6. Method of Sections. When the forces in a few members of a truss are to be determined, then the method of section is mostly used. This method is very quick as it does not involve the solution of other joints of the truss.
In this method, a section line is passed through the members, in which forces are to be determined as shown in Fig. 2.70. The section line should be drawn in such a way that it does not cut more than three members in which the forces are unknown. The part of the truss, on anyone side of the section line, is treated as a free body in equilibrium under the action of external forces on that part and forces in the members cut by the section line. The unknown forces in the members are then determined by using equations of equilibrium as
ƩF_{x} = 0 ƩF_{y} = 0 and ƩM = O.
Fig. 2.70
If the magnitude of the forces, in the members cut by a section line, is positive then the assumed direction is correct. If magnitude of a force is negative, then reverse the direction of that force.
Problem 2.41. Find the forces in the members AB and AC of the truss shown in Fig. 2.71 using method of section.
Sol. First determine the reaction R_{B} and R_{C}
The distance of line of action of 20 kN from point B is
Now draw a section line (11), cutting the members AB and BC in which forces are to be determined. Now consider the equilibrium of the left part of the truss. This part is shown in Fig. 2.71 (a).
Let the directions of F_{BA} and F_{BC} are assumed as shown in Fig. 2.71 (a).
Now taking the moments of all the forces acting on the left part about point C, we get
15 × 5 + (F_{BA} × AC)* = 0
The perpendicular distance between the line of action of F_{BA} and point C is equal to AC)
or 75 + F_{BA} × 5 × cos 30° = 0
The negative sign shows that F_{BA} is acting in the opposite direction (i.e., towards points B). Hence force F_{BA} will be a compressive force.
∴ F_{BA} = 17.32 kN (Compressive). Ans.
Again taking the moments of all the forces acting on the left part about point A, we get 15 × Perpendicular distance between the line of action of
15 kN and point C = F_{BC} × Perpendicular distance between F_{BC} and point A.
15 × 2.5 × cos 60° = F_{BC} × 2.5 × sin 60°
= 8.66 kN (Tensile). Ans.
These forces are same as obtained in Problem 2.31.
* The moment of the force F_{BA} about point C, is also taken by resolving the force F_{BA} into vertical and horizontal components at point B. The moment of the horizontal component about C is zero, whereas the moment of vertical component will be (F_{BA} × sin 60°) × 5 = F_{BA} × 5 × sin 60° or F_{BA} × 5 × cos 30°.
Problem 2.42. A truss of span 5 m is loaded as shown in Fig. 2.72. Find the reactions and forces in the members marked 4, 5 and 7 using method of section.
Sol. Let us first determine the reactions R_{A} and R_{B}
Triangle ABD is a rightangled triangle having angle
ADB = 90°
AD == AB cos 60° = 5 × 0.5 = 2.5 m
The distance of line of action the vertical load 10 kN from point A will be AD cos 60° or 2.5 × 0.5 = 1.25 m.
From triangle ACD, we have
AC = AD = 2.5m
BC = 5 – 2.5 = 2.5 m
In rightangled triangle CEB, we have
Now draw a section line (11), cutting the members 4, 5 and 7 in which forces are to be determined. Consider the equilibrium of the right part of the truss (because it is smaller than the left part).
This part is shown in Fig. 2.72 (a). Let F_{4}, F_{5} and F_{7} are the forces in members 4, 5 and 7. Let their directions are assumed as shown in Fig. 2.72 (a).
Now taking the moments of all the forces acting on the right part about point E, we get
R_{B} × BE cos 30° = F_{4} × (BE × sin 30°)
12 × BE cos 30° + F_{5} × BE = 0
or 12 × cos 30° + F_{5} = 0
F_{5} = – 12 × cos 30° = – 10.392 kN
 ve sign indicates that F_{5} is compressive.
F_{5} = 10.392 kN (Compressive). Ans.
Now taking the moments about point C of all the forces acting on the right parts, we get
12 × (2.5 BE cos 30°) = F_{7} × CE + R_{B} × BC
Negative sign shows that F_{7} is compressive.
F_{7} = 14 kN (Compressive). Ans.
These forces are same as obtained in Problem 2.33.
Problem 2.43. A truss of span 9 m is loaded as shown in Fig. 2.73. Find the reactions and forces in the members marked 1, 2 and 3.
Sol. Let us first calculate the reactions R_{A} and R_{B}.
Taking moments about A, we get
R_{B} × 9 = 9 × 3 + 12 × 6 = 27 + 72 = 99
Now draw a section (1.1), cutting the members 1,2 and 3 in which forces are to be determined. Consider the equilibrium of the left part of the truss (because it is smaller than the right part). This part is shown in Fig.2.73 (a). Let F_{1}, F_{2} and F_{3} are the forces members 1,2, and 3 respectively. Let their directions are assumed as shown in Fig. 2.73 (a)
Taking moments of all the forces acting on the left part about point D, we get
10 × 3 = F_{2} × 4
Now taking the moments of all the forces acting on the left part about point G, we get
10 × 3 + F_{1} × 4 = 0
Negative sign shows that force F_{1} is compressive.
F_{1} = 7.5 kN (Compressive) Ans.
Now taking the moments about the point C, we get
F_{2} × 3 – 9 × 3 + F_{3} × 4 = 0
or F_{2} × 3 – 27 + 7.5 × 4 = 0
Negative sign shows that force F_{2} is compressive.
F_{2} = 1.0 kN (Compressive). Ans.
Problem 2.44. A truss of 12 m span is loaded as shown in Fig. 2.74. Determine the forces in the members DG, DF and EF, using method of section.
Sol. The truss is supported on rollers at B and hence RB will be vertical. The truss is
hinged at A and hence the support sections at A will consists of a horizontal section H_{A} and a vertical section R_{A}.
In triangle AEC, AC = AE × cos 30°
= 4 × 0.866 = 3.464 m
Now length AD = 2 × AC = 2 × 3.464 = 6.928 m
