ANALYSIS OF TRUSSES

A Truss is a two force members made up of bars that are

connected at the ends by joints. Every stress element is in either tension
or compression. Trusses can be classified as plane truss and space truss.

Plane truss is one where the plane of the structure remain in

plane even after the application of loads

While space truss plane will not be in a same plane

Fig shows 2d truss structure and each node has two degrees of freedom.
The only difference between bar element and truss element is that in
bars both local and global coordinate systems are same where in truss
these are different.

There are always assumptions associated with every finite element

analysis. If all the assumptions below are all valid for a given situation,
then truss element will yield an exact solution. Some of the
assumptions are:

Truss element is only a prismatic member ie cross sectional

area is uniform along its length
It should be a isotropic material
Constant load ie load is independent of time
Homogenous material

50
 A load on a truss can only be applied at the joints (nodes)
Due to the load applied each bar of a truss is either induced
with tensile/compressive forces
The joints in a truss are assumed to be frictionless pin joints
Self weight of the bars are neglected

Consider one truss element as shown that has nodes 1 and 2 .The
coordinate system that passes along the element (x l axis) is called
local coordinate and X-Y system is called as global coordinate
system. After the loads applied let the element takes new position
say locally node 1 has displaced by an amount q 1l and node2 has
moved by an amount equal to q2l.As each node has 2 dof in
global coordinate system .let node 1 has displacements q1 and q2
along x and y axis respectively similarly q 3 and q4 at node 2.

Resolving the components q1, q2, q3 and q4 along the bar we get two
equations as

51
 Or

Writing the same equation into the matrix form

Where L is called transformation matrix that is used for local global

correspondence.

Strain energy for a bar element we have

U = ½ qTKq
For a truss element we can write

U = ½ qlT K ql
Where ql = L q and q1T = LT qT

52
Therefore

U = ½ qlT K ql

Where KT is the stiffness matrix of truss element

Taking the product of all these matrix we have stiffness matrix for truss
element which is given as

53
Stress component for truss element

The stress in a truss element is given by

= E

But strain = B ql and ql = T q

Therefore

How to calculate direction cosines

Consider a element that has node 1 and node 2 inclined by an angle
as shown .let (x1, y1) be the coordinate of node 1 and (x2,y2) be the
coordinates at node 2.

54
When orientation of an element is know we use this angle to calculate l
and m as:

l = cos m = cos (90 - ) = sin

and by using nodal coordinates we can calculate using the relation

We can calculate length of the element as

55
 3
2

Solution: For given structure if node numbering is not given we have to

number them which depend on user. Each node has 2 dof say q1 q2 be
the displacement at node 1, q3 & q4 be displacement at node 2, q5 &q6
at node 3.

Tabulate the following parameters as shown

For element 1 can be calculate by using tan = 500/700 ie = 33.6,

length of the element is

= 901.3 mm
Similarly calculate all the parameters for element 2 and tabulate
56
Calculate stiffness matrix for both the elements

Element 1 has displacements q1, q2, q3, q4. Hence numbering scheme
for the first stiffness matrix (K1) as 1 2 3 4 similarly for K2 3 4 5 & 6 as
shown above.

Global stiffness matrix: the structure has 3 nodes at each node 3 dof
hence size of global stiffness matrix will be 3 X 2 = 6
ie 6 X 6

57
From the equation KQ = F we have the following matrix. Since node 1
is fixed q1=q2=0 and also at node 3 q5 = q6 = 0 .At node 2 q3 & q4 are
free hence has displacements.
In the load vector applied force is at node 2 ie F4 = 50KN rest other
forces zero.

By elimination method the matrix reduces to 2 X 2 and solving we get

Q3= 0.28mm and Q4 = -1.03mm. With these displacements we
calculate stresses in each element.

58
Solution: Node numbering and element numbering is followed for the
given structure if not specified, as shown below

Let Q1, Q Q8 be displacements from node 1 to node 4 and

Tabulate the following parameters

Determine the stiffness matrix for all the elements

59
Global stiffness matrix: the structure has 4 nodes at each node 3 dof
hence size of global stiffness matrix will be 4 X 2 = 8
ie 8 X 8

From the equation KQ = F we have the following matrix. Since node 1

is fixed q1=q2=0 and also at node 4 q7 = q8 = 0 .At node 2 because of
roller support q3=0 & q4 is free hence has displacements. q5 and q6
also have displacement as they are free to move.
In the load vector applied force is at node 2 ie F3 = 20KN and at node 3
F6 = 25KN, rest other forces zero.

