framed structures - i.k. international publishing house pvt.ltd. 40 • basic structural analysis...

General Introduction • 33

Gate 87

(k)

Gate 97

(l)

Gate 92

(m)

Framed Structures 2Objectives:

Differentiate between perfect, imperfect and redundant frames. To compute the member forces ina frame by graphical method. To compute the forces in a truss by method of joints. To computethe forces in a truss by method of sections. To compute the forces in a truss by method of tensioncoefficients.

2.1 PLANE FRAMESThe plane trusses are here by termed as plane frames. In this chapter, pin jointed plane frameswhich are statically determinate are considered. A statically determinate frame can be completelyanalysed by using statics. The number of unknown forces is the same as the number of equationsobtained from static equilibrium. The main methods in analysing statically determinate pin jointedplane frames are (i) Graphical solution – Force diagram (ii) Method of resolution at joints (iii)Method of sections and (iv) Tension coefficient method. The first three methods are used in planeframes (trusses) and the fourth method is used for analysing the space frame.

A truss is an assemblage of three or more members which are hinged or pinned. A load appliedon the truss is transmitted to all joints so that the members are in pure compression or tension.

Consider a simple truss made up of three members hinged at the ends to form a triangle. A loadW is acting at the apex of the triangle and due to symmetry, the reactions are W/2 at each support.

W

W/2W/2

B

A C

FIG. 2.1

Due to the application of the load, the joint A and C pulls the member out and for equilibrium atjoint A there should be an equal and opposite force should move away from joint A. In otherwords,member AC in tension. Due to the downward load W, the joint B is pushed vertically downwards.The forces in the members AB and BC are in compression as the joint B is pushed. A force in the

36 • Basic Structural Analysis

Framed Structures • 35

member AB should counteract this force and hence this member force act towards the joint. Hence,the effect of the application of the load results in pure compression or tension.

W

W/2W/2

B

A C

FIG. 2.2

It should be noted that the member forces marked in the above figure are the internal forces. It canbe remembered as, when the force is acting towards the joint it is compressive in nature. When theforce is acting away from the joint, it is tensile in nature. The compression members are called asstrut while tension members are called as tie.

2.2 PERFECT, IMPERFECT AND REDUNDANT PIN JOINTED FRAMES

2.2.1 Perfect FrameA perfect frame has just sufficient number of members to prevent the frame being unstable. Thesimplest perfect frame is in the form of a triangle as shown in Fig. 2.2. It has three members andthree joints, namely A, B and C. If two more members are added to form another triangle, then onemore joint has been added. The frame will remain as a perfect frame as long as any number oftriangles are added. Thus, the frame will remain as a perfect frame. In essence, the first three jointsor pins require three bars to connect them. While the remaining ( j−3) joints require 2( j−3) extrabars. If m is the total number of bars or members required to connect j joints together.

m−3 = 2( j−3)m = 2 j−3

i.e., number of members to form a perfect frame = Twice the number of joints–3.

2.2.2 Imperfect FrameAn imperfect frame is one where the number of members is less than (2 j − 3). In Fig. 2.3 thenumber of joints are 4 and the number of members are four and hence



member AB should counteract this force and hence this member force act towards the joint. Hence,the effect of the application of the load results in pure compression or tension.

W

W/2W/2

B

A C

FIG. 2.2

It should be noted that the member forces marked in the above figure are the internal forces. It canbe remembered as, when the force is acting towards the joint it is compressive in nature. When theforce is acting away from the joint, it is tensile in nature. The compression members are called asstrut while tension members are called as tie.

2.2 PERFECT, IMPERFECT AND REDUNDANT PIN JOINTED FRAMES

2.2.1 Perfect FrameA perfect frame has just sufficient number of members to prevent the frame being unstable. Thesimplest perfect frame is in the form of a triangle as shown in Fig. 2.2. It has three members andthree joints, namely A, B and C. If two more members are added to form another triangle, then onemore joint has been added. The frame will remain as a perfect frame as long as any number oftriangles are added. Thus, the frame will remain as a perfect frame. In essence, the first three jointsor pins require three bars to connect them. While the remaining ( j−3) joints require 2( j−3) extrabars. If m is the total number of bars or members required to connect j joints together.

m−3 = 2( j−3)m = 2 j−3

i.e., number of members to form a perfect frame = Twice the number of joints–3.

2.2.2 Imperfect FrameAn imperfect frame is one where the number of members is less than (2 j − 3). In Fig. 2.3 thenumber of joints are 4 and the number of members are four and hence


The combined force diagram is drawn as follows:1. Starting from the force AB, the known forces, viz. AB,BC and CA working clockwise round

the frame, are set down in order and to scale as ab,bc and ca.2. Consider the apex joint ab the centre of the clock, and the letters are read clockwise around

this centre.3. Therefore, from the letter ‘b’ draw a line parallel to B1 and from the letter ‘a’ draw a line

parallel to 1A and both intersect at the point 1. From the force diagram, the magnitude ofthe forces are obtained. The member force A1 = Member force B1 = 5.7× 5 = 28.5 kN.Member force C1 = 2.8×5 = 14 kN.

4. Consider the joint at the left hand support reaction. Read clockwise in the frame diagram.Member A1 is inclined and in the force diagram a to 1 is downwards and hence mark thearrow correspondingly in the force diagram from 1 to C it is towards right mark this directionat that joint.

BA

C25

60°

50 kN

1

60°

25

FIG. 2.6 Frame diagram

5. Consider the apex joint. Read clockwise B1 is inclined member. In the force diagram, b to 1is upwards and hence mark the arrow upward for member B1 at the apex joint. Member 1Ais a sloping member. From the force diagram, 1 to a is upwards. Therefore, mark the arrowupwards at the apex joint. (compression).

5.7

cm

1 Scale 1 cm = 5 kNc

b

a

FIG. 2.7 Force diagram

38 • Basic Structural Analysis42 • Basic Structural Analysis

A B 25 kN50 kN

4 m4 m

6 m

2

1

3

C

D

FIG. 2.16

2

1

3

Reaction at the top hinge

Reaction at the rollerc

a

b

d

FIG. 2.17

Force in kNMember Strut Tie

A2 − 36B3 − 36C3 43 −C1 86 −12 − 41.51D − 50.023 50 −

Reaction at hinge 103 −Reaction at roller 66 −

2.4 METHOD OF JOINTS

In the method of joints, the member forces are determined using the equilibrium conditions atthat particular joint. In this resolution of forces at the joint, the free body diagram at that joint isconsidered. The procedure is explained as follows:

Framed Structures • 39 38 • Basic Structural Analysis

6. Consider the right hand support reaction and again read clockwise. 1B is inclined member.In the force diagram 1 to b is downwards. Therefore, mark the arrow downwards at joint.

7. To determine the member force C1, from the force diagram it is noted that the force is actingfrom c towards left to 1. Mark the arrow C1 in the same direction in frame diagram.

The final forces are listed below.

Forces in kNMember Strut Tie

A1 28.5 −B1 28.5 −C1 − 14.0

2.3.1 Numerical Problems on Symmetrical Frame and Symmetrical Loading

EXAMPLE 2.1: Determine the forces in the members graphically.

B

C

D

E

A3 m3 m 3 mF G

20 kN

10 kN10 kN

FIG. 2.8

SOLUTIONDue to Symmetry:

VA = VB =10+20+10

2= 20 kN

Using the Bow notations

C

DA1 5

2 3 4

B

3 m3 m 3 mE

20 kN

10 kN10 kN

20 kN 20 kN

FIG. 2.9


A B 25 kN50 kN

4 m4 m

6 m

2

1

3

C

D

FIG. 2.16

2

1

3

Reaction at the top hinge

Reaction at the rollerc

a

b

d

FIG. 2.17


A2 − 36B3 − 36C3 43 −C1 86 −12 − 41.51D − 50.023 50 −

Reaction at hinge 103 −Reaction at roller 66 −

2.4 METHOD OF JOINTS

In the method of joints, the member forces are determined using the equilibrium conditions atthat particular joint. In this resolution of forces at the joint, the free body diagram at that joint isconsidered. The procedure is explained as follows:

40 • Basic Structural Analysis Framed Structures • 39

The combined force diagram is drawn as follows:1. The loads AB, BC, CD, DE and EA are marked to scale.2. Start with a joint of the left hand reaction. Draw a line through the point ‘a’ a line parallel to

A1 and from the point ‘e’ draw a horizontal line parallel to 1E.They intersect at a point and is marked as 1.

3. Move to the next joint where 10 kN load is acting; Through ‘b’ draw a line parallel to B2 andfrom 1 draw a line parallel to 12. These two lines intersect at the point 2.

4. After locating point 2 in the force diagram. Consider the joint where the members 1−2,2−3, 3−E and E − 1 meet. The point 3 is located on intersection of line e1 and a line drawnthrough point 2 and parallel to 2−3 of the frame diagram.

5. The point 4 is located by drawing a line through 3 and parallel to 3−4 in the load diagramwhich intersects the line drawn from ‘C’ in the force diagram and parallel to C−4.

6. The point 5 is marked from point 4. Draw a line parallel to 4−5 of the frame diagram frompoint 4 and this cuts the horizontal line through ‘e’.

7. Using the force diagram, the magnitude of the forces and the directions are obtained.8. It is to be remembered that the arrows indicate not what is being done to the member but

what the member is doing at the joint at each end. Hence, if the arrow is acting towards thejoint it is compression and if the arrow is acting away from the joint then it is tensile force.

Force in kNMember Strut TieA1 D5 36.5 −B2 C4 31.5 −E1 E5 − 31.012 45 8.25 −23 34 − 8.25

C

DA1 5

2 3 4

B

3 m3 m 3 m

E

20 kN

10 kN10 kN


5,1

2

3

4

a

b

e

c

d


Framed Structures • 41 Framed Structures • 39

The combined force diagram is drawn as follows:1. The loads AB, BC, CD, DE and EA are marked to scale.2. Start with a joint of the left hand reaction. Draw a line through the point ‘a’ a line parallel to

A1 and from the point ‘e’ draw a horizontal line parallel to 1E.They intersect at a point and is marked as 1.

3. Move to the next joint where 10 kN load is acting; Through ‘b’ draw a line parallel to B2 andfrom 1 draw a line parallel to 12. These two lines intersect at the point 2.

4. After locating point 2 in the force diagram. Consider the joint where the members 1−2,2−3, 3−E and E − 1 meet. The point 3 is located on intersection of line e1 and a line drawnthrough point 2 and parallel to 2−3 of the frame diagram.

5. The point 4 is located by drawing a line through 3 and parallel to 3−4 in the load diagramwhich intersects the line drawn from ‘C’ in the force diagram and parallel to C−4.

6. The point 5 is marked from point 4. Draw a line parallel to 4−5 of the frame diagram frompoint 4 and this cuts the horizontal line through ‘e’.

7. Using the force diagram, the magnitude of the forces and the directions are obtained.8. It is to be remembered that the arrows indicate not what is being done to the member but

what the member is doing at the joint at each end. Hence, if the arrow is acting towards thejoint it is compression and if the arrow is acting away from the joint then it is tensile force.

