The document discusses analysis of doubly reinforced concrete beams. It begins by explaining how compression reinforcement allows less concrete to resist tension, moving the neutral axis up. It then provides the equations for analyzing strain compatibility and equilibrium in doubly reinforced sections. The document discusses finding the compression reinforcement strain and stress through iteration. It provides reasons for using compression reinforcement, including reducing deflection and increasing ductility. Finally, it includes an example problem demonstrating the full analysis process.
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.
Geotechnical Engineering-II [Lec #19: General Bearing Capacity Equation]Muhammad Irfan
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.
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.
Geotechnical Engineering-II [Lec #19: General Bearing Capacity Equation]Muhammad Irfan
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.
Earthquake Load Calculation (base shear method)
The 3-story standard office building is located in Los Angeles situated on stiff soil. The
structure of the building is steel special moment frame. All moment-resisting frames are
located at the perimeter of the building. Determine the earthquake force on each story in
North-South direction.
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.
Prepared by madam rafia firdous. She is a lecturer and instructor in subject of Plain and Reinforcement concrete at University of South Asia LAHORE,PAKISTAN.
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.
Lec.2 statically determinate structures & statically indeterminate struct...Muthanna Abbu
The student will learn the determination of internal forces in different structures and the
kind of forces distribution due to external & internal effects .He will also learn about the
structures deformation due to these effects .
Earthquake Load Calculation (base shear method)
The 3-story standard office building is located in Los Angeles situated on stiff soil. The
structure of the building is steel special moment frame. All moment-resisting frames are
located at the perimeter of the building. Determine the earthquake force on each story in
North-South direction.
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.
Prepared by madam rafia firdous. She is a lecturer and instructor in subject of Plain and Reinforcement concrete at University of South Asia LAHORE,PAKISTAN.
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.
Lec.2 statically determinate structures & statically indeterminate struct...Muthanna Abbu
The student will learn the determination of internal forces in different structures and the
kind of forces distribution due to external & internal effects .He will also learn about the
structures deformation due to these effects .
check it out: http://goo.gl/vqNk7m
CADmantra Technologies pvt. Ltd. is a CAD Training institute specilized in producing quality and high standard education and training. We are providing a perfact institute for the students intersted in CAD courses CADmantra is established by a group of engineers to devlop good training system in the field of CAD/CAM/CAE, these courses are widely accepted worldwide.
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Aircraft Structures for Engineering Students 5th Edition Megson Solutions ManualRigeler
Full donwload : http://alibabadownload.com/product/aircraft-structures-for-engineering-students-5th-edition-megson-solutions-manual/ Aircraft Structures for Engineering Students 5th Edition Megson Solutions Manual
Sheryar Bismil
Student of Mirpur University of Science & Technology(MUST).
Student of Final Year Civil Engineering Department Main campus Mirpur.
Here we Gonna to learn about the basic to depth wise study of Plan Reinforced Concrete-i.
From basis terminology to wide information about the analysis and design of Concrete member like column,Beam,Slab,etc.
6th International Conference on Machine Learning & Applications (CMLA 2024)ClaraZara1
6th International Conference on Machine Learning & Applications (CMLA 2024) will provide an excellent international forum for sharing knowledge and results in theory, methodology and applications of on Machine Learning & Applications.
NUMERICAL SIMULATIONS OF HEAT AND MASS TRANSFER IN CONDENSING HEAT EXCHANGERS...ssuser7dcef0
Power plants release a large amount of water vapor into the
atmosphere through the stack. The flue gas can be a potential
source for obtaining much needed cooling water for a power
plant. If a power plant could recover and reuse a portion of this
moisture, it could reduce its total cooling water intake
requirement. One of the most practical way to recover water
from flue gas is to use a condensing heat exchanger. The power
plant could also recover latent heat due to condensation as well
as sensible heat due to lowering the flue gas exit temperature.
Additionally, harmful acids released from the stack can be
reduced in a condensing heat exchanger by acid condensation. reduced in a condensing heat exchanger by acid condensation.
Condensation of vapors in flue gas is a complicated
phenomenon since heat and mass transfer of water vapor and
various acids simultaneously occur in the presence of noncondensable
gases such as nitrogen and oxygen. Design of a
condenser depends on the knowledge and understanding of the
heat and mass transfer processes. A computer program for
numerical simulations of water (H2O) and sulfuric acid (H2SO4)
condensation in a flue gas condensing heat exchanger was
developed using MATLAB. Governing equations based on
mass and energy balances for the system were derived to
predict variables such as flue gas exit temperature, cooling
water outlet temperature, mole fraction and condensation rates
of water and sulfuric acid vapors. The equations were solved
using an iterative solution technique with calculations of heat
and mass transfer coefficients and physical properties.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
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.
2. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
SectionsSections
Effect of Compression Reinforcement on the Strength
and Behavior
Less concrete is needed to
resist the T and thereby
moving the neutral axis
(NA) up.
TC
fAT
=
= ys
3. Analysis of DoublyAnalysis of Doubly
Reinforced SectionsReinforced Sections
Effect of Compression Reinforcement on the Strength
and Behavior
( )12
2
sc
1
c
and
2
;C
ReinforcedDoubly
2
;
ReinforcedSingly
aa
a
dfAMCC
a
dfAMCC
ysn
ysn
<
−=′+=
⇒
−==
⇒
5. Reasons for ProvidingReasons for Providing
Compression ReinforcementCompression Reinforcement
Effective of compression reinforcement on sustained
load deflections.
6. Reasons for ProvidingReasons for Providing
Compression ReinforcementCompression Reinforcement
Increased Ductility
reduced
stress block
depth
increase in steel strain
larger curvature are
obtained.
7. Reasons for ProvidingReasons for Providing
Compression ReinforcementCompression Reinforcement
Effect of compression reinforcement on strength and
ductility of under reinforced beams.
ρ < ρb
8. Reasons for ProvidingReasons for Providing
Compression ReinforcementCompression Reinforcement
Change failure mode from compression
to tension. When ρ > ρbal addition of As
strengthens.
Effective reinforcement ratio = (ρ − ρ’)
Compression
zone
allows tension steel to
yield before crushing of
concrete.
9. Reasons for ProvidingReasons for Providing
Compression ReinforcementCompression Reinforcement
Eases in Fabrication -
Use corner bars to hold & anchor stirrups.
10. Effect of CompressionEffect of Compression
ReinforcementReinforcement
Compare the strain distribution in two beams
with the same As
11. Effect of CompressionEffect of Compression
ReinforcementReinforcement
Section 1: Section 2:
Addition of A’s strengthens compression zone so that less
concrete is needed to resist a given value of T. NA
goes up (c2 <c1) and εs increases (εs2 >εs1).
1c
ss
1
11cc1c
ss
85.0
85.085.0
β
β
bf
fA
c
cbfbafCT
fAT
′
=
′=′==
=
1c
ssss
2
21css
2css
1cs
ss
85.0
85.0
85.0
β
β
bf
fAfA
c
cbffA
baffA
CCT
fAT
′
′′−
=
′+′′=
′+′′=
+′=
=
12. Doubly Reinforced BeamsDoubly Reinforced Beams
Under reinforced Failure
( Case 1 ) Compression and tension steel yields
( Case 2 ) Only tension steel yields
Over reinforced Failure
( Case 3 ) Only compression steel yields
( Case 4 ) No yielding Concrete crushes
Four Possible Modes of Failure
13. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Strain Compatibility Check
Assume εs’ using similar
triangles
( )
( )
s
s
0.003
'
'
*0.003
c d c
c d
c
ε
ε
′
= ⇒
−
−
′ =
14. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Strain Compatibility
Using equilibrium and find a
( )
( )
( )
( )
( )
s s y
c s
c
s s y y
1 1 c 1 c
0.85
'
0.85 0.85
A A f
T C C a
f b
A A f d fa
c
f b f
ρ ρ
β β β
′−
′ ′= + ⇒ =
′
′− −
= = =
′ ′
15. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Strain Compatibility
The strain in the compression
steel is
( )
( )
s cu
1 c
y
1
0.85
1 0.003
'
d
c
f d
d f
ε ε
β
ρ ρ
′ ′ = − ÷
′ ′
= − ÷ ÷−
16. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Strain Compatibility
Confirm
;
E
ys
s
y
ys εεεε ≥=≥′
f
( )
( )
y y1 c
s 3
y s
0.85
1 0.003
' E 29 x 10 ksi
f ff d
d f
β
ε
ρ ρ
′ ′
′ = − ≥ = ÷ ÷−
17. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Strain Compatibility
Confirm
( )
( )
( )
( )
y1 c
y
1 c
y y
870.85
' 87
0.85 87
'
87
ff d
d f
f d
d f f
β
ρ ρ
β
ρ ρ
′ ′ −
− ≥
−
′ ′
− ≥ ÷ ÷ ÷ ÷−
18. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Find c
confirm that the tension steel has yielded
( )
s y c s y
ss s y
1
c 1
0.85
0.85
A f f ba A f
A A f
c a c
f b
β
β
′ ′+ =
′−
= ⇒ =
′
y
s cu y
sE
fd c
c
ε ε ε
−
= ≥ = ÷
19. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
If the statement is true than
else the strain in the compression steel
( ) ( )n s s y s y
2
a
M A A f d A f d d
′ ′ ′= − − + − ÷
s sf Eε ′=
20. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Return to the original equilibrium equation
s y s s c
s s s c 1
s s cu c 1
0.85
0.85
1 0.85
A f A f f ba
A E f b c
d
A E f b c
c
ε β
ε β
′= +
′′= +
′ ′= − + ÷
21. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Rearrange the equation and find a quadratic equation
Solve the quadratic and find c.
( )
s y s s cu c 1
2
c 1 s s cu s y s s cu
1 0.85
0.85 0
d
A f A E f b c
c
f b c A E A f c A E d
ε β
β ε ε
′ ′= − + ÷
′ ′ ′⇒ + − − =
22. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Find the fs’
Check the tension steel.
s s cu1 1 87 ksi
d d
f E
c c
ε
′ ′ ′= − = − ÷ ÷
y
s cu y
sE
fd c
c
ε ε ε
−
= ≥ = ÷
23. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Another option is to compute the stress in the
compression steel using an iterative method.
( )
( )
1 c3
s
y
0.85
29 x 10 1 0.003
'
f d
f
d f
β
ρ ρ
′ ′
′= − ÷ ÷−
24. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Go back and calculate the equilibrium with fs’
( )s y s s
c s
c
1
s
0.85
1 87 ksi
A f A f
T C C a
f b
a
c
d
f
c
β
′−
′ ′= + ⇒ =
′
=
′
′= − ÷
Iterate until the c value is
adjusted for the fs’ until the
stress converges.
25. Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
Rectangular SectionsRectangular Sections
Compute the moment capacity of the beam
( ) ( )n s y s s s s
2
a
M A f A f d A f d d
′ ′ ′ ′ ′= − − + − ÷
26. Limitations on ReinforcementLimitations on Reinforcement
Ratio for Doubly ReinforcedRatio for Doubly Reinforced
beamsbeams
Lower limit on ρ
same as for single reinforce beams.
yy
c
min
2003
ff
f
≥
′
=ρ (ACI 10.5)
27. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Given:
f’c= 4000 psi fy = 60 ksi
A’s = 2 #5 As = 4 #7
d’= 2.5 in. d = 15.5 in
h=18 in. b =12 in.
Calculate Mn for the section for the given
compression steel.
28. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Compute the reinforcement coefficients, the
area of the bars #7 (0.6 in2
) and #5 (0.31 in2
)
( )
( )
( ) ( )
( ) ( )
2 2
s
2 2
s
2
s
2
s
4 0.6 in 2.4 in
2 0.31 in 0.62 in
2.4 in
0.0129
12 in. 15.5 in.
0.62 in
0.0033
12 in. 15.5 in.
A
A
A
bd
A
bd
ρ
ρ
= =
′ = =
= = =
′
′ = = =
29. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Compute the effective reinforcement ratio and
minimum ρ
y
c
y
min
0.0129 0.0033 0.00957
200 200
0.00333
60000
3 3 4000
or 0.00316
60000
0.0129 0.00333 OK!
eff
f
f
f
ρ ρ ρ
ρ
ρ ρ
′= − = − =
= = =
= =
≥ ⇒ ≥
30. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Compute the effective reinforcement ratio and
minimum ρ
( )
( )
( )( )( )
( )
1 c
y y
0.85 87
'
87
0.85 0.85 4 ksi 2.5 in. 87
0.0398
60 ksi 15.5 in. 87 60
f d
d f f
β
ρ ρ
′ ′
− ≥ ÷ ÷ ÷ ÷−
≥ = ÷ ÷ ÷ −
0.00957 0.0398≥ Compression steel has not
yielded.
31. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Instead of iterating the equation use the quadratic
method
( )
( ) ( ) ( )
( )( ) ( ) ( )( )( )
( )( ) ( ) ( )
2
c 1 s s cu s y s s cu
2
2 2
2
2
2
0.85 0
0.85 4 ksi 12 in. 0.85
0.62 in 29000 ksi 0.003 2.4 in 60 ksi
0.62 in 29000 ksi 0.003 2.5 in. 0
34.68 90.06 134.85 0
2.5969 3.8884 0
f b c A E A f c A E d
c
c
c c
c c
β ε ε′ ′ ′+ − − =
+
+ −
− =
− − =
− − =
32. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Solve using the quadratic formula
( ) ( )
2
2
2.5969 3.8884 0
2.5969 2.5969 4 3.8884
2
3.6595 in.
c c
c
c
− − =
± − − −
=
=
33. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Find the fs’
Check the tension steel.
s s cu
2.5 in.
1 1 87 ksi
3.659 in.
27.565 ksi
d
f E
c
ε
′ ′= − = − ÷ ÷
=
s
15.5 in. 3.659 in.
0.003 0.00971 0.00207
3.659 in.
ε
−
= = ≥ ÷
34. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Check to see if c works
( )( ) ( )( )
( ) ( ) ( )
2 2
s y s s
c 1
2.4 in 60 ksi 0.62 in 27.565 ksi
0.85 0.85 4 ksi 0.85 12 in.
3.659 in.
A f A f
c
f b
c
β
′ ′ −−
= =
=
The problem worked
35. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
Compute the moment capacity of the beam
( ) ( )
( )( )
( )( )
( )
( )( ) ( )
s y s s s s
2
2
2
2
2.4 in 60 ksi 0.85 3.659 in.
15.5 in.
20.62 in 27.565 ksi
0.62 in 27.565 ksi 15.5 in. 2.5 in.
1991.9 k - in. 166 k - ft
n
a
M A f A f d A f d d
′ ′ ′ ′ ′= − − + − ÷
÷= − ÷
÷−
+ −
= ⇒
36. Example: Doubly ReinforcedExample: Doubly Reinforced
SectionSection
If you want to find the Mu for the problem
( )u u
3.66 in.
0.236
15.5 in.
0.375 0.9
0.9 166 k -ft
149.4 k -ft
c
d
c
d
M M
φ
φ
= =
> ⇒ =
= =
=
From ACI (figure R9.3.2)or figure (pg 100 in your
text)
The resulting ultimate moment is
37. Analysis of FlangedAnalysis of Flanged
SectionSection
Floor systems with slabs and beams are placed
in monolithic pour.
Slab acts as a top flange to the beam; T-
beams, and Inverted L(Spandrel) Beams.
39. Analysis of FlangedAnalysis of Flanged
SectionsSections
If the neutral axis falls
within the slab depth
analyze the beam as a
rectangular beam,
otherwise as a T-beam.
40. Analysis of FlangedAnalysis of Flanged
SectionsSections
Effective Flange Width
Portions near the webs are more highly stressed than
areas away from the web.
41. Analysis of FlangedAnalysis of Flanged
SectionsSections
Effective width (beff)
beff is width that is stressed uniformly to give the same
compression force actually developed in compression
zone of width b(actual)
42. ACI Code Provisions forACI Code Provisions for
Estimating bEstimating beffeff
From ACI 318, Section 8.10.2
T Beam Flange:
eff
f w
actual
4
16
L
b
h b
b
≤
≤ +
≤
43. ACI Code Provisions forACI Code Provisions for
Estimating bEstimating beffeff
From ACI 318, Section 8.10.3
Inverted L Shape Flange
( )
eff w
f w
actual w
12
6
0.5* clear distance to next web
L
b b
h b
b b
≤ +
≤ +
≤ = +
44. ACI Code Provisions forACI Code Provisions for
Estimating bEstimating beffeff
From ACI 318, Section 8.10
Isolated T-Beams
weff
w
f
4
2
bb
b
h
≤
≥