60
Solving the matrix gives the value of q3, q5 and q6.

61
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Globa s t s maliz

K-.]-[t.]+[t«/+lk
2 3 4 S

O -0-66 O
0-66

O
2
- o.l44 o 112
0-04 0.12
-0-66
3
-0 S O 12 -0-2S6
0-136

- o 66 O
O o-66
O

O -6 S +0.S O 6
O

-0-144 o-192 -0-66 0-8 oh - 0-192

O O -o-192 0-2S6
O o-112 -0-

ovealX E" E

R
R

S
O
 (124)
P p g ekanminaon mksod B . c & dl nd
5 Ro colun we hoNe.
3
0.8o4 ,, t o tO =
4o KIO
O-66 s 0

O493S mm
stAnsen s & stlas in ehnut O)

-R -m
e M

O
-

049

6, -(o.49s)
3 oD

E . ool 6 sg

0 E E = 33t.69 m
8 t
M

E, M

Re
6499s
O - O
oo

ECat

Qes-[ -1
.M

- M
 -06 8 o- - o 8
0-495
Soo

(-b-6 -o 6 xo-494s)
Soa

E S.14x lo-

O E =-|19.4 Nm

To fnd f a seaconn
o , - o-66 = R
Kt -3 2,83sN

Kv = -

o.112, + o94V,- o4+ ol2,-02f67,

KHas -o:l4
4+ 0- 192, o.s61 -

t
o-8g, o1
-

K H64 w

R = o-112,- 0i,-o21, oí -
 lert -Deli ine
s h ow in
6 A 4 bas euns
istlns n each ele
NOda dioplacmet
a toL
2
al tta Supprt A=loo mm
ik)Raclm
2SKN E:2x1oS Nm
(030)
3 3
v (ue,3e)

om)L

3omm

21
20 kN
400) (uoye)
4 omm

Node dala

ede nno

O O
2

3 30
=40mmy
4 30

le o-to+ Co-3)
Qe33omM
Re go-o)t(So--)*
Qes Sonn
Re o-io) -+ (30- 3o)
hoMm
Emt tasle
Emt 2
M
ho de node

2 O

2 2 3 30
3
3 So 0.8 -

o-6 0-48 640 36.

3
-

sLkmans malix elnetO.

2 --m
K AE
2
m

-n
n

2 3 4
loox2x(o
O
2

O
o 4

3
s -5 o

2
 3

O 3

0 0 -
30
0
-1

3 S

O O o 3
2= l o
o 6.67 o-66

O O
O -6.63 o 66

-64 o-48 O.64 -0.48

K3 lox2x 10
0-48 o 36 -0.48 -036 2
30
-o-64 -o4 o.64 0-48s
-o48 -0-36 o48 0 36

o s -42
2 441.12 -14
25 -192 R-S6 .92
-192 92 1-44

2x1ox100 S
SO
O
10
- O
o o7 -S0 S o
O o 8

o o o o*
 (30
sEPhnens malix

x- [.]-1t«J+[E<,].lk«]
2 S 6 8

12 -Ss O -2-S6 192 0 O

O O 0-1.12 -144 o O
2
k i
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OO 6-6 -6-69 0 O

S6-1-2 O O
O s6 1.92 -S
S

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O -66 192 o oO 6.
O O - os O
OS

O O O O 8

OUeRalQ.9

Fo

lo S

F 2oxio

Fs=o

Fe -2SKio
Fy= o
FsO
nodel-6t kaad Sppat o,,=o.

hode2' Polle Sppott o (y dspkemt 2u)

hodeh higed Suppot o , u #0 .

O
2oxi
10 O 9s6 192
-42 8 - I | |V
-2Skio

9s6is + 192 = 0 .2
-12/s +8:1 -0-24 -56

S9- V= 0-2S/o
-4-33x 1o

-o.0833 m

stRus n a c elornert

elant O

-ola
O

o O +0t2xio+o
O2o0N
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&0-6 -o.8-o 6 o.00833

So -0 03271,

0-S2 N

m
 3
PRve lannuta0 st smaCa as E Load Uelts f
-Db emel Undes
ARady decud Part A
Sac TRackon fotce
Load vela
Bod fotce veelA

Derve StAai Mal & steLs

ML f cST
larnent (2-D lemt)

(n

Cx)

(x
)
Cons dex a ota csT nat as slho.
Let ,2,vs be te
nodadplacemel ako -dtis
Y-d".

