Y

30 kN
Stiffness Method (Frame Example)
3
Analyze the frame shown using Stiffness method. 3m
2
E is constant for all members. 20 kN
Section properties: 2

member Area Moment of

Inertia* 1 5m
1 A I 1
2 1.5A 3.5I 1 X

* A=0.04 m2 , I=0.0004/3 m4  A=300I 4m

(I) Member stiffness Matrices in Global coordinates:

(1) Member 1: i-node=1, j-node=2, L= 5 m, 5

j= 2 4
6

1
1
2
i= 1 1
3

Coord. #
1 2 3 4 5 6
1
2
3
4 (See Stiffness Handout)
5
6
 (2) Member 2: i-node=2, j-node=3, L= 5 m,

8
7
9
j= 3
2
5
1
i= 2 Coord. #
4
6
4 5 6 7 8 9
 K 11  a c 2  d s 2  57.7210EI   57.7210 43.0387 - 0.5040 - 57.721 - 43.0387 - 0.5040  4
   43.0387
 K 12  (a  d )c s  43.0387   32.6150 0.6720 43.0387 - 32.6150 0.6720 
5
 K 13  e s  0.5040EI   - 0.5040 0.6720 2.8000 0.5040 - 0.6720 1.4000 
   [ K ] 2  EI   6
 - 57.7210 - 43.0387 0.5040 57.7210 43.0387 0.5040 
2 2
 K 22  a s  d c  32.6150EI 
   - 43.0387 - 32.6150 - 0.6720 43.0387 32.6150 - 0.6720 7
 K 23  e c  0.6720   
 K 33  f  2.8EI   - 0.5040 0.6720 1.4000 0.5040 - 0.6720 2.8000 8

(II) Structure Stiffness Matrix (Global Stiffness Matrix):

1 2 3 4 5 6 7 8 9
0.096 0 -0.24 -0.096 0 -0.24 1
0 60 0 0 -60 0 2
-0.24 0 0.8 0.24 0 0.4 3
-0.096 0 0.24 0.096+ 0+ 0.24+
4
57.721 43.0387 -0.504 -57.721 -43.0387 -0.504
=EI  0 -60 0 0+ 60+ 0+
5
43.0387 32.615 0.672 -43.0387 -32.615 0.672
-0.24 0 0.4 0.24+ 0+ 0.8+
6
-0.504 0.672 2.8 0.504 -0.672 1.4
-57.721 -43.0387 0.504 57.721 43.0387 0.504 7
-43.0387 -32.615 -0.672 43.0387 32.615 -0.672 8
-0.504 0.672 1.4 0.504 -0.672 2.8 9
or

(III) Force Vector {F}

{F}91 ={F0}-{FE}
where

{F0} = vector of nodal forces ,

and
{FE} = vector of equivalent end forces (fixed end forces and moments due to span loadings.)

For our problem;

{F0}={F1, F2, …,F9}T = {Fx1, Fy1, Mz1, 20, 0 , 0, Fx3, Fy3, Mz3}T

Fixed End Moments and Forces:

Span 2 is loaded with concentrated load acting at its middle. The fixed end forces in member lcs are
calculated as shown below:

fe6 =-15 kN.m fe4 =9 kN

30 kN
24 kN
18 kN

fe5 =12 kN

fe3 =15 kN.m

fe1 =9 kN
fe2 =12 kN
Member 2

Coord. #
Fixed End Forces In global coordinates:
 4
5
6
7
8
9

The Global load vector becomes:

Unkown forces and

moents (Reactions)

Unkown d.o.f.

The global disp. vector

The global stiffness equation: [K]{}={F}

1 2 3 4 5 6 7 8 9

1
2
3
4
5
6
7
8
9

(Eq. I)

Solution of Stiffness Equations

After applying the boundary conditions, the stiffness matrix equation in free coordinates is
which have the solution

Support Reactions are found by back substituting in equation in (Eq. I) above:

Local effect (end shear forces and Moments in each member) can be obtained by substituting the
displacement results in the member stiffness equations.

Typical Computer Output:

NODAL DISPLACEMENTS
node x-disp. y-disp. z-rotation
1 000 000 000
2 6.517e-001 -4.355e-001 -4.038e+000
3 000 000 000

SUPPORT REACTIONS
supp. x-reac y-reac z-reac
1 9.065e-001 2.613e+001 -1.459e+000
3 -2.091e+001 3.870e+000 -2.127e+001

MEMBER FORCES:
MEM. END SHEAR FORCES END MOMENTS END AXIAL FORCES
I-NODE J-NODE I-NODE J-NODE I-NODE J-NODE
1 -9.065e-001 9.065e-001 -1.459e+000 -3.074e+000 2.613e+001 -2.613e+001
2 8.360e+000 1.564e+001 3.074e+000 -2.127e+001 3.240e+001 -1.440e+001