Civil Engineering (CE) Exam > Civil Engineering (CE) Notes > Structural Analysis > The Direct Stiffness Method: Plane Frames - 2
The Direct Stiffness Method: Plane Frames - 2 | Structural Analysis - Civil Engineering (CE) PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
Page 2
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
After transformation, the assembly of member stiffness matrices is carried out in
a similar procedure as discussed for truss. Finally the global load-displacement
equation is written as in the case of continuous beam. Few numerical problems
are solved by direct stiffness method to illustrate the procedure discussed.
Example 30.1
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method.
Assume and . The flexural rigidity
4 4
10 33 . 1 ; 200 m I GPa E
ZZ
-
× = =
2
04 . 0 m A =
EI and axial rigidity EA are the same for both the beams.
Solution:
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The
numbering of joints and members are also shown in Fig. 30.3b. Each node has
three degrees of freedom. Degrees of freedom at all nodes are also shown in the
figure. Also the local degrees of freedom of beam element are shown in the
figure as inset.
Page 3
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
After transformation, the assembly of member stiffness matrices is carried out in
a similar procedure as discussed for truss. Finally the global load-displacement
equation is written as in the case of continuous beam. Few numerical problems
are solved by direct stiffness method to illustrate the procedure discussed.
Example 30.1
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method.
Assume and . The flexural rigidity
4 4
10 33 . 1 ; 200 m I GPa E
ZZ
-
× = =
2
04 . 0 m A =
EI and axial rigidity EA are the same for both the beams.
Solution:
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The
numbering of joints and members are also shown in Fig. 30.3b. Each node has
three degrees of freedom. Degrees of freedom at all nodes are also shown in the
figure. Also the local degrees of freedom of beam element are shown in the
figure as inset.
Formulate the element stiffness matrix in local co-ordinate system and then
transform it to global co-ordinate system. The origin of the global co-ordinate
system is taken at node 1. Here the element stiffness matrix in global co-
ordinates is only given.
Member 1: ° = = 90 ; 6 ? m L node points 1-2; 0 = l and 1 = m .
Page 4
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
After transformation, the assembly of member stiffness matrices is carried out in
a similar procedure as discussed for truss. Finally the global load-displacement
equation is written as in the case of continuous beam. Few numerical problems
are solved by direct stiffness method to illustrate the procedure discussed.
Example 30.1
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method.
Assume and . The flexural rigidity
4 4
10 33 . 1 ; 200 m I GPa E
ZZ
-
× = =
2
04 . 0 m A =
EI and axial rigidity EA are the same for both the beams.
Solution:
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The
numbering of joints and members are also shown in Fig. 30.3b. Each node has
three degrees of freedom. Degrees of freedom at all nodes are also shown in the
figure. Also the local degrees of freedom of beam element are shown in the
figure as inset.
Formulate the element stiffness matrix in local co-ordinate system and then
transform it to global co-ordinate system. The origin of the global co-ordinate
system is taken at node 1. Here the element stiffness matrix in global co-
ordinates is only given.
Member 1: ° = = 90 ; 6 ? m L node points 1-2; 0 = l and 1 = m .
[ ] [ ] [ ]
1
'
T
kT k T ?? =
??
3333
66
333
1
3333
66
333
1 2 3 4 5 6
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 17.78 10 4.44 10 0 8.88 10
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 8.88 10 4.44 10 0 17.
k
××××
×-×
××× ×
?? =
??
××××
-× ×
×××
3
1
2
3
4
5
6 78 10
??
??
??
??
??
??
??
??
× ??
??
3
(1)
Member 2: ° = = 0 ; 4 ? m L ; node points 2-3 ; 1 = l and 0 = m .
[] [] [ ][ ]
9
8
7
6
5
4
10 66 . 26 10 10 0 10 88 . 8 10 10 0
10 10 10 5 0 10 10 10 5 0
0 0 10 0 . 2 0 0 10 0 . 2
10 88 . 8 10 10 0 10 66 . 26 10 10 0
10 10 10 5 0 10 10 10 5 0
0 0 10 0 . 2 0 0 10 0 . 2
9 8 7 6 5 4
'
3 3 3 3
3 3 3 3
6 6
3 3 3 3
3 3 3 3
6 6
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
× × - × ×
× - × × - × -
× × -
× × - × ×
× × - × ×
× - ×
=
= T k T k
T
(2)
The assembled global stiffness matrix [ ] K is of the order 9 9 × . Carrying out
assembly in the usual manner, we get,
Page 5
3 3
' q p = (30.5c)
Thus the forces in global coordinate system can be related to forces in local
coordinate system by
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
1 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1 0 0
0 0 0 0
0 0 0 0
q
q
q
q
q
q
l m
m l
l m
m l
p
p
p
p
p
p
(30.6a)
Where, ? cos = l and ? sin = m .