Force in kNMember Strut TieA1 D5 36.5 −B2 C4 31.5 −E1 E5 − 31.012 45 8.25 −23 34 − 8.25

C

DA1 5

2 3 4

B

3 m3 m 3 m

E

20 kN

10 kN10 kN


5,1

2

3

4

a

b

e

c

d




A1 12.5 −B3 12.5 −12 11.8 −23 11.8 −1E − 24.02D − 18.53C − 24.0

2.3.3 Numerical Example on Cantilever FramesEXAMPLE 2.3: Use the graphical method and determine the member forces and the reaction atthe supports.

B

25 kN50 kN

4 m4 m

6 m

FIG. 2.15

In the force diagram of the cantilever truss, these is no need of reactions before starting of the same.

1. The load line is drawn as in the previous examples. a−b, b− c, start with a joint at the freeend and reading clockwise, draw line from b parallel to B3 of the frame diagram and frompoint C draw line parallel to C3 and the intersection of the above lines give point 3.

2. The point 2 is located as follows. From the point 3, draw a line parallel to 23 and from ‘a’draw a line parallel to A2. The intersection gives the point 2.

3. The point 1 is obtained as follows. A line is drawn from point 3, parallel to 3−C and frompoint 2 draw line parallel to 21 and the intersection of 1.

4. After marking the points 1, 2 and 3 the member forces and their nature are tabulated here.


2.3.2 Numerical Example on Frame with Loads Suspended from the Bottom Chordof the Frame in Addition to Loads on the Top Chord

EXAMPLE 2.2: Find the forces in all the members of the truss graphically

3 m

60° 60°

10 kN

20 kN

3 m

3 m

10 kN

FIG. 2.12

1. The loads AB, BC, CD, DE and EA are marked to scale.2. Start with the left support joint. Read clockwise and draw. Draw a line from ‘a’ parallel to

A−1 in the load diagram. Draw another line e from the load diagram and parallel to the 1Eof the frame diagram. They intersect at the point 1.

3. The point 2 is located by considering the joint adjacent to the left support. Draw a line frompoint 1 parallel to 1−2 of the frame diagram. From the load diagram, draw a horizontal linethrough d and the line intersect at point 2.

4. From the point 2 draw a line parallel to 2−3 of the frame diagram and from point ‘e’ drawa horizontal line and the intersection of above two lines give point 3.

5. Determine the magnitude and nature of the forces from the forces diagram and tabulate.

A

10

20 kN

20

B

D

CE

10

201

2

3


a,c

b,e

d

1

2

3



1. Check the stability and assess its determinacy of the truss.2. If the truss is of cantilever type, the reactions need not be computed in general. If the truss

is stable and determinate where one support is hinge and the other support is on rollers;compute the reactions at the supports.

3. Draw the free body diagram at each joint and analyse the member forces at a joint whereonly two members meet. Then, consider the adjacent joint where only two unknown forcesto be determined. This process is repeated till the analysis of all joints are completed.

4. The results are tabulated along with magnitude of member forces and the nature of forces.The forces are tensile if they are pulling (acting away) the joint. The forces are compressivein nature if they are pushing (acting towards) the joint.

NUMERICAL EXAMPLEEXAMPLE 2.4: Analyse the truss shown in Fig. 2.18 by method of joints. (May 2010, RVCE)

2

70 kN

1

4 65

20

H4

V4

3

3 m

3 m3 m

V6

FIG. 2.18

SOLUTIONThe reactions at the supports are found out by summing up the forces in horizontal and verticaldirections and also by taking moments of applied forces about the hinge support.

∑H = 0; 20−H4 = 0

H4 = 20 kN

∑V = 0; V4 +V6 = 70 kN

∑M4 = 0; 20×3+70×3−6V6 = 0

V6 = 45 kN

∴ V4 = 25 kN


1. Check the stability and assess its determinacy of the truss.2. If the truss is of cantilever type, the reactions need not be computed in general. If the truss

is stable and determinate where one support is hinge and the other support is on rollers;compute the reactions at the supports.

3. Draw the free body diagram at each joint and analyse the member forces at a joint whereonly two members meet. Then, consider the adjacent joint where only two unknown forcesto be determined. This process is repeated till the analysis of all joints are completed.

4. The results are tabulated along with magnitude of member forces and the nature of forces.The forces are tensile if they are pulling (acting away) the joint. The forces are compressivein nature if they are pushing (acting towards) the joint.

NUMERICAL EXAMPLEEXAMPLE 2.4: Analyse the truss shown in Fig. 2.18 by method of joints. (May 2010, RVCE)

2

70 kN

1

4 65

20

H4

V4

3

3 m

3 m3 m

V6

FIG. 2.18

SOLUTIONThe reactions at the supports are found out by summing up the forces in horizontal and verticaldirections and also by taking moments of applied forces about the hinge support.

∑H = 0; 20−H4 = 0

H4 = 20 kN

∑V = 0; V4 +V6 = 70 kN

∑M4 = 0; 20×3+70×3−6V6 = 0

V6 = 45 kN

∴ V4 = 25 kN


nm < 2n j −34 < (2×4)−34 < 5

B

A C

D

FIG. 2.3

n j = 4nm = 4

2.2.3 Redundant FrameA redundant frame is one where the number of member or members are more than (2 j − 3). InFig. 2.4, the number of joints are

nm > (2n j −3)6 > (2×4)−36 > 5

B

A C

D

FIG. 2.4

n j = 4nm = 6

i.e.,’ redundant frame is having more member/members necessary to produce stability.

2.3 GRAPHICAL SOLUTION-FORCE DIAGRAMSConsider the perfect frame in Fig. 2.5. The forces include the applied load and the reactions atP and Q.

BA

P QC

50 kN

1

FIG. 2.5

Due to symmetry the reactions are 25 kN at joint P and Q respectively. In graphical method theloads and reactions are read clockwise. They are represented by capital letters written on either sideof the force, commonly known as ‘BOW’S Notation’. They are denoted with letters A,B,C and thespace inside the member is denoted by numbers. Note that the letters A,B,C are marked in the mid-dle length of the members and not at the joints. The load at the apex 50 kN is denoted as ‘load AB’.The reaction at the right support is denoted as ‘load BC’. The reaction at the left support 25 kN isdenoted as ‘load CA’. The member force in the horizontal member is denoted as ‘force 1C’.


Joint 4

∑H = 0;F45 −20 = 0F45 = 20 kN

∑V = 0;−F14 +25 = 0F14 = 25 kN

4

F14

F45

25 kN

20

Joint 1

∑V = 0;25−F15 sin45 = 0F15 = 35.4 kN

∑H = 0;20−F12 +F15 cos45 = 020−F12 +35.4cos45 = 0F12 = 45.03 kN

1

F15

F12

25 kN

20 kN

45°

Joint 2

45.03−F23 = 0F23 = 45.03 kN

∑V = 0; F25 = 70.0 kN F25

F23

70 kN

45.03

2

Joint 5

F53

70 kN

20

35.4

5

45°

45°

∑V = 0;F53 cos45+35.4sin45−70 = 0F53 = 63.61 kN


nm < 2n j −34 < (2×4)−34 < 5

B

A C

D

FIG. 2.3

n j = 4nm = 4

2.2.3 Redundant FrameA redundant frame is one where the number of member or members are more than (2 j − 3). InFig. 2.4, the number of joints are

nm > (2n j −3)6 > (2×4)−36 > 5

B

A C

D

FIG. 2.4

n j = 4nm = 6

i.e.,’ redundant frame is having more member/members necessary to produce stability.

2.3 GRAPHICAL SOLUTION-FORCE DIAGRAMSConsider the perfect frame in Fig. 2.5. The forces include the applied load and the reactions atP and Q.

BA

P QC

50 kN

1

FIG. 2.5

Due to symmetry the reactions are 25 kN at joint P and Q respectively. In graphical method theloads and reactions are read clockwise. They are represented by capital letters written on either sideof the force, commonly known as ‘BOW’S Notation’. They are denoted with letters A,B,C and thespace inside the member is denoted by numbers. Note that the letters A,B,C are marked in the mid-dle length of the members and not at the joints. The load at the apex 50 kN is denoted as ‘load AB’.The reaction at the right support is denoted as ‘load BC’. The reaction at the left support 25 kN isdenoted as ‘load CA’. The member force in the horizontal member is denoted as ‘force 1C’.


Hence the final forces are

2

70 kN

1

4 65

20 kN

20

3

3 m

20

25

45 45

4563.61

7035.4

FIG. 2.19

EXAMPLE 2.5: Analyse the truss shown in Fig. 2.20 by method of joints. Indicate the memberforces on a neat sketch of the truss. (MSRIT, Jan. 2010)

BD

θ1

FA

4 m

E3 m 3 m

20 kN

C

4 m

FIG. 2.20

SOLUTIONTaking moment about the hinge at A;

∑MA = 0; 20×6−8HC = 0HC = 15 kN

∑H = 0; ∴ HA = 15 kN

∑V = 0; VA = 20 kN


∑MA = 0;2(1.5)+4(4.5)−6VD = 0VD = 3.5 kNVA = 2.5 kN

Consider Joint A

FAB

FAE

2.5 kN

A60°

∑V = 0FAB sin60 = 2.5FAB = 2.89 kN

∑H = 0FAE = FAB cos60= 2.89cos60= 1.45 kN

Joint D

FCD

FDE

3.5 kN

60°D

∑V = 0FCD sin60 = 3.5FCD = 4.04 kN

∑H = 0FCD cos60−FDE = 0FDE = 4.04cos60 = 2.02 kN

Joint E

FEB

FEC

1.45 kN E 2.02 kN

60° 60°

∑V = 0FEC sin60+FEB sin60 = 0FEC = −FEB

∑H = 02.02+FEC cos60−1.45−FEB cos60 = 02.02−FEB cos60−1.45−FEB cos60 = 00.57 = 2FEB cos60FEB = 0.57; FEC = −0.57


Joint C

25 kN

Cθ

FBC

FCD

∑V = 0FCD sinθ = 25

FCD =25

sinθ=

250.6

= 41.67 kN

∑H = 0−FBC +FCD cosθ = 0−FBC +41.67×0.8 = 0

FBC = 33.34 kN

Joint B

50 kN

33.34

FBD

FAB

B

∑H = 0−FAB +33.34 = 0

FAB = 33.34 kN

∑V = 0FBD −50 = 0FBD = 50 kN

Joint D

50 kN 41.67

FDE

FAD

θ

θ

θ1

D

cosθ = 4/5 = 0.8sinθ = 3/5 = 0.6

sinθ1 = 4/5 = 0.8cosθ1 = 3/5 = 0.6

∑H = 0

−41.67cosθ+FDE cosθ−FAD sinθ1 = 0−41.67(0.8)+FDE(0.8)−0.8 FAD = 0FDE −FAD = 41.67 (2.1)


