structural analysis using both manual calculations"slope deflection" and software"Robot" for a frame.Term protect solution.See https://www.slideshare.net/OmarDaher6/cvle-312-term-project.

1. 0
Beirut Arab University
Faculty of Engineering
Department of Civil and Environmental
Engineering
CVLE 312
Structural Analysis II
Spring 2017/2018
Term Project
Group: F
Members:
Omar Daher 201702351
Omar Talih Fawaz 201600407
Date of submission:11/05/2018.
Submitted to:
1. Dr. Wael Slika 2.Ing. Meriana Othman

2. 1
Table of contents:
1. Introduction: P:2
2. Chapter 1: manual calculation of the reactions at the
supports: P:4
3. Chapter 2: Software calculation and analysis: P:13
4. Conclusion: P:17

3. 2
Introduction:
Definition of frames:
A Frame structure is a structure having the combination of beam, column and slab to
resist the lateral and gravity loads. These structures usually used to overcome the large
moments developing due to the applied loading.
Frame geometry and parameters:
Parameters for group F:
W
(KN/m)
P1
(KN)
P2
(KN)
P3
(KN)
P4
(KN)
a
(m)
b
(m)
c
(m)
I
(V.E)
I
(H.E)
Group F 20 6 12 14 10 5 3 3 3I 4I
Figure 1-1: The Required Structural Frame

4. 3
Manual method that is chosen and its importance:
The manual method chosen is slope deflection because it is easy and not confused. And if
we want to use moment distribution it is confused and takes a lot of time and the error is
high.
Importance: George A. Maney introduced the slope-deflection method for the analysis of
indeterminate beams and frames in 1915. The method takes into account only the bending
deformations of structures. Although ,the slope-deflection method is itself considered to
be a useful tool for analyzing indeterminate beams and frames, an understanding of the
fundamentals of this method provides a valuable introduction to the matrix sti¤ness
method, which forms the basis of most computer software currently used for structural
analysis.
Software used:
Civil engineers have one of the world's most important jobs, and to achieve this you have
to be precise.
The software that is used is two soft wares:
1. Autodesk Robot Structural Analysis Professional 2014: We choose it because it is easy
to use and it is precise.
Where we can find this type of frames:
This frame can be found in many structures. However, it is used mostly in places where
we have slender columns like airports, and mostly used in steel structures where the
design of hinge more probable can be executed.
The Results Requirements For group F:
Show the values of the reactions in tabulated format.

5. 4
Chapter 1: manual calculation of the reactions at the supports:
Figure 2-1: Applied loads on the structure and the dimensions
Figure 2-2: Deformed Shape -Elastic Shape

6. 5
Assumptions:
φD’=EIφD
φE’=EIφE
φF’=EIφF
ψ’=EIψ
Sign convention:
Member AD:
FEM: Fixed end moments.
MAD=
2∗𝐸∗(3∗𝐼)
9
(0+φD-3ψ)-20
MAD=
2
3
φ’D-2𝜓′
− 20 1
MDA=
4
3
φ’D-2𝜓′
+ 16 2
Member DE: FEM: Fixed end moments.

7. 6
MDE=
2∗𝐸∗(4𝐼)
5
(2 φD+ φE)-
125
3
MDE=
16
5
φ’D+
8
5
φ’E-
125
3
3
MED=
8
5
φ’D+
16
5
φ’E+
125
3
4
Member EB: FEM=0 no applied loads.
MEB=
3∗𝐸∗(3𝐼)
9
(φE-ψ)
MEB= φ’E-ψ’5
Member EF:
FEM: fixed end moments.
MEF=
2∗𝐸∗(3𝐼)
9
(2* φE+ φF)-
21
2
=
8
3
* φ’E+
4
3
φ’F-
21
2
=MEF6
MFE=
4
3
φ’E+
8
3
φ’F+
21
2
7
Member FC:
FEM=0.
MFC=
2∗𝐸∗(3𝐼)
9
(2 φF-3ψ)
MFC=
4
3
φ’F-2ψ’8
MCF=
2
3
φ’F-2ψ’9
Maintaining equilibrium at the nodes:

8. 7
At node D:
MDE+MDA=0
4
3
φ’D-2*ψ’+16+3.2 φ’D+1.6 φ’E-
125
3
=0
①
68
15
* φ’D+1.6 φ’E+0-2ψ’=
77
3
At Node E:
MED+MEB+MEF=0 solving the equality……
② 1.6* φ’D+
103
15
* φ’E+
4
3
φ’F-ψ’=
−187
6
AT node F:
MFE+MFC=o
…..calculation we will get:
③0 +
4
3
* φ’E+4* φ’F-2ψ’=
−21
2

9. 8
Need of fourth equation:
Bring VA and VB and Vc:
VA=???
∑ 𝑀@𝐷=0.
-MAD-MDA+6*3+12*6+9*VA=0

10. 9
VA=
𝑀𝐴𝐷+𝑀𝐷𝐴−90
9
=
2φ’D−4ψ′−94
9
VB=???
VB=
𝑀𝐸𝐵
9
=
φ’E−4ψ′
9
.
Vc=???.

11. 10
Vc=
𝑀𝐶𝐹+𝑀𝐹𝐶
9
=
2∗φ’F−4∗ψ′
9
VA+VB+Vc+6+12-10=0
④
2
9
* φ’D+
1
9
* φ’E+
2
9
φ’F-ψ’=
22
9
Final equations:
①
68
15
* φ’D+1.6 φ’E+0-2ψ’=
77
3
② 1.6* φ’D+
103
15
* φ’E+
4
3
φ’F-ψ’=
−187
6
③0 +
4
3
* φ’E+4* φ’F-2ψ’=
−21
2
Go to mat lab to bring the solutions:

12. 11
Finally,
φD’=6.9864
φE’=-6.148
φF’=-1.5336
ψ’=-1.9158
Substitute each value in equations 1 2 3 4 5 6 7 8 9.
MAD=-11.508KN.m.
MDA=29.1468KN.m.
MDE=-29.1468KN.m.
MED=33.1713KN.m.
MEB=-4.2322KN.m.
MEF=-28.9395KN.m.
MFE=-1.78693KN.m.
MFC=1.7868KN.m.
MCF=2.809KN.m.
VA=-8.0404KN.
VB=-0.47024KN.
Vc=0.52067KN.
Normal forces.

13. 12
∑ 𝑀@𝐷=0
-100*2.5+29.1469-33.1713+NE=0
ND=50.80488KN.
NE1=49.19512KN.
∑ 𝑀@𝐹=0
-NE2*6+14*3+28.3395+1.7868=0
NE2=12.1205KN.
NE=61.31617KN.
NF=1.879895KN.
Therefore:
NA=50.80488KN.
NB=61.31617KN.
Nc=1.879895KN.

14. 13
Chapter 2: Software calculation and analysis:
Using robot:
Step1:
Go to tools  Job preferences put all the units in metric
In addition, go to modifications and put all the densities
zero as not the self-weight to be included.
Step 2:

15. 14
Draw the 5 bars and assign at nodes 1, 6 as fixed support and at node 4 as hinge.
Put the vertical properties of the sections as rectangular with (h=25cm and b=30cm)and
horizontal members as (h=25cm and b=4cm) to make it by default as 3I and 4I.
Step 3:loads.
Put all loads as 1 live loads(to take one case) and apply each one on the members and it is
assigned to us.
Step 3: run the calculations:
Bring the moments and forces from the table.

16. 15
Bending moment diagrams:

17. 16
Normal force diagrams:
Shear force diagrams:

18. 17
Conclusion:
manual Software Error(%)
Support Shear Normal Moment Shear Normal Moment Shear Normal Moment
A -8.0404 50.8048
-
11.5108 -8.02 49.27 -11.43 0.253719 3.020974 0.701949
B
-
0.47024 61.31617 0 -0.47 62.78 0 0.051038 2.387347 -
C 0.52067 1.879895 2.89092 0.49 1.95 2.78 5.890487 3.729198 3.836841
In this project, we had learned how the software facilitates our field in civil and environmental
engineer and the error between the manual calculations and electronic calculations.
Thanks for BAU University for teaching us this important courses and thanks for Dr. Wael Slika
and Ing. Meriana Othman wishing them to hire better degrees.