This may be compactly written as,
{} [] { } ' q T p
T
= (30.6b)
30.3.3 Member global stiffness matrix
From equation (30.1b), we have
{} []{} ' ' ' u k q =
Substituting the above value of { } ' q in equation (30.6b) results in,
{} [] [ ] { } ' ' u k T p
T
= (30.7)
Making use of equation (30.3b), the above equation may be written as
{} [] [ ] [ ] { } u T k T p
T
' = (30.8)
or
{} [] { } u k p = (30.9)
The equation (30.9) represents the member load-displacement relation in global
coordinate system. The global member stiffness matrix [ ] k is given by,
[] [ ] [ ] [ ] T k T k
T
' = (30.10)
After transformation, the assembly of member stiffness matrices is carried out in
a similar procedure as discussed for truss. Finally the global load-displacement
equation is written as in the case of continuous beam. Few numerical problems
are solved by direct stiffness method to illustrate the procedure discussed.
Example 30.1
Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method.
Assume and . The flexural rigidity
4 4
10 33 . 1 ; 200 m I GPa E
ZZ
-
× = =
2
04 . 0 m A =
EI and axial rigidity EA are the same for both the beams.
Solution:
The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The
numbering of joints and members are also shown in Fig. 30.3b. Each node has
three degrees of freedom. Degrees of freedom at all nodes are also shown in the
figure. Also the local degrees of freedom of beam element are shown in the
figure as inset.
Formulate the element stiffness matrix in local co-ordinate system and then
transform it to global co-ordinate system. The origin of the global co-ordinate
system is taken at node 1. Here the element stiffness matrix in global co-
ordinates is only given.
Member 1: ° = = 90 ; 6 ? m L node points 1-2; 0 = l and 1 = m .
[ ] [ ] [ ]
1
'
T
kT k T ?? =
??
3333
66
333
1
3333
66
333
1 2 3 4 5 6
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 17.78 10 4.44 10 0 8.88 10
1.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 0
4.44 10 0 8.88 10 4.44 10 0 17.
k
××××
×-×
××× ×
?? =
??
××××
-× ×
×××
3
1
2
3
4
5
6 78 10
??
??
??
??
??
??
??
??
× ??
??
3
(1)
Member 2: ° = = 0 ; 4 ? m L ; node points 2-3 ; 1 = l and 0 = m .
[] [] [ ][ ]
9
8
7
6
5
4
10 66 . 26 10 10 0 10 88 . 8 10 10 0
10 10 10 5 0 10 10 10 5 0
0 0 10 0 . 2 0 0 10 0 . 2
10 88 . 8 10 10 0 10 66 . 26 10 10 0
10 10 10 5 0 10 10 10 5 0
0 0 10 0 . 2 0 0 10 0 . 2
9 8 7 6 5 4
'
3 3 3 3
3 3 3 3
6 6
3 3 3 3
3 3 3 3
6 6
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
× × - × ×
× - × × - × -
× × -
× × - × ×
× × - × ×
× - ×
=
= T k T k
T
(2)
The assembled global stiffness matrix [ ] K is of the order 9 9 × . Carrying out
assembly in the usual manner, we get,
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
- - -
-
- -
- -
- -
-
-
- - -
=
66 . 26 10 0 33 . 13 10 0 0 0 0
10 5 0 10 5 0 0 0 0
0 0 2000 0 0 2000 0 0 0
33 . 13 10 0 44 . 44 10 44 . 4 88 . 8 0 44 . 4
10 5 0 10 3 . 1338 0 0 3 . 1333 0
0 0 2000 44 . 4 0 5 . 2001 44 . 4 0 48 . 1
0 0 0 88 . 8 0 44 . 4 77 . 17 0 44 . 4
0 0 0 0 3 . 1333 0 0 3 . 1333 0
0 0 0 44 . 4 0 48 . 1 44 . 4 0 48 . 1
K (3)
Read More
|
34 videos|140 docs|31 tests
|
FAQs on The Direct Stiffness Method: Plane Frames - 2 - Structural Analysis - Civil Engineering (CE)
| 1. What is the direct stiffness method in structural analysis? |  |
Ans. The direct stiffness method is a numerical technique used in structural analysis to determine the displacements, forces, and moments in a structure. It involves representing the structure as an interconnected system of elements, such as beams or trusses, and using the stiffness properties of these elements to calculate the response of the entire structure.
| 2. How is the direct stiffness method applied to plane frames? | |
Ans. In the direct stiffness method for plane frames, the structure is divided into individual members (beams or columns) and nodes (points where members are connected). The stiffness matrix for each member is determined based on its geometry and material properties. Then, the global stiffness matrix for the entire frame is assembled by combining the individual member stiffness matrices. By applying appropriate boundary conditions and solving the resulting system of equations, the displacements and forces in the frame can be determined.
| 3. What are the advantages of using the direct stiffness method in structural analysis? | |
Ans. The direct stiffness method offers several advantages in structural analysis. Firstly, it provides an accurate and efficient way to analyze complex structures, including those with irregular geometries or loadings. Secondly, it allows for easy incorporation of different types of boundary conditions, such as fixed supports or pinned connections. Additionally, it can handle both linear and nonlinear behavior of materials and supports, making it versatile for a wide range of structural problems.