Now taking the moments about A, we get
RB × 12 = 2 × AC + 1 × AD + 1 × AE
= 2 × 3.464 + 1 × 6.928 + 1 × 4 = 17.856
Now draw the section line (11), passing through members DG, DF and EF in which the forces are to be determined. Consider the equilibrium of the right part of the truss. This part is shown in Fig. 2.74 (a). Let F_{DG}, F_{FD} and F_{EF} are the forces in members DG, FD and EF respectively. Let their directions are assumed as shown in Fig. 2.74 (a). Taking moments of all forces acting on right part about point F, we get
R_{B} × 4 + F_{DG} × FG = 0
or 1.49 × 4 + F_{DG} × (4 × sin 30°) = 0
ve sign shows that the force F_{DG} is compressive.
F_{DG} = 2.98 kN (Compressive). Ans.
Now taking the moments about point D, we get
R_{B} × BD cos 30° = F_{FE} × BD × sin 30°
R_{B} × cos 30° = F_{FE} × sin 30°
= 2.58 kN (Tensile). Ans.
Now taking the moments of all forces acting on the right part about B, we get
F_{FD} × distance between F_{FD} and B = 0
F_{FD} = 0. Ans. (distance between F_{FD} and B is not zero)
2.12.7. Graphical Method. The force in a perfect frame can also be determined by a graphical method. The analytical methods (such as method of joints and method of sections) give absolutely correct results, but sometimes it is not possible to get the results from analytical method. Then a graphical method can be used conveniently to get the results. The graphical method also provides reasonable accurate results.
The naming of the various members of a frame are done according to Bow’s notations. According to this notation of force is designated by two capital letters which are written on either side of the line of action of the force. A force with letters A and B on either side of the line of action is shown in Fig. 2.75. This force will be called AB.
The following steps are necessary for obtaining a graphical solution of a frame.
(i) Making a space diagram
(ii) Constructing a vector diagram
(iii) Preparing a force table.
 Making a space diagram. The given truss or frame is drawn accurately according to some linear scale. The loads and support reactions in magnitude and directions are also shown in the frame. Then the various members of the frame arc named according to Bow’s notation. Fig. 2.76 (a) shows a given truss and the forces in the members AB, BC and AC are to be determined. Fig. 2.76 (b) shows the space diagram to same linear scale. The member AB is named as PS and so on.
Fig. 2.76 (a) Given Diagram
Fig. 2.76 (b) Space Diagram
Fig. 2.76 (c) Vector Diagram
2. Constructing a vector diagram. Fig. 2.76 (c) shows a vector diagram, which is drawn as given below:
i. Take any point p and draw pq parallel to PQ vertically downwards. Cut pq = 4 kN to same scale.
ii. Now from q draw qr parallel to QR vertically upwards and cut qr = 2 kN to the same scale.
iii. From r draw rp parallel to RP vertically upwards and cut rp = 2 kN to the same scale.
iv. Now from p, draw a line ps parallel to PS and from r, draw a line rs parallel to RS, meeting the first line at s. This is vector diagram for joint (A). Similarly the vector diagrams for joint (B) and (C) can be drawn.
3. Preparing a force table. The magnitude of a force in a member is known by the length of the vector diagram for the corresponding member, i.e. the length ps of the vector diagram will give the magnitude of force in the member PS of the frame.
Nature of the force (i.e., tensile or compressive) is determined according to the following procedure:
 In the space diagram, consider any joint. Move round that joint in a clockwise direction. Note the order of two capital letters by which the members are named. For example, the members at the joint (A) in space diagram Fig; 2.76 (b) are named as PS, SR and RP.
 Now consider the vector diagram. Move on the vector diagram in the order of the letters (i.e., ps, sr and rp).
 Now mark the arrows on the members of the space diagram of that joint (here joint A).
 Similarly, all the joints can be considered and arrows can be marked.
 If the arrow is pointing towards the joint, then the force in the member will be compressive whereas if the arrow is away from the joint, then the force in the member will be tensile
Problem.2.45. Find the forces in the members AB, AC and BC of the truss shown in Fig. 2.77.
Sol. First determine the reactions R_{B} and R_{C}
Now draw the space diagram for the truss along with load of 20 kN and the reaction R_{B} and R_{C} equal to 15 kN and 5 kN respectively as shown in Fig. 2.77 (b). Name the members AB, AC and BC according to Bow’s notations as PR, QR and RS respectively. Now construct the vector diagram as shown in Fig. 2.77 (c) and as explained below:
i. Take any point p and draw a vertical line pq downward equal to 20 kN to some suitable scale. From q draw a vertical line qs upward equal to 5 kN to the same scale to represent the reaction at C. Then sp will represent the reaction R_{B} to the scale.
Fig. 2.77 (c) Vector Diagram
ii. Now draw the vector diagram for the joint (B) From p, draw a line pr parallel to PR and from s draw a line sr parallel to SR, meeting the first line at r. Now prs is the vector diagram for the joint (B). Now mark the arrows on the joint B. The arrow in member PR will be towards to joint B, whereas the arrow in the member RS will be away from the joint. B as shown in Fig. 2.77 (b).
iii.Similarly draw the vector diagrams for joint A and C. Mark the arrows on these joints in space diagram.
Now measure the various sides of the vector diagram. The forces are obtained by multiplying the scale factor. The forces in the members are given a tabular form as.
Member 
Force in member 
Nature of force 