Joint C

FCE

FBC

4 kN

FCD

= 4.04 kN

60° 60°

C

∑H = 0;FBC −FCE cos60−FCD cos60 = 0FBC −FCE cos60 = 4.04cos60

∑V = 0;−4−FCE sin60+4.04sin60 = 0FCE = −0.57 kN

Substituting FCE in the above equation

FBC = 1.73 kN

2 kN

1.45

1.73

0.572.89 0.57 4.04

2.02

4 kN

A

B C

DE

FIG. 2.23

EXAMPLE 2.7: Determine the magnitude and nature of forces in all the number of the pin jointedplane truss shown in Fig. 2.24 by method of joints. (VTU, June 08)

6 m

A

E

B C

D

4 m

25 kN 50 kN 25 kN

4 m

FIG. 2.24

52 • Basic Structural AnalysisFramed Structures • 51

∑V = 0;

−50−41.67sinθ+FAD cosθ1 +FDE sinθ = 0

−50−41.67(0.6)+0.6FAD +0.6FDE = 0

0.6FAD +0.6FDE = 75

FDE +FAD = 125 (2.2)

Solving Eqns. (1) and (2);FDE = 83.3 kN, FAD = 41.67 kN

Joint E

83.3

FAE

θ2

E

tanθ2 = 8/6

sinθ2 = 0.8

cosθ2 = 0.6

∑V = 0; FAE = 83.3cosθ2 = 50 kN

50

A

E

B C

D

33.3

8.33

41.6741.67 50

25 kN 50 kN 25 kN

33.3

FIG. 2.25


EXAMPLE 2.9: Determine the forces in members and tabulate neatly. Use method of joints.(VTU, Dec. 06)

8 kN

H

8 kN

4 kN

C

F

D EA

8 kN

4 kN2 m

G

Bθ

4 m

3 m 3 m 3 m 3 m

FIG. 2.28

SOLUTIONDue to Symmetry: VA = VB =

2(4)+3(8)2

= 16 kNJoint A

16 kN

4 kN

FAF

FAC

θA A

∑V = 0−4−FAF sinθ+16 = 012−0.555 FAF = 0FAF = 21.62 kN

∑H = 0FAC −FAF cosθ = 0FAC = 21.62×0.832 = 18.0 kN

Joint C

18.0

FCD

C

∑H = 0;FCD = 18.0 kN


∑V = 0;

−50−41.67sinθ+FAD cosθ1 +FDE sinθ = 0

−50−41.67(0.6)+0.6FAD +0.6FDE = 0

0.6FAD +0.6FDE = 75

FDE +FAD = 125 (2.2)

Solving Eqns. (1) and (2);FDE = 83.3 kN, FAD = 41.67 kN

Joint E

83.3

FAE

θ2

E

tanθ2 = 8/6

sinθ2 = 0.8

cosθ2 = 0.6

∑V = 0; FAE = 83.3cosθ2 = 50 kN

50

A

E

B C

D

33.3

8.33

41.6741.67 50

25 kN 50 kN 25 kN

33.3

FIG. 2.25


EXAMPLE 2.9: Determine the forces in members and tabulate neatly. Use method of joints.(VTU, Dec. 06)

8 kN

H

8 kN

4 kN

C

F

D EA

8 kN

4 kN2 m

G

Bθ

4 m

3 m 3 m 3 m 3 m

FIG. 2.28

SOLUTIONDue to Symmetry: VA = VB =

2(4)+3(8)2

= 16 kNJoint A

16 kN

4 kN

FAF

FAC

θA A

∑V = 0−4−FAF sinθ+16 = 012−0.555 FAF = 0FAF = 21.62 kN

∑H = 0FAC −FAF cosθ = 0FAC = 21.62×0.832 = 18.0 kN

Joint C

18.0

FCD

C

∑H = 0;FCD = 18.0 kN


Joint D

θ1

30 kN

D120

FDB

FDE

∑V = 0FDB sinθ1 −30 = 0

FDB =30

0.707= 42.43 kN

∑H = 0−120+FDE +42.43cosθ1 = 0

FDE = −42.43×0.707+120 = −30 kN+12090 kN

1.25 m

2.50 m

134.17

134.17

30 kN 30 kN

60 kN

90120 DA

B

C

HG

E F

42.4330

FIG. 2.30

EXAMPLE 2.11: Determine the forces in all members of all Bollman truss by method of joints.(VTU, July 2005)

20 kN 20 kN 20 kN

10 m 10 m 10 m 10 mA B

HG

C D E

F

θ1θ

1θ

2

θ3

10 m

FIG. 2.31


Joint D

θ1

30 kN

D120

FDB

FDE

∑V = 0FDB sinθ1 −30 = 0

FDB =30

0.707= 42.43 kN

∑H = 0−120+FDE +42.43cosθ1 = 0

FDE = −42.43×0.707+120 = −30 kN+12090 kN

1.25 m

2.50 m

134.17

134.17

30 kN 30 kN

60 kN

90120 DA

B

C

HG

E F

42.4330

FIG. 2.30


20 kN 20 kN 20 kN

10 m 10 m 10 m 10 mA B

HG

C D E

F

θ1θ

1θ

2

θ3

10 m

FIG. 2.31


SOLUTIONDue to Symmetry

VA = VB =3(20)

2= 30 kN

From geometry

tanθ1 =1030

tanθ2 = 10/20 tanθ3 = 10/10sinθ1 = 0.316 sinθ2 = 0.447 sinθ3 = 0.707cosθ1 = 0.949 cosθ2 = 0.895 cosθ3 = 0.707

Joint C,D,EFCF = 20 kN, FDG = 20 kN, FEH = 20 kN

20 kN

FCF

C

Joint F

θ1

θ3

20 kN

FFB

FFA

F

∑H = 0;−FFA cosθ3 +FFB cosθ1 = 0−0.707FFA +0.949FFB = 0

∑V = 0;−20+FFA sinθ3 +FFB sinθ1 = 00.707FFA +0.316FFB = 20

on solving the above equations

FFA = 21.22 kNFFB = 22.37 kN

Joint G

θ2

θ2

20 kN

FGB

FGA

G

∑H = 0;−FGA cosθ2 +FGB cosθ2 = 0FGA = FGB

∑V = 0;−20+FGA sinθ2 +FGB sinθ2 = 02 FGA sinθ2 = 202 FGA(0.447) = 20FGA = 22.37 kN

Page 23

movement with respect to horizontal direction. Hence, this phenomenon of ground excitation of a tall water tank can be considered as a single degree of freedom/kinematic indeterminacy.Consider another example wherein a rigid body is resting on three springs. The displacement of the springs can be through two degrees of freedom, i.e., the displacement at any point of the body and the rotation also. In real situations like an aeroplane in the sky, can be considered a oating body with no external constraints. It has three displacement components in x, y, z directions and rotation components w

x, w

y and w

z. This rigid body in space has six degrees of freedom. A rigid

body in a coplanar system has three degrees of freedom; two translational displacements in x and y directions and a rotation.

Page 26

When the same system is considered as a deformable system shown in Fig. 1.22(b), every point in the beam moves from its initial position of the undeformed position. Thus, the system possesses in nite number of degrees of freedom. The beam is discretized and the nodes are marked as 1 to 4. The beam has four degrees of freedom neglecting the horizontal displacement and the slopes at the nodes. Figure 1.22(b) refers to the vertical displacements at the nodes.

Page 28

number of equation to be solved are a few, then we can be use Cramer’s rule, Method of leading coef cients, Gauss elimination methods, Gauss Seidel iteration method, Relaxation method, Matrix inversion method, Cholesky’s method and Crout’s method. If the number of equations is unwieldy, we have to use other computing tools, viz., MATLAB package.For a linear structural system, we can apply the principle of superposition. Thus, displacements, stresses and strains can be evaluated.

FFA

= 21.22 kN F

FB = 15.81 kN

Page 59


Joint D

θ1

30 kN

D120

FDB

FDE

∑V = 0FDB sinθ1 −30 = 0

FDB =30

0.707= 42.43 kN

∑H = 0−120+FDE +42.43cosθ1 = 0

FDE = −42.43×0.707+120 = −30 kN+12090 kN

1.25 m

2.50 m

134.17

134.17

30 kN 30 kN

60 kN

90120 DA

B

C

HG

E F

42.4330

FIG. 2.30


20 kN 20 kN 20 kN

10 m 10 m 10 m 10 mA B

HG

C D E

F

θ1θ

1θ

2

θ3

10 m

FIG. 2.31


Joint D

θ1

30 kN

D120

FDB

FDE

∑V = 0FDB sinθ1 −30 = 0

FDB =30

0.707= 42.43 kN

∑H = 0−120+FDE +42.43cosθ1 = 0

FDE = −42.43×0.707+120 = −30 kN+12090 kN

1.25 m

2.50 m

134.17

134.17

30 kN 30 kN

60 kN

90120 DA

B

C

HG

E F

42.4330

FIG. 2.30


20 kN 20 kN 20 kN

10 m 10 m 10 m 10 mA B

HG

C D E

F

θ1θ

1θ

2

θ3

10 m

FIG. 2.31


SOLUTIONDue to Symmetry

VA = VB =3(20)

2= 30 kN

From geometry

tanθ1 =1030

tanθ2 = 10/20 tanθ3 = 10/10sinθ1 = 0.316 sinθ2 = 0.447 sinθ3 = 0.707cosθ1 = 0.949 cosθ2 = 0.895 cosθ3 = 0.707

Joint C,D,EFCF = 20 kN, FDG = 20 kN, FEH = 20 kN

20 kN

FCF

C

Joint F

θ1

θ3

20 kN

FFB

FFA

F

∑H = 0;−FFA cosθ3 +FFB cosθ1 = 0−0.707FFA +0.949FFB = 0

∑V = 0;−20+FFA sinθ3 +FFB sinθ1 = 00.707FFA +0.316FFB = 20

on solving the above equations

FFA = 21.22 kNFFB = 22.37 kN

Joint G

θ2

θ2

20 kN

FGB

FGA

G

∑H = 0;−FGA cosθ2 +FGB cosθ2 = 0FGA = FGB

∑V = 0;−20+FGA sinθ2 +FGB sinθ2 = 02 FGA sinθ2 = 202 FGA(0.447) = 20FGA = 22.37 kN

Page 23

movement with respect to horizontal direction. Hence, this phenomenon of ground excitation of a tall water tank can be considered as a single degree of freedom/kinematic indeterminacy.Consider another example wherein a rigid body is resting on three springs. The displacement of the springs can be through two degrees of freedom, i.e., the displacement at any point of the body and the rotation also. In real situations like an aeroplane in the sky, can be considered a oating body with no external constraints. It has three displacement components in x, y, z directions and rotation components w

x, w

y and w

z. This rigid body in space has six degrees of freedom. A rigid

body in a coplanar system has three degrees of freedom; two translational displacements in x and y directions and a rotation.

Page 26

When the same system is considered as a deformable system shown in Fig. 1.22(b), every point in the beam moves from its initial position of the undeformed position. Thus, the system possesses in nite number of degrees of freedom. The beam is discretized and the nodes are marked as 1 to 4. The beam has four degrees of freedom neglecting the horizontal displacement and the slopes at the nodes. Figure 1.22(b) refers to the vertical displacements at the nodes.

