This presentation has include the methodology to analysis of plane and space truss.

1. Unit-5
Tension Coefficient Method
Presented by : -DIVYA VISHNOI

2. Analysis of frame

3. Tension Coefficient Method
• Prof. R.V. Southwell introduce the Tension
Coefficient Method.
• It is systematic presentation of the method
of joints.
• This is specially useful for space frame.
• This method is valid for perfect frame.

4. Tension Coefficient
• The Tension Coefficient for a member of a
frame is defined as the pull or tension in
that member divided by its length.
t= Tension coefficient for the member
T= Pull in member
L= Length of the member
L
T
t =

5. Tension Coefficient

6. Tension Coefficient
• Consider a member AB of a pin jointed
perfect frame in equilibrium under a given
system of external forces (or reactions)
acting at the joint.
• be the resulting pull in the member.TAB

7. Tension Coefficient
Let (xA, yA) be the coordinate of A and (xB, yB) be the
coordinate of joint B
In X- direction,
TAB Cos = TAB
= TAB
= tAB*(xB- xA)

8. Tension Coefficient
In Y- direction,
TAB Sin = TAB
= TAB
= tAB*(yB- yA)
Note : -A member which is compression will have –ve tension coefficeint.

9. Analysis of Plane Frame
• Let us consider a joint A, where a no. of
members AB,AC,AD…….. etc are
meeting.
• Let PA is the external force acting at the
joint A and let xA & yA be the component of
this force P in X- direction and Y-direction.

10. Analysis of Plane Frame
Under equilibrium condition,
∑H=0
tAB×(xB - xA)+ tAC×(xC - xA)+ tAD×(xD - xA)+………..+XA =0
∑V=0
tAB×(yB - yA)+ tAC×(yC - yA)+ tAD×(yD- yA)+………..+YA=0
So,
∑t (xF - xN )+XA =0
∑t (yF - yN )+YA =0
(xF, yF )= Coordinate of far end
(xN, yN )= Coordinate of near end

11. Problem
• A plane frame consists of two members
AB and CB, hinged at A and C to the wall,
as shown in fig. Determine the forces in
the two members due to vertical force P
applied at joint B.

12. Problem

13. Problem
Let us consider origin at joint at C and CX and CY be the axes of
reference.
The coordinates of three joints are C(0,0) ; B(2,0) ; A(0,1.5).
There are only two members AB and CB.
So,
2.0 m (given)

14. Problem
At joint B,
∑H=0
tBA×(xA – xB)+ tBC×(xC – xB)+ 0=0
tBA×(0-2) + tBC× (0-2)=0
tBA+ tBC =0 ………………(1)
∑V=0
tBA×(yA – yB)+ tBC×(yC – yB)+ P=0
tBA× (1.5-0)+ tBC×(0-0)-P=0
1.5 tBA =P KN/m
From equ. (1) KN/m (-ve, means BC in Compression)

15. Problem
• Force in member BA = TAB = tBA×LBA
• Force in member BC = TBC = tBC×LBC

16. Analysis of Space Frame
• All concepts are similar as plane frame but
the coordinate system is in three
directions (X,Y,Z).
• So, the Formulas are;
∑t (xF - xN )+XA =0
∑t (yF - yN )+YA =0
∑t (zF - zN )+ZA =0
(xF, yF ,zF)= Coordinate of far end
(xN, yN, zN )= Coordinate of near end

17. Problem
• A space frame shown in fig. is supported
at A,B,C and D in a horizontal plane,
through pin joints. The member EF is
horizontal, and is at a height of 3m above
the base. The loads at the joints E and F,
shown un the fig. act in a horizontal
plane. Find the forces in all the members
of the frame.

18. Problem

19. Problem
Point X Y Z
A 0 6 0
B 0 0 0
C 7 0 0
D 7 6 0
E 2 3 3
F 5 3 3
Let the origin be at B with X and Y axes along BC and
BA respectively and let Z axis be directed vertically.
The coordinates of various points are as under:

20. Problem
Length of various members

21. Problem
Joint E: The three equations at joint E as follow:
tEA×(xA - xE)+ tEB×(xB - xE)+ tEF×(xF - xE)+XE =0
tEA×(0-2)+ tEB×(0-2)+ tEF×(5-2)+5 =0
-2tEA-2tEB+ 3tEF+5 =0 ……………….(1)
tEA×(yA - yE)+ tEB×(yB - yE)+ tEF×(yF - yE)+YE =0
tEA×(6-3)+ tEB×(0-3)+ tEF×(3-3)+10 =0
3tEA-3tEB+10 =0 ………………..(2)

22. Problem
tEA×(zA - zE)+ tEB×(zB - zE)+ tEF×(zF - zE)+ZE =0
tEA×(0-3)+ tEB×(0-3)+ tEF×(3-3) =0
-3tEA-3tEB =0 ………………..(3)
From (3) tEA=-tEB
From (2)
OR
From (1)

23. Problem
Joint F: The three equations at joint E as follow:
tFD×(xD – xF)+ tFC×(xC – xF)+ tFB×(xB – xF)+ tFE×(xE – xF)+XF =0
tFD×(7-5)+ tFC×(7-5)+ tFB×(0-5)+ tFE×(2-5) =0
2tFD+2tFC-5tFB-3 tFE =0 ……………….(4)
tFD×(yD – yF)+ tFC×(yC – yF)+ tFB×(yB – yF)+ tFE×(yE – yF)+YF =0
tFD×(6-3)+ tFC×(0-3)+ tFB×(0-3)+ tFE×(3-3)+15 =0
3tFD-3tFC-3tFB +15=0 ………………..(5)

24. Problem
tFD×(zD – zF)+ tFC×(zC – zF)+ tFB×(zB – zF)+ tFE×(zE – zF)+ZF =0
tFD×(0-3)+ tFC×(0-3)+ tFB×(0-3)+ tFE×(3-3) =0
-3tFD-3tFC-3tFB =0 ………………..(6)
By solving the equations (4), (5) and (6)
tFD = -2.5 KN/m
tFC = +1.7857 KN/m
tFB = +0.7143 KN/m

25. Problem
Member Length t T (KN)
EA 4.6904 -1.67 -7.817
EB 4.6904 +1.67 +7.817
EF 3.0 -1.67 +5.0
FC 4.6904 +1.7857 +8.376
FD 4.6904 -2.50 -11.726
FB 6.5574 +0.7143 +4.684

26. References
• SMTS-II, Theory of Structures, by “B.C.
Punmia” , ISBN:8170086183.
• Structural and Stress analysis, by “T.H.G.
Megson”, ISBN: 97-8008-099-936-4.
• www.iitg.ac.in/kd/Lecture
%20Notes/ME101-Lecture07-KD.pdf