U- N, ,tN r
Vs
V N, V, f N, n,+ N, V
T shape 4 a Lij
N, &, N2 N-a
In Li as Slape , Au stlay in t u a n e t

C G onstat. Henee Sueh a tASanzla ee

shet stAain s Cantat, C i i , h elanmel Call

CsT
E &Lec
U: & n (1--a) +

+2l4,-«)+

-
V &. + 2% + ( - £ - ) .

=&(-) +1(-«) +u
Conateut

& mobes

u O
N O
N O
NT

V
L t mat 36
P te ay pant wilTi, Can be
expksosed
tuns sLape f gnodad o-odaa

N*+
*+N*2 t NX3
Y N4+ , 2+ N,

4 &1t 4. +
(1-E-n)Y
4 () e +
(Y»-43) +
P
- *i
&.

ecnea
3 t 27? + , 3

nS & enl
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&

he x

Call tAann feemalan nalE

Joco bian the TAanofrnalon T)

T iUse fo tonskotnn nala
maliir
Co-oidn ai
Corlien o-dinr. {l Cale2
 3

T i n d a n g eP e i a g a li e z

&cha s itoXe
malua.
a

M23 SR

T -323

on

Co- ciss s T
. mn
= ) 42, 2
X23 23
2+ 1
2 -
3
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T
ven as

23-

u
IT
.

Adarn E0«O
& -(1,
= ( 6).
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) (1,-s)

23 -g(-e)
TI
- )tig (-u)
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Staain matix 6

+v

hom )&1 thon phanious detvaon

Ysa-) t2 (-ts)

- ,-) + a (s)+» (*«)|

122 - 3-3
=--+
2=412
-*It%
(x-x)

1,( s)- ha e)
3)
i)-22(-«) - x1(y-)

x)- ,-9,)+ Xs (,-)+,(1,-e) Yia(-«) -

= 2 , + (*-a)+1+ a 1
+(17-41 -g
X2, +
X2, +X13 ,+ 4,2 +412 % i7 -

16 T
32
32 O 2

O.
L8]- nt stain
Ht desplacnlmalx
taen stkai selaz
to Six nodad deplocam
nodalNod doplacmet nmaliie.
sERain Mai
DeRve sttain &sttns
elent(20-nat) acobian mabix)

,)
3

L
(x)

x(o)
Consider a
loca uadala@zia a t
,2, 3 E4
hanui node
Lt ,vVs V2 te nodal diplaamat ae -d
4-drn
Cowni der
desplocmnt (uv) at a poit (x,) tRi
He
t n t tan be Aeptastalas in tatnn ok t
shape
ps , 4 oda dsplacimne .

U= N,, t
N, + N, s t
N,4,
'V

GL
 o
ere NJ shape f" aki x.
Noda dipkummut malx

The Co-isa Po
be Aapiperlrd in teenms oape Prs & nodal a-ddi nla
N ,X+N2 X2 +N, 2 +N y
N 4, +N, 4 +N,9+ N, Y

Assun a
) o ta f"
(En)

US paulad deva UsCa ala pae

dvaene

f
Jx

iCaiiad jacoban malkx. shape P ki,eas

 N ()(-)

x )(-a)x, (tes)-)k,+(«&Ken) oc
+

d EE t

(1-&)x, -(1ea)* +(te &) x,+ (1-s )x.}-

l-i(1-)t (l-n)'% +(tt) 1a

-(t+) ©
Let
T
then"OLconw

TT

Cabete T =7
Tobian malx.

T2-T2
I7I-T T

d d1=T| d&d. ).
osove E Can Le Ud ft
devin e
stnans ma
Dve stAaini lacnnt ulaI Bn
-al
fo uadslala elet

stain diplacak rALon 2D

By Considss f-u *" ) >pAwzous duzu on

T-71
ITT J

I7

Pt fv E" (1 PAan20we deeiuaGn.

T 2
 ((42
,,& w

-T

wL 4" O, & maliix fr&usaig O

-S2 n
O T
T

coher -J
O -J

-Tar

U N, ,+ N,, +
N, Vs t N, V, 7
V N tNL V, + N, V t Ny ve
= -(-a)4, - (#), +(ea) y +(1-a),>0-

-(-a) +(1-, +(trm)- (a)u

N , +
dnz + Ng +
dN
=
-(-a), -(148J, +
(1+&)%+(-)%
& 0,0,0 maliix m

--)o () (*n) o
(tr) o
-(1-E.) (1) o
(&)
= o -C-)
o
(-) o
o
(1) O
(ttn)o (
o - -+)
O
Sushe s"

ARoo StAuss
fr-J-fe}

Numerca TL on
ALd neulkal plo blmo & devaton has ben
donn in Pant A ( chaptes )