Page 28

number of equation to be solved are a few, then we can be use Cramer’s rule, Method of leading coef cients, Gauss elimination methods, Gauss Seidel iteration method, Relaxation method, Matrix inversion method, Cholesky’s method and Crout’s method. If the number of equations is unwieldy, we have to use other computing tools, viz., MATLAB package.For a linear structural system, we can apply the principle of superposition. Thus, displacements, stresses and strains can be evaluated.

FFA

= 21.22 kN F

FB = 15.81 kN

Page 59


Joint C

60

60 kN

C

∑H = 0∴ FBC = 0

CB

D

60.0 kN

60.0 kN78.1 kN

78.1 kN

50 kN

E

A

FIG. 2.33

EXAMPLE 2.13: Find the forces in the members of the truss in Fig. 2.34. Tabulate your resultsneatly. (VTU, July 2004)

10 k

N

10 k

N

20 k

N

4 m

H

4 m4 m

30° 60°

E

C

A B

D

G

F10 kN

FIG. 2.34


SOLUTIONThe reactions are found using the equilibrium equations.Summing up all the forces in the vertical direction;∑V = 0;

VA +VF = 50+40VA +VF = 90 (1)

∑MA = 0;

50(6)+40(9)−6VF = 0VF = 110 kN

∴ VA = 90−110 = −20 kN

This means that the direction of VA is downwards.Joint A

FAH

FAB

20 kN

A

θ

Considering the joint A; a vertical downward force of 20 kN is acting down. To balance this thereshould be an upward force of 20 kN to balance. This is due to FAH since FAB is a horizontal forcewhich do not give a vertical component.

∴ FAH sinθ = 20

FAH =20

sinθ= 36 kN

Resolving the forces horizontally at the joint A;

FAB = FAH cosθ= 36×0.832= 29.95 kN

Joint E

tanθ1 = 4/3sinθ1 = 0.8cosθ1 = 0.6

40 kNF

DE

FEF

θ1 E


S.No Member Force (kN) Nature1 AC 42.20 Compressive2 CD 42.20 Compressive3 DG 25.77 Compressive4 GB 29.76 Compressive5 BF 25.77 Tensile6 FE 25.77 Tensile7 AE 51.55 Tensile8 CE 20.00 Compressive9 ED 29.24 Tensile

10 FD 0.00 —11 FG 0.00 —

EXAMPLE 2.14: Analyse the truss shown in the Fig. 2.35 by method of joints check the forcesin the members AC, AD & BD by the method of sections. (VTU, Aug. 2000)

F

DC

E

A B4 m

3 m

3 m

20 kN

10 kN

FIG. 2.35

Joint E

10 kN FEF

FEC

E ∑H = 0 ∑V = 010−FEF = 0 FEC = 0FEF = 10 kN


Joint C

FCD

FCE

60°

60°

30°

30°

20 kN

42.2

C

∑H = 020cos60−FCD cos30−FCE cos60+42.2cos30 = 0FCD cos30+FCE cos60 = 46.550.866 FCD +0.5 FCE = 46.55

∑V = 0−20sin60−FCD sin30+FCE sin60+42.2sin30 = 0−0.5 FCD +0.866 FCE = −3.78FCD = 42.20 kNFCE = 20.00 kN

Joint G

30°

60°30°

FGB

= 29.76

FGD

FGF

G

∑H = 0FGD cos30−FGB cos30+FGF cos60 = 00.866 FGD −0.866 FGB +0.5 FGF = 00.866 FGD +0.5 FGF = 0.866×29.760.866 FGD +0.5 FGF = 25.77

∑V = 0−FGD sin30+FGF sin60+29.76sin30 = 0−0.5 FGD +0.866 FGF = −14.88FGD = 25.77 kN, FGF = 0

Joint D

60°

60°

60°

30°30°

25.77

10 kN

42.20

FDE

D

∑V = 0;−10sin60+42.20sin30−FDE sin60+25.77sin30 = 0FDE = 29.24 kN

Page 64

Page 65

EXAMPLE 2.14: Analyse the truss shown in Fig. 2.35 by method of joints. (VTU, Aug. 2000)

Page 64

Page 65


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Page 65


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Page 65


66 • Basic Structural Analysis64 • Basic Structural Analysis Framed Structures • 63

Joint C

FCD

FCE

60°

60°

30°

30°

20 kN

42.2

C

∑H = 020cos60−FCD cos30−FCE cos60+42.2cos30 = 0FCD cos30+FCE cos60 = 46.550.866 FCD +0.5 FCE = 46.55

∑V = 0−20sin60−FCD sin30+FCE sin60+42.2sin30 = 0−0.5 FCD +0.866 FCE = −3.78FCD = 42.20 kNFCE = 20.00 kN

Joint G

30°

60°30°

FGB

= 29.76

FGD

FGF

G

∑H = 0FGD cos30−FGB cos30+FGF cos60 = 00.866 FGD −0.866 FGB +0.5 FGF = 00.866 FGD +0.5 FGF = 0.866×29.760.866 FGD +0.5 FGF = 25.77

∑V = 0−FGD sin30+FGF sin60+29.76sin30 = 0−0.5 FGD +0.866 FGF = −14.88FGD = 25.77 kN, FGF = 0

Joint D

60°

60°

60°

30°30°

25.77

10 kN

42.20

FDE

D

∑V = 0;−10sin60+42.20sin30−FDE sin60+25.77sin30 = 0FDE = 29.24 kN

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Resolving vertically,

FEF sinθ1 = 40

FEF =400.8

= 50 kN

Resolving all the forces in the horizontal direction

FDE = FEF cosθ1 = 50×0.6 = 30 kN

Joint FThe truss is supported on rollers at the joint F . Only one reaction will be acting perpendicular tothe base of the roller. Hence,

110 kN

FDE

FGF

FFE

= 50 kN

F

θ1

θ

∑H = 0;FGF cosθ−50cosθ1 = 0FGF = 50×0.6/0.832FGF = 36.06 kN

∑V = 0;110−36.06sinθ−50sinθ1 +FDF = 0FDF = 36.06(0.555)+50(0.8)−110= −50 kN

The −ve sign indicates that we have to change the nature of the force in FDF and hence FDF iscompressive.

Joint B

FBH

29.95 kN FBCB

B

Resolving all the forces meeting at the joint in the vertical direction and as no forces are acting inthe vertical direction; FBH = 0. Resolving the forces at the joint B along the horizontal directionFBC = 29.95 kN


Resolving vertically,

FEF sinθ1 = 40

FEF =400.8

= 50 kN

Resolving all the forces in the horizontal direction

FDE = FEF cosθ1 = 50×0.6 = 30 kN

Joint FThe truss is supported on rollers at the joint F . Only one reaction will be acting perpendicular tothe base of the roller. Hence,

110 kN

FDE

FGF

FFE

= 50 kN

F

θ1

θ

∑H = 0;FGF cosθ−50cosθ1 = 0FGF = 50×0.6/0.832FGF = 36.06 kN

∑V = 0;110−36.06sinθ−50sinθ1 +FDF = 0FDF = 36.06(0.555)+50(0.8)−110= −50 kN

The −ve sign indicates that we have to change the nature of the force in FDF and hence FDF iscompressive.

Joint B

FBH

29.95 kN FBCB

B

Resolving all the forces meeting at the joint in the vertical direction and as no forces are acting inthe vertical direction; FBH = 0. Resolving the forces at the joint B along the horizontal directionFBC = 29.95 kN


Joint H

FCH

FHC

36 kN

H

A, H and G are on the same line. Hence, resolving the forces meeting at the joint H, A HGResolving all the forces along the line, FHG = 36 kNJoint C

FCD

FCG

29.95 kN C

Resolving all the forces in the vertical direction: FCG = 0Resolving all the forces in the horizontal direction: FCD = 29.95 kNJoint G

FGD

FGF

36 kN

G

Resolving all the forces meeting at the joint G, along the line HGF and perpendicular that line;

FGD = 0;FGF = 36 kN


The computed forces are shown in Fig. 2.40.

A C D

H

G

F

E30 kN30 kN

36 kN

36 kN

36 kN

30 kN 30 kN

50 kN50 kN

40 kN50 kN

4 m

B

FIG. 2.40

2.5 METHOD OF SECTIONS

A statically determinate frame can be completely analysed by static methods. The number ofunknowns is the same as the number of equations obtained from static equilibrium conditions.

In method of sections, we isolate a portion of a frame by a section. This section causes the inter-nal force appear to act as external forces in the isolated portion. The unknown forces are evaluatedby using equations of equilibrium. The three equations of equilibrium are used to evaluate theunknowns. Thus, the section should cut only three members. (which include the one whose forceis to be determined) and take moments about a point through which the lines of action of other twomembers intersect. The work is made simple by judicious of choice of such sections.This method is not desirable if used to determine all the forces in the members of a frame. However,it is readily useful to determine the forces in selected members of the truss.

2.5.1 Application of Method of SectionsConsider a truss shown in Fig. 2.41. It is required to calculate the force in the member GF of thebottom boom.

8 m 8 m 8 m

4 m

A

B C D

F EG

X

X

FIG. 2.41


1. Imagine the truss to be cut completely through the section X −X passing through the mem-bers BC,CG and GF .

2. Assume that the right portion of the truss X −X to be removed. The portion to the left ofX −X would then collapse, because three forces in the members BC, GC and GF whichwere necessary to retain equilibrium had been removed.

3. If three forces FBC, FGC and FGF are now applied to the portion of the frame concerned asshown in Fig. 2.42 then the portion of the frame will be in equilibrium under the action ofthe reaction 66.7 kN, applied 50 kN load and the forces FBC, FGC and FGF .

4. These three forces are yet knowns in magnitude and direction. The member force FGF is tobe determined by knowing that the other two member forces FCB and FCG meet at point Cand that they have no moment about that point.

8 m

66.7 50 kN

45°

4 m

A

B C

G 45°

FIG. 2.42

5. Thus taking moments about the point C. The moment of the reaction is a clockwise (66.7×12 = 800.4). The moment of the applied load is anticlockwise (50 × 4), i.e., 200 kNmFor equilibrium, the anticlockwise moment of FGF is (+4×GF) and this must be equal tothe clockwise moment of (800.4− 200) = 600.4 kNm. Therefore, 4FGF = 600.4, Hence,FGF = 150.1 kN. This force is tension (pulling away from the joint G).

6. To determine the force in FBC, take moments of all the forces about the point G, where othertwo members of the cut truss FGC and FGF meet.

(a) The moment of the reaction is (66.7×8), i.e., 533.6 kN, (Clockwise)(b) The moment of the forces of 50 kN, FGC and FGF meet at the point G and do not give

any moments.(c) The moment due to the required member force is (FBC ×4), i.e., 4FBC (anticlockwise)(d) 4FBC = 533.6 and FBC = 133.4 kN.

This force is compressive (Acting towards the joint B).

7. To determine the force in GC member, i.e., FGC; resolve all the forces in the vertical directionof the cut truss considering the left part only. The cut members BC and GF are horizontal andhence they do not give a vertical component; it is easier to consider the vertical equilibrium.


