This document presents the slope deflection method for analyzing statically indeterminate beams and frames. It discusses the assumptions of the method, which include rigid joints and neglecting distortions from axial and shear stresses. Examples are provided to demonstrate solving for unknown joint rotations and displacements using equilibrium equations, then determining final end moments and drawing shear and bending moment diagrams. Portal frames are discussed, including how side sway can be accounted for in the analysis.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Slope and Deflection Method ,The Moment Distribution Method ,Strain Energy Me...Aayushi5
An introduction to Different method is given here. The strain energy method actually solves these equations by the method of successive approximations.
Problems are solved to illustrate the Strain energy method as applied element.
Force Force and Displacement Matrix MethodAayushi5
The analysis of a structure by the matrix method may be described by the following steps:
1. Problem statement
2. Selection of released structure
3. Analysis of released structure under loads
4. Analysis of released structure for other causes
5. Analysis of released structure for unit values of redundant
6. Determination of redundants through the superposition equations.
7. Determine the other displacements and actions. The following are the four flexibility matrix equations for calculating redundants member end actions, reactions and joint displacements
where for the released structure
8.All matrices used in the matrix method are summarized in the following tables
Correlations between Undrained Shear Strength and Standard Penetration Test SPT N Values.
(After Terzaghi and Peck, 1967 and Shower, 1979)
Correlations beetwen SPT N and Friction Angle to Estimate Overburden Pressure.
(De Mello's 1971)
Analysis of indeterminate beam by slopeand deflection methodnawalesantosh35
Slope-deflection method ,Slope-deflection equations, equilibrium equation of
method, application to beams with and without joint translation and rotation, Sinking or yielding of support,
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
Slope and Deflection Method ,The Moment Distribution Method ,Strain Energy Me...Aayushi5
An introduction to Different method is given here. The strain energy method actually solves these equations by the method of successive approximations.
Problems are solved to illustrate the Strain energy method as applied element.
Force Force and Displacement Matrix MethodAayushi5
The analysis of a structure by the matrix method may be described by the following steps:
1. Problem statement
2. Selection of released structure
3. Analysis of released structure under loads
4. Analysis of released structure for other causes
5. Analysis of released structure for unit values of redundant
6. Determination of redundants through the superposition equations.
7. Determine the other displacements and actions. The following are the four flexibility matrix equations for calculating redundants member end actions, reactions and joint displacements
where for the released structure
8.All matrices used in the matrix method are summarized in the following tables
Correlations between Undrained Shear Strength and Standard Penetration Test SPT N Values.
(After Terzaghi and Peck, 1967 and Shower, 1979)
Correlations beetwen SPT N and Friction Angle to Estimate Overburden Pressure.
(De Mello's 1971)
Analysis of indeterminate beam by slopeand deflection methodnawalesantosh35
Slope-deflection method ,Slope-deflection equations, equilibrium equation of
method, application to beams with and without joint translation and rotation, Sinking or yielding of support,
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
About
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
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Hierarchical Digital Twin of a Naval Power SystemKerry Sado
A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
1. DEPARTMENT OF CIVIL ENGINEERING
SHIVALIK COLLEGE OF ENGINEERING
Dehradun, Uttrakhand, India, Email: info@sce.org.in, Website: www.sce.org.in
SLOPE DEFLECTION METHOD
ON PORTAL FRAMES
NAME – ASHU KUMAR KUSHWAHA
ROLL NO – 180410107004
BRANCH – B.TECH 3RD YEAR (C.E)
FACULTY NAME – DR.SUJEET KUMAR
DESIGNATION - RISP
2. INTRODUCTION
This method was developed by G.A Maney in
Germany in 1914. This method is applicable to
all types of statically indeterminate beams &
frames and in this method, we solve for
unknown joint rotations, which are expressed in
terms of the applied loads and the bending
moments. Deflections due to shear and axial
stresses are not considered as the effect are
small.
3. In slope deflection method, we shall find out the relation
between moments and slope at the joints. Then total
rotation due to applied loads and settlement of supports
may be found out and then by using the relations between
moments and slopes the net moments induced at the joint
may be found out.
a) The beam is having a constant section between each
pair of supports.
b) The joint in a structure as whole, may rotate or deflect,
but the angles between members meeting at the joint
remain the same.
4. ASSUMPTIONS
This method is based on the following simplified
assumptions:
All the joints of the frame are rigid.
Distortion, due to axial and shear stresses, being
very small, are neglected.
6. SIGN CONVENTION:–
(1) ROTATIONS:– Clockwise joint rotations are
considered as (+ve).
(2) END MOMENTS:– Clockwise end moments
are considered as (+ve).
(3) SETTLEMENT:- When leads to clockwise
rotation considered as (+ve).