According to given truss 
According to Bow’s notation 

AB AC BC 
PR QR RS 
17.3 kN10.0 kN
8.7 kN 
CompressiveCompressive
Tensile 
Problem 2.46. A truss of span 7.5 m carries a point load of 1000 N at joint D as shown
in Fig. 2.78. Find the reactions and forces in the member of the truss.
Sol. First determine the reactions R_{A} and R_{B}
Taking moments about A, we get
R_{B} × 7.5 = 5 × 1000
Now draw the space diagram for the truss along with load of 1000 N and reactions R_{A} and R_{B} equal to 333 N and 667 N respectively as shown in Fig. 2.78 (b). Name the members AC, CB, AD, CD and DB according to Bow’s notations as PR, PQ, RT, QR and QS respectively. Now construct the vector diagram as shown in Fig. 2.78 (c) and as explained below:
i. Take any point s and draw .a vertical line st downward equal to load 1000 N to some suitable scale. From t draw a vertical line tp upward equal to 333 N to the same scale to represent the reaction at A. The ps will represent the reaction RB to the scale.
ii. Now draw the vector diagram for the joint A. From p, draw a line pr parallel to PR and from t draw a line tr parallel to RT, meeting the first line at r. Now prt is the vector diagram for the joint A. Now mark the arrows on the joint A. The arrow in the member PR will be towards the joint A, whereas the arrow in the member RT will be away from the joint A as shown in Fig. 2.78 (b).
iii. Similarly draw the vector diagrams for the joint C, Band D. Mark the arrows on these joints as shown in Fig. 2.78 (b).
Now measure the various sides of the vector diagrams. The forces in the members are obtained by multiplying the scale factor to the corresponding sides of the vector diagram. The forces in members are given in a tabular form as :
Member 
Force in member 
Nature of force 

According to given truss 
According to Bow’s notation 

AC AD CB CD DB 
PR RT PQ QR QS 
666 N
576.7 N 1333 N 1155 N 1555 N 
Compressive
Tensile Compressive Tensile Tensile 
Problem 2.47. Determine the forces in all the members of a cantilever truss shown in Fig. 2.79.
Sol. In this case the vector diagram can be drawn without knowing the reactions. First of all the space diagram for the truss along with loads of 1000 N of joints B and C. Name the members AB, BC, CD, DE, AD and BD according to Bow’s notation as PT, QS, SR, RV, VT and ST respectively. Now construct the vector diagram as shown in Fig. 2.79 (c) and as explained below:
 The vector diagram will be started from joint C where forces in two members are unknown. Take any point q and draw a vertical line qr downward equal to load 1000 N to some suitable scale. From r, draw a line rs parallel to RS and from q draw a line qs parallel QS, meeting the first line at s. Now qrs is the vector diagram for the joint C. Now mark the arrows on the joint C. The arrow in the member RS will be towards the joint C, whereas the arrow in the member SQ will be away from the joint C as shown in Fig. 2.79 (b).
Fig.2.79 (c) Vector Diagram
ii.Now draw the vector diagram for the joints B and D similarly.
Mark the arrows on these joints as shown in Fig. 2.79 (b).
Now measure the various sides of the vector diagram. The forces in the members are given in a tabular form as:
Member 
Force in member 
Nature of force 

According to given truss 
According to Bow’s notation 

AB BC CD DE AD BD 
PT QS SR RV VT ST 
1333 N
1333 N 1666 N 2500 N 833 N 1000 N 
Tensile
Tensile Compressive Compressive Tensile Compressive 
From the vector diagram, the reactions R_{1} and R_{2} at A and E can be determined in magnitude and directions.
Reaction R_{2} = rv = 2500 N. This will be towards point E.
Reaction R_{1} = vp = 2000 N. This will be away from the point A as shown in Fig. 2.79 (b). The reaction R1 is parallel to vp.
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