The reaction at A is 66.7 kN acting upwards and hence it is taken as negative. The appliedload at G is 50 kN and is acting downwards and therefore it is positive. The vertical compo-nent due to FGC is downwards, i.e., FGC sin45. This is negative.Summing up all the above forces in the vertical direction and equating upward forces todownward forces.

FGC sin45+50 = 66.7∴ FGC = 23.62 kN.

This is compressive as the force is acting towards the joint.8. It must be remembered that the arrows must be considered in respect to the nearest point of

that portion of the frame which remains after the cut has been made.

2.5.2 Numerical ProblemsEXAMPLE 2.17: Determine the nature and magnitude of forces in members DE,DIand HI of the truss shown in Fig. 2.45 by using method of sections.

3 kN 6 kN 6 kN 3 kN

DC EB

G

VA VF

6 m

60°A F

6 kN

H

12 kN

I

6 kN

6 m6 m 6 m

X

X

60°

FIG. 2.43

SOLUTION

∑V = 0 VA +VF = 42

∑MA = 0 3(3)+6(9)+6(15)+3(21)+6(6)+12(12)+6(18)−24VF = 0VF = 21 kN∴ VA = 21 kN


EXAMPLE 2.18: Determine the forces in the members BC, FC and FG by method of sections.

30 kN

B

F H

G

60 kN

C

30 kN

D

4.8 4.8 m

EA

4.8 4.8

4.8 m

3.6

FIG. 2.45

SOLUTIONDue to symmetry, the reactions

VA = VE =30+60+30

2= 60 kN

To determine the forces in BC, FC and FG cut a section X-X as shown in Fig. 2.48Consider left part of the truss and analyse the equilibrium.

60 kN

30 kN

θ

θ

θ1

θ

4.8 4.8

3.6

X

X

CA B

F

FIG. 2.46

To determine the force in BC, take moment about ‘F’ where other two members of the cutsection FC and FG meet.

60×4.8 = FBC ×3.6FBC = 80 kN


30 kN

θ

θ1

θ

4.8 m

3.6 m

C

G

A

60 kN

B

F �

A �

F

FIG. 2.47

From geometry A′C = 19.2 m

To determine the force in FC, take moment about A′ where the other two members of the cutsection BC and FG meet.Let A′F ′ be perpendicular to the line CF extended.

sinθ =A′F ′

A′CA′F ′ = 19.2×0.6 = 11.52 m

∑M ′A = 0−60× (A′A)+30 (A′B)+FFC (11.52) = 0−60×9.6+30 (19.2−4.8) = −11.52 FFC∴ FFG = 12.5 kN (Compressive)

To determine the force in FG, take moment about C where the members BC and FC meet.

θ2

6 m

F

C′

C

FIG. 2.48

In Fig. 2.45, FC =√

BC2 +BF2

=√

4.82 +3.62 = 6 m


To determine the forces in DE, DI and HI, cut a section X −X as shown in Fig. 2.44 and considerthe equilibrium of the cut truss.

3 kN

60°

2.6 m

21 kN

VDE

FDI

FHI

6

D

I

X

X

FIG. 2.44

To determine the force FDE; take moment about I where other two members of the cut sectionmeet.

FDE ×2.6+3(3)−21×6 = 0

FDE = 45 kN

To determine the force FDI, resolve all the forces vertically

21−3−6−FDI sin60 = 0

FDI = 13.86 kN

To determine the force FHI, take moment about D where other two members of the cut

section meet.

6×3+3×6−9×21−2.6 FHI = 0

∴ FHI = −58.85 kN

The negative sign indicates that the nature of FHI is tensile.


We know that θ = 36◦52′ and θ1 = 14◦2′

θ2 = ∠CFG = θ+θ1 = 50◦54′

sinθ2 =C ′C

6C ′C = 6sinθ2 = 4.66 m

FFG(4.66)−30(4.8)+60(9.6) = 0

FFG = 92.7 kN (Tensile)

EXAMPLE 2.19: Determine the forces in CD, CG and FG by method of sections.

E

E

C

D

A

B

3 m3 m

30°

HA

VA

3 mF G

2.5 kN

2.5 kN 2.6 m

2.6 m5 kN

FIG. 2.49

SOLUTIONThe reactions at A, i.e., VA and HA and the reaction at E; i.e., VE are found using equilibriumequations.

∑V = 0; VA +VE = 2.5sin60+5sin60+2.5sin60

VA +VE = 8.66

∑H = 0; 2.5cos60+5cos60+2.5cos60−HA = 0

HA = 5 kN

∑MA = 0; 5(2.6)+2.5(5.2)−9VE = 0

VE = 2.88 kNVA = 5.78 kN


Cut a section through the members CD, CG and FG and consider the right part of the truss.

EF

FG

FCD

G 3 m

D

G′

60°

2.88

30°

X

X

C

FCG

FIG. 2.50

tan30 = GD/2.6GD = 2.6tan30 = 1.5 m

To determine the force in CD, take moment of the forces in the cut truss about G where othertwo members of the cut section, viz. CG and FG meet.

1.5 FCD = 2.88×3FCD = 5.76 kN

To determine the force in FG, take moment about C where other two members of the cutportion of the truss, viz. CD and GC meet.

4.5tan30 (FFG) = 2.88×4.5FFG = 4.98 kN

To determine the force in GC, take moment about E where the cut members CD and FG meet.Let EG′ is perpendicular to the line CG extended

∴ FCG ×G′E = 0FCG = 0


EXAMPLE 2.20: Find the forces in the members ED, EF and FG. Use method of sections.(VTU, Aug. 2004)

B

D

CG

A

10 kN

4 m

30° 60°60°

4 m4 m

E FH

10 kN

10 kN

20 kN

FIG. 2.51

SOLUTIONThe length of panel AC, CD are found using geometry.

cos30 = 6/AD

AD = 6/cos30

= 6.93 m

∴ AC = CD = AD/2

= 3.46 m

The reactions VA, HA and VB are determined using the equilibrium equations.

∑H = 0;

10cos60+20cos60+10cos60−HA = 0

HA = 20 kN

∑V = 0;

VA +VB = 10sin60+20sin60+10sin60+10

VA +VB = 44.64 kN

∑MA = 0;

20(3.46)+10(6.92)+10(4)−12VB = 0

VB = 14.87 KN

To determine the forces in the members CD, DE and EF cut a section through these members andfor convenience consider the right part of the truss.


B

D

G

E FFE

FCD

FDE

8 m

F

10 kN

FIG. 2.52

The force in the member CD can be found by taking moment about E.

10×3.46−14.87×8−FCD ×CE = 0

FCD =−84.36

CE=

−84.364sin30

= −42.18

The −ve sign indicates that FCD is compressive.To determine the force in FFE take moment about D where other two members of the cutsection CD and CE meet.

−14.87×6+FFE ×3.46 = 0tan30 = DH/6DH = 3.46 m

FFE = 25.8 kN

To compute the force in FG, cut a section through the members DG, FG and FB.

B

G

FFG

14.87 kNF

X

X

FIG. 2.53


∑MB = 0

FFG ×GB = 0

∴ FFG = 0

EXAMPLE 2.21: Determine the forces in the members FE, FD, CD of the truss shown in Fig.2.54

B

E

F

A3 m 3 m 3 mC D

50 kN

3 m

FIG. 2.54

SOLUTIONThe reactions are found out by using equilibrium equations.

VA +VB = 50

Taking moment about A;

50×3−9VB = 0

VB = 16.7 kN; VA = 33.3 kN

To determine the forces in FE, FD and CD, cut a section through these members and consider theright-hand portion of the truss.

FCD ×3 = 16.7×6

FCD = 33.4 kN

To determine the force FFE; take moment about D; where other two members of the cut truss,FFD and FCD meet.


FCD

FFD

VB

FFE

E

DB

D′

θ

6 m3 m

16.7 kN

F

3 m

FIG. 2.55

tanθ = 3/6i.e. θ = 26◦33′54′′.

In ∆ BDD ′

sinθ =DD ′

DB∴ DD ′ = 3sin26◦33′54′′.

= 1.34 m.FEB ×1.34 = 16.7×3

FEB = 37.4 kN

To determine the force in FD, take moment about B where other two members of the cutsection, viz. FFE and FCD meet.Extend FD downwards to B′ such that BB′ is perpendicular to the projected line of FD. Consideringthe ∆DBB′;

sinθ =BB′

3BB′ = 3sinθ = 3× sin26◦33′54′′.BB′ = 1.34 m

∑MB = 0; FFD ×BB′ = 0FFD = 0


15 kN

θ

FEF

FDE

E

Consider Joint F

FFFG 59.8

FFD

At joint F , there is no vertical force acting. Hence, resolving the forces in the vertical directionFFD = 0. Resolving the forces in the horizontal direction,

∑H = 0; FFG −59.8 = 0FFG = 59.8 kN

Consider Joint D

D

FDC

FDE

= 61.73FDG

At joint D; resolving the forces perpendicular tothe line CDE, FDG = 0 as the forces FDC andFDE cannot give components in the perpendic-ular direction. Resolving the forces along theline of forces. FDE the force FDC is obtained asFDC = 61.73 kN.

Joint G

G

FGC

FAG 59.8

Resolving the forces in the vertical direction atjoint G; FGC = 0 and resolving the forces in thehorizontal direction FAG = 59.8 kN.

86 • Basic Structural Analysis 88 • Basic Structural Analysis

In ∆CGE;

sin60 =CGCE

CE = CG/sin60 = 2.165/sin60CE = 2.5 m

To determine the force in BC, CE and ED cut a section through these members and considerthe right portion of the truss.

30 kNF

CB

25 kN

G

E

C

3.75 mD30°

X

X

FIG. 2.59

FCB is obtained by taking moment about E where other two members of the cut truss meet.

30×1.25−25×5−FCB ×2.5 = 0FCB = −35 kN

-ve sign indicates the force FCB is compressive.To determine the force in DE, take moment of the forces of the cut truss about C.

FDE ×CG = 25×3.75FDE ×2.165 = 25×3.75

FDE = 43.3 kN

To determine the force in FCE, take moment about D.

30×3.75 = FCE ×4.33FCE = 26 kN.

To determine the forces in BE and AE, cut a section through the three members, viz. BC, BEand AE and consider the left portion of the truss.


EXAMPLE 2.23: A truss of 10 m span is loaded as shown in Fig. 2.58 find the forces in themembers of the truss by using method of sections.

C

B

A

30 kN

25 kN

GF E

10 m

D60° 60° 60° 30°

FIG. 2.58

SOLUTION

∑V = 0 sin30 =AB10

VA +VD = 25+30 AB = 10sin30

VA +VD = 55 kN AB = 5 m

∑MA = 0 cos30=BD10

25×5cos60+30×6.25−10VD = 0 BD = 8.66 mVD = 25 kN BC = CD = 4.33 m

VA = 30 kN AF = 5cos60AG = 10−4.33cos60

AG = 6.25 m

In ∆CGD

sin30 =CGCD

CG = CDsin30 = 4.33sin30 = 2.165 m


In ∆CGE;

sin60=CGCE

CE = CG/sin60= 2.165/sin60CE = 2.5 m

To determine the force in BC, CE and ED cut a section through these members and considerthe right portion of the truss.