7. DERIVATION
Considering a continuous beam where A,B are the intermediate supports,
the final end moments developed at support A,B (MAB,MBA) will be due to
effect of
i) Effect of loading
ii) Rotation of joint A
iii) Rotation of joint B
iv) Effect of settlement of support
A B
8. By using superposition theorem the final end moment will be the summation of all the effects.
i) Effect of loading
A B
ii) Rotation at A
A B
ӨA
MFAB
= - PL/8
MFBA
= PL/8
MFAB1
= 4EIӨA/L
MFBA1
= 2EI ӨA/L
P
L/2 L/2
9. iii) Rotation at B
A B
ӨB
iv) Support B sinks by ∆
A B
ӨA
∆
MFAB2
= 2EIӨB/L
MFBA2
= 4EIӨB/L
MFAB3
= -6EI∆/L2
MFBA3
= -6EI∆/L2
10. Final end moments
MAB
= MFAB
+ MFAB1
+ MFAB2
+ MFAB3
MAB
= MFAB
+ 4EIӨA + 2EIӨB - 6EI∆
L L L2
MAB
= MFAB
+ 2EI (2ӨA +ӨB - 3∆ )
L L
Similarly,
MBA
= MFBA
+ MFBA1
+ MFBA2
+ MFBA3
MBA
= MFBA
+ 2EI (ӨA +2ӨB - 3∆ )
L L
11. PROCEDURE
i) Find fixed end moments, considering all span as a fixed beam.
* Do not consider effect of sinking in this step.
ii) Write slope deflection equation, considering the displacement as unknowns (∆,Ө).
iii) Using equilibrium conditions find the unknown displacement
iv) Substitute the values of (Ө,∆) in equation and determine the final end moments.
v) Draw the final shear force diagram, bending moment diagram, elastic curve.
14. EXAMPLE 1 :–
Analyse the structure by slope deflection method. Sketch the bending moment and shear force
diagram.
SOLUTION:-
a) Fixed end moments:
Span AB:
MFAB = MFBA = 0
Span BC:
MFBC = - WL2 = -10x36 = -30 KNm
12 12
MFCB = WL2 = 10x36 = 30 KNm
12 12
Span CD:
MFCD = MFDC = 0
FRAMES WITHOUT SIDE SWAY
15. b) Slope deflection equation:
MAB
= MFAB
+ 2EI (2ӨA +ӨB - 3∆ ) = 0 + 2EI (0 +ӨB + 0) (ӨA = 0)
L L 3
M
AB = 2EI ӨB / 3
MBA
= MFBA
+ 2EI (ӨA +2ӨB - 3∆ ) = 0 + 2EI (0 +2ӨB +0)
L L 3
MBA
= 4EI ӨB / 3
MBC
= MFBC
+ 2EI (2ӨB +ӨC - 3∆ ) = -30 + 2EI (2ӨB - ӨB + 0) (ӨC = -ӨB )
L L 6
M
BC
= = -30 + EI ӨB / 3
16. c) Joint equilibrium equations:
At joint B,
∑M=0 MBA + MBC = 0
4EI ӨB – 30 + EI ӨB = 0
3 3
ӨB = -30 x 3 = 18 ӨC = - ӨB = -18
5EI EI EI
d) Final moments:
M
AB = 2EI (18) = 12 KNm
3 EI
M
BA 4EI (18) = 24 KNm
3 EI
M
BC = -30 + EI (18) = -24 KNm
3 EI
M
CB = 24 KNm
M
CD = -24 KNm
M
DC = -12 KNm
17. e) To draw shear force diagram:
Span AB:
Taking moment about A,
-FBA (3) + MAB + MBA = 0
24 + 12 = FBA (3)
FBA = 12 KN
FAB = 12 KN
Span BC:
RC = 10 x 6 = 30 KN
2
RB = 30 KN
Span CD:
FCD = 12 KN
FDC = 12 KN
18. f) Draw bending moment diagram:
Simply supported bending moment in BC = WL2 = 10 x 62 = 45 KNm
8 8
19. EXAMPLE 2:–
Analyse the structure by slope deflection method. Sketch the bending
moment and shear force diagram.
