STRUCTURAL ANALYSIS II CE 313

Turja Deb Department of Civil Engineering Leading University, Sylhet.

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Question:

Determine the reaction and draw the shear and bending moment diagrams for the beam

shown in figure below by using moment-distribution method.

Figure: 01

Solution: The value of 퐸 = 200 퐺푝푎 = 200 × 10 푀푝푎 and 퐼 = 500 × 10 푚푚 , and for center span 퐼 = 2퐼.

The distribution factor at each joint must be computed first. The stiffness factors for the member are,

The stiffness factor: 퐾 =

=4 × 200 × 10 × 500 × 10

1000

= 4 × 10 푁 −푚푚.

퐾 =

=4 × 200 × 10 × 2 × 500 × 10

1000

= 8 × 10 푁 −푚푚.

퐾 =

=4 × 200 × 10 × 500 × 10

800

= 5 × 10 푁 −푚푚.

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The distribution factor (DF):

DF = = 1

DF = = ×( × ) ( × ) = 0.34

퐷퐹 = = ×( × ) ( × ) = 0.66

퐷퐹 = = ×( × ) ( × ) = 0.62

퐷퐹 = = ×( × ) ( × ) = 0.38

퐷퐹 = = 1

Calculation of fixed end moment (FEM):

퐹퐸퐹퐸푀 = − = −115.2 퐾푁 −푚.

퐹퐸푀 = = 172.8 퐾푁 −푚.

퐹퐸퐹퐸푀 = − = −115.2 퐾푁 −푚.

퐹퐸푀 = = 172.8 퐾푁 −푚.

퐹퐸푀 = − = −150 퐾푁 −푚.

퐹퐸푀 = = 150 퐾푁 −푚.

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Moment distribution table:

풋풐풊풏풕 푨 푪 푬 푮

푴풆풎풃풆풓 퐴퐶 퐶퐴 퐶퐸 퐸퐶 퐸퐺 퐺퐸

푫푭 1 0.34 0.66 0.62 0.38 1

푭푬푴

−115.2 172.8 −115.2 172.8 −150 150

푫풊풔풕 115.2 −19.584 −38.016 −14.136 −8.664 −150

푪푶 −9.792 57.6 −7.068 −19.008 75 −4.332

푫풊풔풕 9.792 −17.18 −33.35115 58.285 35.723 4.332

푪푶 −8.59 4.896 29.1425 −16.676 2.166 17.8615

푫풊풔풕 8.59 −11.573 −22.465 8.9962 5.5138 −17.8615

푪푶 −5.7865 4.295 4.4981 −11.2325 −8.93 2.7569

푫풊풔풕 5.7869 −2.99 −5.8 12.5 7.66175 −2.7569

푪푶 −1.495 2.89325 6.25 −2.9 −1.37845 3.83

푫풊풔풕 1.495 −3.1 −6 2.652639 1.625811 −3.83

푪푶 −1.55 0.7475 1.32632 −3 −1.915 0.129

푫풊풔풕 1.55 −0.7 −1.37 3.0473 1.8677 −0.129

푴 0 188.1 −188.1 191.329 −191.329 0

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Computation of shear force

I 2 I I6 m 4 m 6 m 4 m 4 m 4 m

1 2 0 K N 1 2 0 K N 1 5 0 K N

AB D

C EF

G

1 8 8 . 1 K N - m 1 8 8 . 1 K N - m 1 9 1 . 3 2 9 K N - m 1 9 1 . 3 2 9 K N - m

Figure: 02 the internal moment in the beam joint.

Calculate the reaction at support:

6m 4m

120KN

188.1KN-m

RA RC

Figure: 03

2I

6m 4m

120KN

DC E

RC RE

188.1KN-m 191.329KN-m

Figure 04

푀푅퐴 = 0

→ 120 × 6 − 10푅퐶 + 188.1 = 0

→ 푅퐶 = 90.81 퐾푁

푀푅퐶 = 0

→ 10푅퐴 − 120 × 4 + 188.1 = 0

→ 푅퐴 = 29.19 퐾푁

Taking moment in reaction RA in figure 3,

Again taking moment in reaction RC,

푀푅퐶 = 0

→ 120 × 6− 10푅퐸 − 188.1 + 191.329 = 0

→ 푅퐸 = 72.3229 퐾푁

푀푅퐸 = 0

→ 10푅퐶 − 120 × 4− 188.1 + 191.239 = 0

→ 푅퐶 = 47.677퐾푁

Now form figure 04

And,

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I

4m 4m

150KN

191.329KN-m

EF

G

RE RG

Figure: 05

6m 4m

120KN

188.1KN-m

29.19KN 90.81KN2I

6m 4m

120KN191.329KN-m188.1KN -m

DC E

47.6771KN 72.3229KNI

4m 4m

150KN

191.329KN-m

EF

G

98.9161KN 51.0839kN

Figure: 06 Reaction at support.

푀푅퐸 = 0

→ 150 × 4 − 8푅퐶 − 191.329 = 0

→ 푅퐺 = 51.0839 퐾푁

푀푅퐶 = 0

→ −150 × 4 + 8푅퐸 − 191.329 = 0

→ 푅퐴 = 98.9161 퐾푁

From figure 05,

Again taking moment in reaction RC,

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Shear Force and Bending Moment diagram:

6m 4m 6m 4m 4m 4m

120KN 120KN 150KN

AB D

C EF

G

29.19KN 183.4871KN 171.239KN 51.0839KN

29.19

-90.81

47.6771

98.9161

-51.0839

175.14

-188.1

97.9626

-191.329

204.3354

0 V(KN)

0 M(KN-m)

-72.3229

(Solved)

Filename: Assignment on CE 313,Turja Deb Mitun id 1301060028 .docx Directory: C:\Users\Turja\Documents Template: C:\Users\Turja\AppData\Roaming\Microsoft\Templates\Normal.dotm Title: Structural Analysis ii CE 313 Subject: Author: Turja Deb Keywords: ID:1301060028 Comments: Creation Date: 9/21/2015 7:28:00 PM Change Number: 46 Last Saved On: 10/4/2015 5:51:00 PM Last Saved By: Turja Deb Total Editing Time: 1,018 Minutes Last Printed On: 10/5/2015 8:44:00 AM As of Last Complete Printing Number of Pages: 7 Number of Words: 559 (approx.) Number of Characters: 3,188 (approx.)