30 kNFCB

25 kN

G

E

C

3.75 mD30°

X

X

FIG. 2.59

FCB is obtained by taking moment about E where other two members of the cut truss meet.

30×1.25−25×5−FCB ×2.5 = 0FCB = −35 kN

-ve sign indicates the force FCB is compressive.To determine the force in DE, take moment of the forces of the cut truss about C.

FDE ×CG = 25×3.75FDE ×2.165= 25×3.75

FDE = 43.3 kN

To determine the force in FCE, take moment about D.

30×3.75 = FCE ×4.33FCE = 26 kN.

To determine the forces in BE and AE, cut a section through the three members, viz. BC, BEand AE and consider the left portion of the truss.

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30 kN

25 kN

5 m

E D

A

B

D'

60°

FIG. 2.60

sin60 =DD′

EDDD′ = 5sin60 = 4.33

∑MD = 030×10−25×7.5−FBE ×4.33 = 0FBE = 26 kN

To determine the force in AE, take moment about B.

30×2.5−FAE ×4.33 = 0FAE = 17.32 kN

Joint A

60°

30 kN

FAE

FAB

A

∑V = 0−FAB sin60+30 = 0

FAB = 34.64 kN


30 kN

25 kN

5 m

E D

A

B

D'

60°

FIG. 2.60

sin60 =DD′

EDDD′ = 5sin60 = 4.33

∑MD = 030×10−25×7.5−FBE ×4.33 = 0FBE = 26 kN

To determine the force in AE, take moment about B.

30×2.5−FAE ×4.33 = 0FAE = 17.32 kN

Joint A

60°

30 kN

FAE

FAB

A

∑V = 0−FAB sin60+30 = 0

FAB = 34.64 kN


Joint D

30°

25 kN

FCD

FED

D

∑V = 0

25 = FCD sin30

FCD = 50 kN

C

B

A

30 kN

35

26 kN

25 kN

43.3 kN

34.6 kN

50

26 kN

17.32 kN E

10 m

D60° 60° 60°

FIG. 2.61

EXAMPLE 2.24: Figure 2.62 shows a pinjointed truss supported by a hinge at A and roller at G.Determine the force in each of the five members meeting at joint B.

B C D E F

H GIJK180 kN4 m

LA45º 45º

125 kN

6 @ 4 m

4 m

FIG. 2.62


The reactions at A and G are found by using vertical equilibrium equations and moment equilibrium.

∑V = 0; VA +VG = 180+125VA +VG = 305

∑MA = 0; 180×4+125×8−24VG = 0VG = 71.7 kNVA = 233.3 kN

At the joint B, there are five members meeting, viz. BA, BL, BK, BJ and BC. The method ofobtaining the forces is by using method of joints as well as method of sections. The forces BA andBL are obtained by using method of joints at A and L. The force in BJ is determined by method ofsections. Then the member forces BC and BK are obtained by resolution forces at the joint B.

Joint A

45°

233.3 kN

FAL

FAB

A

Resolving the forces verticallyFAB sin45 = 233.3

FAB = 329.9 kNResolving the forces horizontally

FAL = FAB cos 45

FAL = 233.3 kN

Joint L

FLB

180 kN

L

Resolving the forces verticallyFLB −180 = 0

FLB = 180 kN


The reactions at A and G are found by using vertical equilibrium equations and moment equilibrium.

∑V = 0; VA +VG = 180+125VA +VG = 305

∑MA = 0; 180×4+125×8−24VG = 0VG = 71.7 kNVA = 233.3 kN

At the joint B, there are five members meeting, viz. BA, BL, BK, BJ and BC. The method ofobtaining the forces is by using method of joints as well as method of sections. The forces BA andBL are obtained by using method of joints at A and L. The force in BJ is determined by method ofsections. Then the member forces BC and BK are obtained by resolution forces at the joint B.

Joint A

45°

233.3 kN

FAL

FAB

A

Resolving the forces verticallyFAB sin45 = 233.3

FAB = 329.9 kNResolving the forces horizontally

FAL = FAB cos 45

FAL = 233.3 kN

Joint L

FLB

180 kN

L

Resolving the forces verticallyFLB −180 = 0

FLB = 180 kN


To determine the force in BJ, cut a section through CD, BJ and KJ.

B C

AK

180 kN233.3

L

J

FBJ

125 kN

45° θ

X

X

FIG. 2.63

tanθ = 4/8sinθ = 0.4472

In the cut truss, two cut members, viz. CD and KJ are horizontal. They do not give any verticalcomponent of the force. Hence to maintain equilibrium in the vertical direction, the force FBJ whichis indirect at an angle θ should give a vertical component to balance the effect due to the verticalreaction at A and the applied downward loads. Hence resolving vertically.

FBJ sinθ+233.3−180−125 = 0.FBJ = 160.33 kN

Using method of joints at joint B (as shown in Fig. 2.64)

B

45°

329.9

180

106.33

FBC

FBK

θ

FIG. 2.64

∑H = 0; 329.99 cos45−FBC +160.33 cosθ−FBK cos45 = 0FBC +0.707 FBK = 376.74

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∑V = 0; 329.99 sin45−180+FBK sin45−163.33 sinθ = 0FBK = 26.07 kNFBC = 358.31 kN

EXAMPLE 2.25: Determine the forces in the members BC, GC and GF of the truss shown inFig. 2.65 by method of sections. VTU, Dec. 2006

30°

3 @ 3 m = 9 m

60°

10 kN

G

B

AE

C

D

F

60°

60°

60°

20 kN

10 kN

FIG. 2.65

SOLUTIONThe reactions are found using equilibrium equations

∑H = 0; HE −10cos60−20cos60−10cos60 = 0HE = 20 kN

∑V = 0; VA +VE −10sin60−20sin60−10sin60 = 0VA +VE = 40sin60VA +VE = 34.64 kN

∑ME = 0; 20(2.6)+10(5.2)−9VA = 0

VA = 11.55 kNVE = 23.09 kN

To determine the forces in FBC, FCG and FGF cut a section X −X as shown in Fig. 2.66.The force FGF is obtained by taking moment about C.

11.55×4.5 = FGF ×4.5tan30

To determine the force in GC, take moment about A

FGC Z = 0


∑V = 0; 329.99 sin45−180+FBK sin45−163.33 sinθ = 0FBK = 26.07 kNFBC = 358.31 kN

EXAMPLE 2.25: Determine the forces in the members BC, GC and GF of the truss shown inFig. 2.65 by method of sections. VTU, Dec. 2006

30°

3 @ 3 m = 9 m

60°

10 kN

G

B

AE

C

D

F

60°

60°

60°

20 kN

10 kN

FIG. 2.65

SOLUTIONThe reactions are found using equilibrium equations

∑H = 0; HE −10cos60−20cos60−10cos60 = 0HE = 20 kN

∑V = 0; VA +VE −10sin60−20sin60−10sin60 = 0VA +VE = 40sin60VA +VE = 34.64 kN

∑ME = 0; 20(2.6)+10(5.2)−9VA = 0

VA = 11.55 kNVE = 23.09 kN

To determine the forces in FBC, FCG and FGF cut a section X −X as shown in Fig. 2.66.The force FGF is obtained by taking moment about C.

11.55×4.5 = FGF ×4.5tan30

To determine the force in GC, take moment about A

FGC Z = 0


30° 60°

G

Z

H

4.56 tan 30

11.55

4.5 m

B

C

AF

GF

X

X

FIG. 2.66

∴ FGC = 0

To determine the force in BC, take moment about G.

FBC ×BG = 11.55×3FBC × (3sin30) = 11.55×3

FBC = 23.1 kN

2.6 METHOD OF TENSION COEFFICIENTS

The method of tension coefficients was developed by Southwell (1920) and is applicable for planeand space frames. Contemporily this was developed by Muller Breslau independently.

A

z

x

y

B

TAB

( xA ,y

A ,z

A )

( xB

,yB

,zB

)

FIG. 2.67


Let AB be any member of a truss having length LAB and the tensile force be TAB. The force TAB canbe expressed as

TAB = tABLAB

where tAB is a tension coefficient, LAB is the length of member AB. The component of TAB inthe direction of x axis is tAB (xB −xA) and similarly the components in the direction of y axis istAB (yB −yA) and in z direction it is tAB (zB = −zA) respectively.

If at the joint A, let the components of the external loads be XA,YA and ZA acting in x,y and zdirections respective then for equilibrium of the joint where the members AB, AC and AQ meet

tAB(xB −xA)+ tAC(xC −xA)+ · · ·+ tAQ(xQ −xA)+XA = 0tAB(yB −yA)+ tAC(yC −yA)+ · · ·+ tAQ(yQ −yA)+YA = 0tAB(zB −zA)+ tAC(zC −zA)+ · · ·+ tAQ(zQ −zA)+ZA = 0

At each joint, a similar set of three equations are formed for each joint and the tension coefficientsare determined. The force in the member is obtained by multiplying this tension coefficient withthe corresponding length of the members.This is known as method of tension coefficients. The procedure is as follows:

1. Mark the positive directions of x,y and z axis.2. Assume all the members are in tension, i.e., the force is moving away from the joint.3. Write down the equilibrium equations at each joint.4. Solve the equations for unknown tension coefficients.5. Calculate TAB = LAB tAB.

2.6.1 Numerical ProblemsEXAMPLE 2.26: The Fig. 2.68 shows a Warren type cantilever truss along with the imposedloads. Determine the forces in all the members using the method of joints / tension coefficients.(Anna-Univ, Dec. 2008)

3 kN

2 kN 2 kN

3 kN

AB

E D

C

60° 60°

60°

FIG. 2.68


Let AB be any member of a truss having length LAB and the tensile force be TAB. The force TAB canbe expressed as

TAB = tABLAB

where tAB is a tension coefficient, LAB is the length of member AB. The component of TAB inthe direction of x axis is tAB (xB −xA) and similarly the components in the direction of y axis istAB (yB −yA) and in z direction it is tAB (zB = −zA) respectively.

If at the joint A, let the components of the external loads be XA,YA and ZA acting in x,y and zdirections respective then for equilibrium of the joint where the members AB, AC and AQ meet

tAB(xB −xA)+ tAC(xC −xA)+ · · ·+ tAQ(xQ −xA)+XA = 0tAB(yB −yA)+ tAC(yC −yA)+ · · ·+ tAQ(yQ −yA)+YA = 0tAB(zB −zA)+ tAC(zC −zA)+ · · ·+ tAQ(zQ −zA)+ZA = 0

At each joint, a similar set of three equations are formed for each joint and the tension coefficientsare determined. The force in the member is obtained by multiplying this tension coefficient withthe corresponding length of the members.This is known as method of tension coefficients. The procedure is as follows:

1. Mark the positive directions of x,y and z axis.2. Assume all the members are in tension, i.e., the force is moving away from the joint.3. Write down the equilibrium equations at each joint.4. Solve the equations for unknown tension coefficients.5. Calculate TAB = LAB tAB.

2.6.1 Numerical ProblemsEXAMPLE 2.26: The Fig. 2.68 shows a Warren type cantilever truss along with the imposedloads. Determine the forces in all the members using the method of joints / tension coefficients.(Anna-Univ, Dec. 2008)

3 kN

2 kN 2 kN

3 kN

AB

E D

C

60° 60°

60°

FIG. 2.68


Joint D

60°

FDE

FDC

2 kN

D

∑V = 0FDC sin60−2 = 0

FDC =2

sin60= 2.31

∑H = 0FDE −FDC cos60 = 0FDE = 2.31cos60FDE = 1.16 kN

Joint C

60° 60°

3

FCB

FCE

FCD

= 2.31

C∑V = 0

−3−2.31sin60+FCE sin60 = 0FCE = 5.77 kN

∑H = 0−FCB +5.77cos 60+FCD cos 60 = 0FCB = 4.04 kN

Joint E

60° 60°

2 kN

FEB

FEF

5.77

1.16E

∑V = 0FEB sin60−5.77sin60−2 = 0FEB = 8.08 kN

∑H = 0FEF −8.08cos60−5.77cos60−1.16 = 0FEF = 8.08 kN

Joint B

60° 60°

3 kN

FAB

FFB

8.08

4.04B

∑V = 0−3−8.08sin60+FFB sin60 = 0FFB = 11.54 kN

∑H = 04.04+8.08cos60+11.54cos60−FAB = 0FAB = 13.85 kN


3 kN

2 kN 2 kN

8.08

4.0413.85

8.08

11.54 5.77

1.16

2.31

3 kN

A B C

DEF

FIG. 2.69

Using tension coefficient method

F(16,0)

3 kN

2 kN

E

2 kN

x

3 kN

(8,0)

A (16,6.93) B (12,6.93) C (4,6.93)

D (0,0)

FIG. 2.70

At Joint D

∑H = 0, tDC(xC − xD)+ tDE(xE − xD) = 0tDC(4−0)+ tDE(8−0) = 0

4tDC +8tDE = 0

∑V = 0, tDC(yC − yD)+ tDE(yE − yD)−2 = 06.93tDC +0 = 2

∴ tDC = 0.2886tDE = −0.1443

Joint C

∑H = 0; tCD(xD − xC)+ tCE(xE − xC)+ tCB(xB − xC) = 00.2886(0−4)+ tCE(8−4)+ tCB(12−4) = 0

4 tCE +8 tCB = 1.1544


3 kN

2 kN 2 kN

8.08

4.0413.85

8.08

11.54 5.77

1.16

2.31

3 kN

A B C

DEF

FIG. 2.69

Using tension coefficient method

F(16,0)

3 kN

2 kN

E

2 kN

x

3 kN

(8,0)

A (16,6.93) B (12,6.93) C (4,6.93)

D (0,0)

FIG. 2.70

At Joint D

∑H = 0, tDC(xC − xD)+ tDE(xE − xD) = 0tDC(4−0)+ tDE(8−0) = 0

4tDC +8tDE = 0

∑V = 0, tDC(yC − yD)+ tDE(yE − yD)−2 = 06.93tDC +0 = 2

∴ tDC = 0.2886tDE = −0.1443

Joint C

∑H = 0; tCD(xD − xC)+ tCE(xE − xC)+ tCB(xB − xC) = 00.2886(0−4)+ tCE(8−4)+ tCB(12−4) = 0

4 tCE +8 tCB = 1.1544


∑V = 0; tCD(yD − yC)+ tCE(yE − yC)+ tCB(yB − yC)−3 = 00.2886(0−6.93)+ tCE(0−6.93)+ tCB(6.93−6.93) = 3

tCE = −0.7215tCB = +0.5051

Joint E

∑H = 0; tEC(xC − xE)+ tEB(xB − xE)+ tEF(xF − xE)+ tED(xD − xE) = 0−0.7215(4−8)+ tEB(12−8)+ tEF(16−8)−0.1443(0−8) = 0

4tEB +8tEF = −4.04

∑V = 0; tEC(yC − yE)+ tEB(yB − yE)+ tEF(yF − yE)+ tED(yD − yE)−2 = 0tEC(6.93−0)+ tEB(6.93−0)+ tEF(0−0)+ tED(0−0) = 2

6.93 tEB +6.93 tEC = 2.006.93 tEB +6.93×0.7215 = 2.00

Solving the above two equations:tEB = 1.0101tEF = −1.0100

Joint B

∑H = 0, tBA(xB − xA)+ tBF(xF − xB)+ tBE(xE − xB)+ tBC(xC − xB) = 0tBA(16−12)+ tBF(16−12)+ tBE(8−12)+ tBC(4−12) = 0

4tBA +4tBF −4tBE −8tBC = 04tBA +4tBF = 4(1.0101)+8×0.5051

tBA + tBF = 2.0203

∑V = 0, tBA(yA − yB)+ tBF(yF − yB)+ tBE(yE − yB)+ tBC(yC − yB)−3 = 0tBA(6.93−6.93)+ tBF(0−6.93)+ tBE(0−6.93)+ tBC(6.93−6.93)−3 = 0

−6.93tBF −6.93(1.0101) = 3tBF = −1.443tBA = 3.463

TAB = 3.463×4 = 13.85 TFB = −1.443×8 = −11.54TBC = 0.5051×8 = 4.04 TBE = +1.0101×8 = +8.08TCD = 0.2886×8 = 2.31 TCE = −0.7215×8 = −5.77TDE = −0.1443×8 = −1.15 TEF = −1.0101×8 = −8.08


3 kN

2 kN 2 kN

8.08

4.0413.85

8.08

11.54 5.77

1.15

2.31

3 kN

A B C

DE

F

FIG. 2.71

EXAMPLE 2.27: Analyse the space truss shown in Fig. 2.72 using tension coefficient method.(Anna Univ, 2009).

y

x

200 kN

8.3 m

4.5 m

1 m

3 m

50 kN

7 m

1 m

A

C

B

Dz

FIG. 2.72

Joint x y zA 0 0 0B 8.3 0 0C 3.0 7 1D 3 0 4.5


3 kN

2 kN 2 kN

8.08

4.0413.85

8.08

11.54 5.77

1.15

2.31

3 kN

A B C

DE

F

FIG. 2.71

EXAMPLE 2.27: Analyse the space truss shown in Fig. 2.72 using tension coefficient method.(Anna Univ, 2009).

y

x

200 kN

8.3 m

4.5 m

1 m

3 m

50 kN

7 m

1 m

A

C

B

Dz

FIG. 2.72

Joint x y zA 0 0 0B 8.3 0 0C 3.0 7 1D 3 0 4.5


Resolving the forces at joint C in x direction

tCA(xA − xC)+ tCD(xD − xC)+ tCB(xB − xC)+50 = 0tCA(0−3)+ tCD(3−3)+ tCB(8.3−3)+50 = 0

−3tCA +5.3tCB = −50 (1)

Resolving the forces at joint C in y direction

tCA(yA − yC)+ tCD(yD − yC)+ tCB(yB − yC)−200 = 0tCA(0−7)+ tCD(0−7)+ tCB(0−7) = 200

−7tCA −7tCB −7tCD = 200 (2)

Resolving the forces at the joint C in z direction

tCA(zA − zC)+ tCD(zD − zC)+ tCB(zB − zC) = 0

tCA(0−1)+ tCD(4.5−1)+ tCB(0−1) = 0

−tCA +3.5tCD − tCB = 0 (3)

∴

−3 5.3 0−7 −7.0 −7.0−1 −1.0 +3.5

tCAtCBtCD

=

−50200

0

tCA = −8.166 tCB = −14.056 tCD = −6.349

LCA =√

(3−0)2 +(7−0)2 +(1−0)2 = 7.68 m.

LCB =√

(3−8.3)2 +(7−0)2 +(1−0)2 = 4.68 m.

LCD =√

(3−3)2 +(7−0)2 +(1−4.5)2 = 7.83 m.

TCA = tCA LCA = −62.71 (Compression)TCB = tCB LCB = −65.78 (Compression)TCD = tCD LCD = −49.71 (Compression)

EXAMPLE 2.28: Figure 2.73 shows an elevation and plan of tripod having legs of unequallengths resting without slipping on a sloping plane, if the pinjointed apex carries a vertical loadof 100 kN. Calculate the force in each leg.


2.7 m

C

DB

A

0.7 m

2 m

1.3 m

1.35 m

C

DB

A

1.3 m

1.3 m 1.3 my

x

FIG. 2.73

Joint x y zA 0 0 0B −2.7 +1.3 −2C 0 −1.3 −3.3D 1.35 +1.3 1.6

Resolving the forces at the joint A is x direction

tAB(xB − xA)+ tAC(xC − xA)+ tAD(xD − xA) = 0tAB(−2.7−0)+ tAC(0−0)+ tAD(1.35−0) = 0

−2.7tAB +1.35tAD = 0


2.7 m

C

DB

A

0.7 m

2 m

1.3 m

1.35 m

C

DB

A

1.3 m

1.3 m 1.3 my

x

FIG. 2.73

Joint x y zA 0 0 0B −2.7 +1.3 −2C 0 −1.3 −3.3D 1.35 +1.3 1.6

Resolving the forces at the joint A is x direction

tAB(xB − xA)+ tAC(xC − xA)+ tAD(xD − xA) = 0tAB(−2.7−0)+ tAC(0−0)+ tAD(1.35−0) = 0

−2.7tAB +1.35tAD = 0


Resolving the forces at the joint A in y direction

tAB(yB − yA)+ tAC(yC − yA)+ tAD(yD − yA) = 0tAB(1.3−0)+ tAC(−1.3−0)+ tAD(1.30−0) = 0

1.3tAB −1.3tAC +1.3tAD = 0

Resolving the forces at the joint A in z direction

tAB(zB − zA)+ tAC(zC − zA)+ tAD(zD − zA) = 0tAB(−2−0)+ tAC(−3.3−0)+ tAD(2.6−0)−100 = 0

−2 tAB −3.3 tAC +2.6 tAD = 100

−2.7 0 1.35+1.3 −1.3 1.3−2 −3.3 +2.6

tABtACtAD

=

00

100

tAB = −14.925; tAC = −44.776; tAD = −29.851

LAB =√

2.72 +1.32 +22 = 3.602 m

LAC =√

02 +1.32 +3.32 = 3.547 m

LAD =√

1.352 +1.32 +2.62 = 3.205 mTAB = tABLAB = 53.76 (Compressive)TAC = tACLAC = 158.82 (Compressive)TAD = tADLAD = 95.67 (Compressive)

EXAMPLE 2.29: Figure 2.74 shows plan and elevation of a symmetrical statically determinatespace truss supporting a vertical load of 75 kN at D. Determine the forces in the members DE andEF .

3 m

x

75 kNElevation

A

D

E,F

z

B,C


3 m

x

y

C

A

E

F

D

B

3 m

3 m 3 m

FIG. 2.74

Joint x y z

A +6 0 +3B +6 +3 0C +6 −3 0D 0 0 0E +3 −3 +3F +3 +3 +3

Resolving all the forces meeting at joint D in z direction as

tDE(zE − zD)+ tDC(zC − zD)+ tDB(zB − zD)+ tDF(zF − zD)−75 = 0tDE(3−0)+ tDC(0−0)+ tDB(0−0)+ tDF(3−0) = 75

3tDE +3tDF = 75tDE + tDF = 25

As the structure and loading are symmetrical

tDE = tDF

and hencetDE = 12.5


3 m

x

y

C

A

E

F

D

B

3 m

3 m 3 m

FIG. 2.74

Joint x y z

A +6 0 +3B +6 +3 0C +6 −3 0D 0 0 0E +3 −3 +3F +3 +3 +3

Resolving all the forces meeting at joint D in z direction as

tDE(zE − zD)+ tDC(zC − zD)+ tDB(zB − zD)+ tDF(zF − zD)−75 = 0tDE(3−0)+ tDC(0−0)+ tDB(0−0)+ tDF(3−0) = 75

3tDE +3tDF = 75tDE + tDF = 25

As the structure and loading are symmetrical

tDE = tDF

and hencetDE = 12.5


Resolving all the forces in the directions of y axis at the joint E

tEC(yC − yE)+ tEA(yA − yE)+ tEF(yF − yE)+ tED(yD − yE) = 0tEC(−3+3)+ tEA(0+3)+ tEF(3+3)+ tED(0−3) = 0

3tEA +6tEF −3×12.5 = 03tEA +6tEF = 37.5

Resolving all the forces in the direction of z-axis

tEC(zC − zE)+ tEA(zA − zE)+ tEF(zF − zE)+ tED(zD − zE) = 0tEC(0−3)+ tEA(3−3)+ tEF(3−3)+ tED(0−3) = 0

−3tEC −3tED = 0tEC = tED

Resolving all the forces in the direction of x-axis

tEC(xC − xE)+ tEA(xA − xE)+ tEF(xF − xE)+ tED(xD − xE) = 0tEC(6−3)+ tEA(6−3)+ tEF(3−3)+ tED(0−3) = 0

3tEC +3tEA −3tDE = 03(−12.5)+3tEA −3×12.5 = 0

3tEA = 75

tEA = 25

Substituting in the equation:3×25+6 tEF = 37.5

tEF = −6.25

LDE =√

(xE − xD)2 +(yE − yD)2 +(ZE −ZD)2

=√

32 +32 +32 = 5.2m

LEF =√

(xF − xE)2 +(yF − yE)2 +(ZF −ZE)2

=√

(3−3)2 +(3+3)2 +(3−3)2 = 6m

∴ TDE = tDELDE = 12.5×5.2 = 65 kN (Tensile)

TEF = tEFLEF = −6.25×6 = −37.5 kN (Compressive)


EXAMPLE 2.30: A space frame shown in figure is supported at A, B, C and D in a horizontalplane through ball joints. The member EF is horizontal and is at a height of 3 m above the base.The loads at the joints E and F shown in Fig. 2.75 act in a horizontal plane. Find the forces in allof the members of the frame. (Anna Univ, 2004)

C

A

E F

D

B

3 m

2 m 2 m

y

x3 m

3 m

10

15 kN 20

FIG. 2.75

Joint x y zA 0 6 0B 0 0 0C 7 0 0D 7 6 0E 2 3 3F 5 3 3

Resolving all the forces at joint E in x direction

tEA(xA − xE)+ tEB(xB − xE)+ tEF(xF − xE)+10 = 0tEA(0−2)+ tEB(0−2)+ tEF(5−2)+10 = 0

Resolving all the forces in y direction at joint E

tEA (yA − yE)+ tEB (yB − yE)+ tEF (yF − yE)+15 = 0tEA(6−3)+ tEB(0−3)+ tEF(3−3)+15 = 0

3tEA −3tEB = −15

Resolving all the forces in z direction at joint E

tEA (zA − zE)+ tEB (zB − zE)+ tEF (zF − zE)+0 = 0tEA(0−3)+ tEB(0−3)+ tEF(3−3) = 0

−3tEA −3tEB = 0


EXAMPLE 2.30: A space frame shown in figure is supported at A, B, C and D in a horizontalplane through ball joints. The member EF is horizontal and is at a height of 3 m above the base.The loads at the joints E and F shown in Fig. 2.75 act in a horizontal plane. Find the forces in allof the members of the frame. (Anna Univ, 2004)

C

A

E F

D

B

3 m

2 m 2 m

y

x3 m

3 m

10

15 kN 20

FIG. 2.75

Joint x y zA 0 6 0B 0 0 0C 7 0 0D 7 6 0E 2 3 3F 5 3 3

Resolving all the forces at joint E in x direction

tEA(xA − xE)+ tEB(xB − xE)+ tEF(xF − xE)+10 = 0tEA(0−2)+ tEB(0−2)+ tEF(5−2)+10 = 0

Resolving all the forces in y direction at joint E

tEA (yA − yE)+ tEB (yB − yE)+ tEF (yF − yE)+15 = 0tEA(6−3)+ tEB(0−3)+ tEF(3−3)+15 = 0

3tEA −3tEB = −15

Resolving all the forces in z direction at joint E

tEA (zA − zE)+ tEB (zB − zE)+ tEF (zF − zE)+0 = 0tEA(0−3)+ tEB(0−3)+ tEF(3−3) = 0

−3tEA −3tEB = 0


Solving the equations

−2 −2 +33 −3 0

−3 −3 0

tAEtEBtEF

=

−10−15

0

tAE = −2.5, tEB = +2.5, tEF = −3.33

LAE =√

(0−2)2 +(6−3)2 +(0−3)2 = 4.69 m

LEB =√

(0−2)2 +(0−3)2 +(0−3)2 = 4.69 m

LEF =√

(5−2)2 +(3−3)2 +(3−3)2 = 3 m

TAE = tAELAE = −2.5×4.69 = −11.725 (Compressive)

TEB = tEBLEB = +2.5×4.69 = +11.725 (Tensile)

TEF = tEFLEF = −3.33×3.00 = −9.99 (Compressive)

Resolving all the forces in x direction at joint F

tFE(xE − xF)+ tFD (xD − xF)+ tFC (xC − xF) = 0−3.33(2−5)+ tFD(7−5)+ tFC(7−5) = 0

2tFD +2tFC = −10tFD + tFC = −5

Resolving all the forces in y direction at the joint F

tFE(yE − yF)+ tFD (yD − yF)+ tFC (yC − yF)+20 = 0−3.33(3−3)+ tFD(6−3)+ tFC(0−3) = 0

3tFD −3tFC = 0

Solving the above

tFD = −2.5, tFC = −2.5

LFD =√

(7−5)2 +(6−3)2 +(0−3)2 = 4.69 m

LFC =√

(7−5)2 +(0−3)2 +(0−3)2 = 4.69 mTFD = tFDLFD = −11.73 kNTFC = tFCLFC = −11.73 kN


REVIEW QUESTIONS

Remembrance:

2.1 Define plane and space truss?2.2 What are the different types of analysis of trusses?2.3 List the assumptions made in truss analysis?2.4 Explain the steps involved in method of joints?2.5 Explain the steps involved in method of sections?2.6 Explain the method of tension coefficients?2.7 What type of analysis is used in determining the forces of a space truss?2.8 For what kind of trusses can be analysed by method of joints and method of sections?2.9 Define Tension coefficient?2.10 Which method is preferable to find out the forces in a few members of a truss?2.11 What is the primary function of a truss?2.12 What is the minimum numbers of elements to make a simple truss?

Understanding:

2.1 Distinguish between a simple truss, compound truss and complex truss?2.2 What kind of stresses developed if the loads are not applied at the joints?2.3 What are the limitations of method of joints?2.4 Identify the truss members having zero forces joints?


REVIEW QUESTIONS

Remembrance:

2.1 Define plane and space truss?2.2 What are the different types of analysis of trusses?2.3 List the assumptions made in truss analysis?2.4 Explain the steps involved in method of joints?2.5 Explain the steps involved in method of sections?2.6 Explain the method of tension coefficients?2.7 What type of analysis is used in determining the forces of a space truss?2.8 For what kind of trusses can be analysed by method of joints and method of sections?2.9 Define Tension coefficient?2.10 Which method is preferable to find out the forces in a few members of a truss?2.11 What is the primary function of a truss?2.12 What is the minimum numbers of elements to make a simple truss?

Understanding:

2.1 Distinguish between a simple truss, compound truss and complex truss?2.2 What kind of stresses developed if the loads are not applied at the joints?2.3 What are the limitations of method of joints?2.4 Identify the truss members having zero forces joints?

108•

BasicStructuralAnalysis

10 kN

F DE

A B C

3 m

3 m3 m

10 kN 80 kN

A BF

E D C

4 m4 m

3 m3 m

40 kN 40 kN

10 kNCB

A D

3 m

C

I H G

D

E

FB

A

3 m

3 m 2 m 2 m 2 m

40 kN P=20 kN

CD

B

A

FE

3 m

3 m 3 m 3 m

A

F

E B D

G

H

C

100 kN

4 m

3 m 3 m 4 m 5 m


EXERCISE PROBLEMS

2.1 Find the forces in all the members of the truss shown in figure. (Anna Univ, April, 2008)

C

AD

θ

2 m2 mB

4 kN

6 kN

1.5 m

(Ans AC = 2.5 kN, CB = −7.5, BD = +6, DA = +6.0, DC = +6.0)

2.2 Determine the forces in the members of the truss as shown in figure by method of joints.(VTU, Dec. 2008)

CAD 4 m4 m

B20 kN

30 kN

3 m

(Ans AD = +30 kN, DC = +30, BD = +30, AB = −12.5, BC = −37.5)

2.3 Determine the nature and magnitude of forces induced in all the members of the frame loadedas shown in figure by method of joints. (VTU, Feb. 2002)


EXERCISE PROBLEMS

2.1 Find the forces in all the members of the truss shown in figure. (Anna Univ, April, 2008)

C

AD

θ

2 m2 mB

4 kN

6 kN

1.5 m

(Ans AC = 2.5 kN, CB = −7.5, BD = +6, DA = +6.0, DC = +6.0)

2.2 Determine the forces in the members of the truss as shown in figure by method of joints.(VTU, Dec. 2008)

CAD 4 m4 m

B20 kN

30 kN

3 m

(Ans AD = +30 kN, DC = +30, BD = +30, AB = −12.5, BC = −37.5)

2.3 Determine the nature and magnitude of forces induced in all the members of the frame loadedas shown in figure by method of joints. (VTU, Feb. 2002)


BA

ED

60° 60°

20 kN

3 m3 m C

10 kN

5 kN

(Ans AD =−22.75, DE =−11.16, EB =−11.9, AC = 11.38, CB = 10.95, DC =−0.43,EC = −0.43)

2.4 Determine the forces in all the members of the following truss by method of joints. (VTU,Feb. 2003)

2 kN

AF

B C

ED

3 m 3 m 3 m

2 kN

45°

(Ans AB =−0.95 kN, BC =−0.658, CD

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