| 4. Are there any limitations or assumptions associated with the direct stiffness method? | |
Ans. Yes, there are certain limitations and assumptions associated with the direct stiffness method. One major assumption is that the structure being analyzed is linearly elastic, meaning it obeys Hooke's law. This assumption may not hold for materials that exhibit significant nonlinear behavior, such as plastic deformation. Additionally, the method assumes that the structure is statically determinate, meaning the number of unknown displacements is equal to the number of equations. If the structure is statically indeterminate, additional techniques, such as the flexibility or stiffness method, need to be used.
| 5. How does the direct stiffness method contribute to the analysis of plane frames in civil engineering? | |
Ans. The direct stiffness method plays a crucial role in the analysis of plane frames in civil engineering. It allows engineers to determine the internal forces and deformations in frame structures, which are essential for assessing their stability, strength, and overall performance. By using this method, engineers can optimize the design of plane frames, ensuring they can withstand the applied loads and meet safety requirements. Additionally, the direct stiffness method facilitates the evaluation of structural modifications or repairs, helping engineers make informed decisions for retrofitting existing frames.
Related Exams
About this Document
|
4.97/5 Rating |
|
Nov 15, 2024 Last updated |
Document Description: The Direct Stiffness Method: Plane Frames - 2 for Civil Engineering (CE) 2024 is part of Structural Analysis preparation.
The notes and questions for The Direct Stiffness Method: Plane Frames - 2 have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The Direct Stiffness Method: Plane Frames - 2 covers topics
like and The Direct Stiffness Method: Plane Frames - 2 Example, for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for The Direct Stiffness Method: Plane Frames - 2.
Introduction of The Direct Stiffness Method: Plane Frames - 2 in English is available as part of our Structural Analysis
for Civil Engineering (CE) & The Direct Stiffness Method: Plane Frames - 2 in Hindi for Structural Analysis course.
Download more important topics related with notes, lectures and mock test series for Civil Engineering (CE)
Exam by signing up for free. Civil Engineering (CE): The Direct Stiffness Method: Plane Frames - 2 | Structural Analysis - Civil Engineering (CE)
Description
Full syllabus notes, lecture & questions for The Direct Stiffness Method: Plane Frames - 2 | Structural Analysis - Civil Engineering (CE) - Civil Engineering (CE) | Plus excerises question with solution to help you revise complete syllabus for Structural Analysis | Best notes, free PDF download
Information about The Direct Stiffness Method: Plane Frames - 2
In this doc you can find the meaning of The Direct Stiffness Method: Plane Frames - 2 defined & explained in the simplest way possible. Besides explaining types of
The Direct Stiffness Method: Plane Frames - 2 theory, EduRev gives you an ample number of questions to practice The Direct Stiffness Method: Plane Frames - 2 tests, examples and also practice Civil Engineering (CE)
tests
|
Explore Courses for Civil Engineering (CE) exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Viva Questions
,ppt
,The Direct Stiffness Method: Plane Frames - 2 | Structural Analysis - Civil Engineering (CE)
,Objective type Questions
,Summary
,Exam
,Previous Year Questions with Solutions
,MCQs
,The Direct Stiffness Method: Plane Frames - 2 | Structural Analysis - Civil Engineering (CE)
,past year papers
,Extra Questions
,Semester Notes
,shortcuts and tricks
,Free
,mock tests for examination
,Important questions
,study material
,The Direct Stiffness Method: Plane Frames - 2 | Structural Analysis - Civil Engineering (CE)
,video lectures
,Sample Paper
,practice quizzes
;
Additional Information about The Direct Stiffness Method: Plane Frames - 2 for Civil Engineering (CE) Preparation
The Direct Stiffness Method: Plane Frames - 2 Free PDF Download
The The Direct Stiffness Method: Plane Frames - 2 is an invaluable resource that delves deep into the core of the Civil Engineering (CE) exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the The Direct Stiffness Method: Plane Frames - 2 now and kickstart your journey towards success in the Civil Engineering (CE) exam.
Importance of The Direct Stiffness Method: Plane Frames - 2
The importance of The Direct Stiffness Method: Plane Frames - 2 cannot be overstated, especially for Civil Engineering (CE) aspirants.
This document holds the key to success in the Civil Engineering (CE) exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
The Direct Stiffness Method: Plane Frames - 2 Notes
The Direct Stiffness Method: Plane Frames - 2 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to The Direct Stiffness Method: Plane Frames - 2.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, The Direct Stiffness Method: Plane Frames - 2 Notes on EduRev are your ultimate resource for success.
The Direct Stiffness Method: Plane Frames - 2 Civil Engineering (CE) Questions
The "The Direct Stiffness Method: Plane Frames - 2 Civil Engineering (CE) Questions" guide is a valuable resource for all aspiring students preparing for the
Civil Engineering (CE) exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study The Direct Stiffness Method: Plane Frames - 2 on the App
Students of Civil Engineering (CE) can study The Direct Stiffness Method: Plane Frames - 2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the The Direct Stiffness Method: Plane Frames - 2,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of The Direct Stiffness Method: Plane Frames - 2 is prepared as per the latest Civil Engineering (CE) syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup