Temesgen Abiy GSR3604/07 Page 1

Assignment 3.1

Analyze the beam and loading shown below using the basic stiffness method. Assume EI is

constant throughout the beam.

Fig 1

Solution

N = 73 P= 180+73 = 253KN

a= 5+0.01(73) = 5.73m w1= 60+0.02(73) = 61.46KN/m

b= 3+0.01(73) = 3.73m w2= 90+0.02(73) = 91.46KN/m

c= 8+0.02(73) = 9.46m M= 120+0.2(73) = 134.6KN-m

EI =Constant

Fig 2

Temesgen Abiy GSR3604/07 Page 2

Step1 Degree of freedom

Kinematic indeterminacy = 3 i.e. ƟB, ƟC and ƟD

Assigning coordinates for corresponding joint which is free to displacement (may be translation

or rotation). Let assign joint B as coordinate 1, C as coordinate 2 and D as coordinate 3.

Fig 3 Coordinates

Fig 4 Giving unit value of ƟB =1

Fig 5 Giving unit value of ƟC =1

Fig 5 Giving unit value of ƟD =1

Temesgen Abiy GSR3604/07 Page 3

Step3 Stiffness calculation

If unit displacement is imposed only once at corresponding coordinates and the stiffness matrix

is formulated as:

K11 = 4EI/LBA +4EI/LBC , LAB =LBC = 9.46m

K11 = 8EI/LBA = 8EI/9.46 , LCD = 9.46m

K12 =K21 = K23= K32=2EI/LBA = 2EI/9.46

K22 = 4EI/LCB + 4EI/LCD = 8EI/9.46

K33 = 4EI/LDC + 4EI/LDE = 8EI/9.46

K13 =K31 = 0

Therefore the stiffness matrix K will be:

11 12 13

21 22 11

31 32 33

K K K

K K K K

K K K

8 2 0

2 8 29.46

0 2 8

EIK

Fixed End Moment (F.E.M)

AB MF

AB =2

2

Pab

L+

22 2

2(6 8 3 )

12

waL aL a

L

=2

2

253*5.73*3.73

9.46+

22 2

2

61.46*5.73(6*9.46 8*5.73*9.46 3*5.73 )

12*9.46 = 604.57 KNm

MF

BA = -2

2

Pa b

L-

3

2(4 3 )

12

waL a

L

= -2

2

253*5.73 *3.73

9.46-

3

2

61.46*5.73(4*9.46 3*5.73)

12*9.46 = -568.56KNm

BC MF

BC = 2

2

Pab

L+

2(b 2 )

Mba

L =

= 2

2

253*5.73*3.73

9.46+

2

134.6*3.73(3.73 2*5.73)

9.46 =182.01KNm

Temesgen Abiy GSR3604/07 Page 4

MF

CB= -2

2

Pa b

L-

2(2 )

Mab a

L

= -2

2

253*5.73 *3.73

9.46-

2

134.6*5.73(2*3.73 5.73)

9.46 =-361.13KNm

CD

MF

CD = 2

30

WL=

291.46*9.46

30 = 272.83KNm

MF

DC = 2

20

WL=

291.46*9.46

20= -409.245KNm

DC

MF

DE = 2

20

WL=

291.46*9.46

20 = 409.245KNm

MF

ED = 2

30

WL=

291.46*9.46

30= -272.83KNm

Step3 Restraining forces

P’1 = MF

BA + MF

BC = -568.56+182.01= -386.55 KNm

P’2 = MF

CB + MF

CD = -361.13+272.83 = -88.3 KNm

P’2 = MF

DC + MF

DE = -409.245+409.245 = 0 KNm -

386.55

' 88.3

0

P KNm

Step4 Determine the actual displacement (rotation in our case).

{D}= [K]-1

({AD} – {ADL}), AD =P = 0 and ADL = P’

[ƟB ƟC ƟD]T = -[K]

-1{ADL} =[ 459.91 -11.27 2.82]

T

Temesgen Abiy GSR3604/07 Page 5

ƟB =

rad ƟC = -

rad ƟD =

rad

Step5 Forces on restrained and in given structure by Slope Deflection

Equation:

2 3(2 )AB A B AB

EIM FEM

LAB LAB

ƟA= ƟE= ∆ = 0

2 3*0*(2*0 459.91 ) 604.57

9.46 LAB

= 701.80KNm

2 3(2 )BA B A BA

EIM FEM

LBA LBA

2 3*0

(2*459.91 0 ) 568.569.46 LBA

= -374.09KNm

2

2 3BC B C BC

BC

EIM FEM

L

Ψ = ∆/L =0 , ∆ =0

2

2*459.91 11.27 3*0 182.019.46

= 374.09 KNm

2

2 3CB C B

BC

EIM FEMCB

L

, Ψ = ∆/L =0 , ∆ =0

2

2*11.27 459.91 3 361.139.46

= -268.66KNm

Temesgen Abiy GSR3604/07 Page 6

2

2 3CD C D CD

CD

EIM FEM

L

Ψ = ∆/L =0 , ∆ =0

2

2* 11.26 2.815 3 272.839.46

=268.66KNm

2

2 3DC D C DC

DC

EIM FEM

L

2

2*2.815 11.25 3*0 409.2459.46

= -410.433KNm

2

2 3DE D E DE

DE

EIM FEM

L

2

2*2.815 0 3*0 409.2459.46

= 410.433KNm

2

2 3ED E D ED

DE

EIM FEM

L

2

2*0 2.815 3*0 272.839.46

= -272.23KNm

Temesgen Abiy GSR3604/07 Page 7

Step6 Determine other support reactions and draw shear force and bending

moment diagram.

ΣMB =0 :- RA(9.46)+701.80-374.09+61.46(5.73*6.595)+253*3.73 = 0

RA = 379.9KN RBL = 225.257KN

ΣMB =0: RCL(9.46)+374.09-268.66-253*5.73-134.6 = 0

RCL = 156.327KN RBR = 96.672KN

Temesgen Abiy GSR3604/07 Page 8

ΣMC =0: RD(9.46) +268.66-409.245-0.5*91.46*9.46*6.30 = 0

RDL = 302.52KN RCR = 130.08KN

ΣMD =0: RE(9.46) -272.83+409.245-0.5*91.46*9.46*6.30 = 0

RE = 130.08KN RDR = 302.52KN

Temesgen Abiy GSR3604/07 Page 9

Reactions

RA = 379.9KN

RB = RBL +RBR = 225.257KN + 96.672KN = 321.929KN

RC = RCL + RCR = 156.327KN + 130.08KN = 286.407KN

RD = RDL + RDR = 302.52KN + 302.52KN = 605.04KN

RE = 130.08KN

Check To check the correctness of the final support reactions obtained, the equilibrium of the

entire beam under vertical forces is considered as follows:

ΣFy =0 , RA+ RB+ RC+ RD+ RE – (w1*a)-P-P-2(0.5* w2*c)= 0

379.9 + 321.929 + 286.407 + 605.04+130.08--61.46*5.73-253-253-2*(0.5*9.46*91.46) = 0

1723.36-1723.36= 0 -------------------------------------- OK

Shear force diagram

Bending moment diagram