SOLUTION:-
a) Fixed end moments:
MFAB = - WL2 = -10x16 = -13.33 KNm
12 12
MFBA = WL2 = 10x16 = 13.33 KNm
12 12
MFBC = -10x2 = -20 KNm
MFBE = -WL = 20x4 = -10 KNm
8 8
MFEB = WL = 20x4 = 10 KNm
8 8
20. b) Slope deflection equation:
MAB
= MFAB
+ 2EI (2ӨA +ӨB - 3∆ ) = -13.33 + 2Ex2I (0 +ӨB + 0)
L L 4
M
AB = -13.33 + EI ӨB
MBA
= MFBA
+ 2EI (ӨA +2ӨB - 3∆ ) = 13.33 + 2Ex2I (0 +2ӨB +0)
L L 4
MBA
= 13.33 + 2EI ӨB
MBE
= MFBE
+ 2EI (2ӨB +ӨE - 3∆ ) = -10 + 2EI (2ӨB +0 + 0)
L L 4
M
BE = -10 + EI ӨB
MEB
= MFEB
+ 2EI (2ӨE +ӨB - 3∆ ) = 10 + 2EI (0 + ӨB + 0)
L L 4
M
BE
= = 10 + 0.5EI ӨB
21. c) Joint equilibrium equations:
At joint B,
∑M=0 MBA + MBC + MBE = 0
13.33 + 2EI ӨB – 10 + EIӨB -20 = 0
3EI ӨB – 16.67 = 0
ӨB = 16.67 ӨC = 5.557
3EIӨC EI
d) Final moments:
M
AB = -13.33 + EI (5.557) = -7.773 KNm
EI
M
BA = 13.33 + 2EI (5.557) = 24.447 KNm
EI
M
BC = -20 KNm
M
BE = -10 + EI (5.557) = -4.443 KNm
EI
M
EB = 10 + EI (5.557) = 12.778 KNm
2 EI
22. e) To draw shear force diagram:
Span AB:
taking moment about B,
RA x 4 – 10 x 4(4/2) - 7.773 + 24.447 = 0
RA = 15.832 KN
RB1 = 10 x 4 – RA = 24.168 KN
Span BE:
Taking moment about B,
-RE (4) – 4.443 + 12.778 + 20(2) = 0
RE = 12.084 KN
RB2 = total load – RE = 7.916 KN
23. f) Draw bending moment diagram:
Simply supported bending moment in AB = WL2 = 10 x 16 = 20 KNm
8 8
Simply supported bending moment in BE = WL = 20 x 4 = 20 KNm
4 4
24. PORTAL FRAMES WITH SIDE SWAY
Portal frames may sway due to following reasons:
1. Eccentric or unsymmetrical loading on the portal frames.
2. Unsymmetrical shape of the frames.
3. Different end conditions of the column of the portal
frame.
4. Non uniform section of the members of the frame.
5. Horizontal loading on the column of the frame.
6. Settlement of the supports of the frame.
25. EXAMPLE 3 :–
Analyse the frame by slope deflection method and draw the bending moment diagram.
SOLUTION:-
a) Fixed end moments:
MFAB = -WL = -8 x 3 = -3 KNm
8 8
MFBA = WL = 8 x 3 = 3 KNm
8 8
MFBC = - WL2 = -9x22 = -3 KNm
12 12
MFCB = WL2 = 9x22 = 3 KNm
12 12
MFCD = MFDC = 0
26. b) Slope deflection equation:
MAB
= MFAB
+ 2EI (2ӨA +ӨB - 3∆ ) = -3 + 2Ex2I (ӨB - 3∆) = -3 + 4EI (ӨB - ∆)
L L 3 3 3
MBA
= MFBA
+ 2EI (ӨA +2ӨB - 3∆ ) = 3 + 2Ex2I (0 +2ӨB - 3∆) = 3 + 4EI (2ӨB - ∆)
L L 3 3 3
MBC
= MFBC
+ 2EI (2ӨB +ӨC - 3∆ ) = -3 + 2EI (2ӨB + ӨC) = -3 + EI (2ӨB + ӨC)
L L 2
MCB
= MFCB
+ 2EI (2ӨC +ӨB - 3∆ ) = 3 + 2EI (2ӨC + ӨB) = 3 + EI (2ӨC + ӨB)
L L 2
MCD
= MFCD
+ 2EI (2ӨC +ӨD - 3∆ ) = 0 + 2E x 1.5I (2ӨC - 3∆) = 1.5EI (2ӨC - 3∆)
L L 2 2 2
MDC
= MFDC
+ 2EI (2ӨD +ӨC - 3∆ ) = 0 + 2E x 1.5I (ӨC -3∆) = 1.5EI (ӨC -3∆)
L L 2 2 2
27. c) Equilibrium and shear equations:
At joint B,
∑M=0 MBA + MBC = 0
3 + 4EI (2ӨB - ∆) – 3 + EI(2ӨB + ӨC ) = 0
3
14 ӨB - 4 ∆ + 3 ӨC = 0
MCB + MCD = 0
3 + EI (2 ӨC + ӨB ) + 1.5EI(2 ӨC - 3∆) = 0
2
ӨB + 5 ӨC - 2.25EI∆ = - 3
EI
MAB + MBA – Ph + MCD + MDC + P = 0
L L
EI ( 8 ӨB + 13.5 ӨC – 18.82∆ ) + 24 = 0
On solving,
ӨB = 0.4148
EI
ӨC = - 0.044
EI
∆ = 1.42
EI
28. d) Final moments:
M
AB = -3 + 4EI (0.4148 – 1.42) = -4.34 KNm
3 EI EI
M
BA = 3 + 4EI (2 x 0.4148 – 1.42) = 2.21 KNm
3 EI EI
M
BC = -3 + EI( 2 x 0.4148 - 0.044) = -2.21 KNm
EI EI
M
CB = 3 + EI (2 x -0.044 – 0.4148) = 3.33 KNm
EI EI
M
CD = 1.5EI (2 x -0.044 – 3 x 1.42) = -3.33 KNm
EI 2 x EI
M
DC = 1.5EI(-0.044 – 3 x 1.42) = -3.26 KNm
EI 2 x EI
29. e) Draw bending moment diagram:
Simply supported bending moment in AB = WL = 8 x 3 = 6 KNm
4 4
Simply supported bending moment in BC = WL2 = 9 x 22 = 4.5 KNm
8 8