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CE 412

Structural Analysis and Design Sessional-II

Department of Civil Engineering Ahsanullah University of Science and Technology

Version 1; November, 2015

AHSANULLAH UNIVERSITY OF SCIENCE &TECHNOLOGY

Page | 2

CE412: Structural Analysis & Design Sessional - II

Preface

This lab handout is intended to give an overview of a Multi storied Building and a

Balanced Cantilever Bridge structural analysis and design. It concentrates on the gravity loading

only. This handout provides a basic guideline for analysis, design and detailing works as well as

reviewing a standard code of practice. To provide the undergraduate students a well-organized,

user-friendly, and easy-to-follow resource, this handout is divided into two major parts. The first

part mainly focuses on the structural analysis and design of Reinforced concrete (RC)

Multistoried Building that includes design of Slab, Beam, Column, Stair, Water reservoir and

Lateral load analysis. The other part deals with the Balanced Cantilever Bridge including an

introduction to Bridge Engineering, details about Balanced Cantilever Bridge, design of Deck

Slab, design of Railing, Post and Curb/Sidewalk, design of Interior Girder considering dead and

live loads only, design of Exterior Girder considering dead and live loads only, design of

Diaphram or Cross Girder and Design of Articulation. Handouts of Dr. Khan Mahmud Amanat,

and Mr. Ruhul Amin, of BUET were helpful as well as suggestions from some faculty members

of the Department of Civil Engineering, AUST.

Zasiah Tafheem

Shafiqul Islam

Department of Civil Engineering

Ahsanullah University of Science and Technology


Page | 3


INDEX

Part I

Structural Analysis and Design of the Multistoried RC building

Sl. No. Name Page no.

1 Introduction 5

2 Design of Stair 10

3 Design of Overhead Water Reservoir (OWR) 14

4 Lateral Load Calculation 23

5 Design of Slab 27

6 Design of Beam only for Gravity Load 31

7 Design of Column 38

8 References 45


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Part II:

Preliminary Design of the Superstructure of a Balanced Cantilever

Bridge for Gravity loading

Sl.

No.

Name Page No.

1 Introduction to Bridge Engineering 49

2 About Balanced Cantilever Bridge 60

3 Design of Deck Slab 69

4 Design of Railing, Post and Sidewalk 71

5 Design of Interior Girder 74

6 Design of Exterior Girder 90

7 Design of Cross Girder/ Diaphram 92

8 Design of Articulation 93

9 References 99


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Part I:Structural Analysis and Design of the Multistoried building

Introduction

Generally the design of any structure (building, bridge etc) can be dividing in two segments,

Foundation design (footing, basement, retaining wall, abutment, underground water

reservoir etc)

Design of superstructure (beam, column, slab, girder, stair etc)

Figure 1: Super structural elements


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Figure 2: Foundation elements


Page | 7


Figure 3: Gravity load distribution


Page | 8


Foundation design: it can also be dividing in two parts,

Determination of soil load bearing capacity, soil stability and the lateral forces introduce

by soil.

The structural design of the elements beneath GL.

# Note: the B.C. of soil and the other parameters like atterberg limit, angle of internal friction,

consolidation can be taken direct from soil report but a general cross check between the relations

among the parameters are required.

Steps of design

Specified the type of structural system like RCC or Steel or Composite, beam supported

or flat plate or braced etc.

Specified the loads considering the type of services like residential or commercial or

institutional etc based on associate codes and judgments.

Prepare a preliminary model of the structure with preliminary sections based on the

judgment.

Analysis the model for desired load combinations according to BNBC in the context of

Bangladesh,

I. DL+LL

II. 1.4DL+1.7LL

III. 0.75[1.4DL+1.7LL±1.7{1.1(EQx or EQy)}] ~ 1.05DL+1.275LL+1.4(EQx or EQy)

IV. 0.75{1.4DL+1.7LL±1.7(Wx or Wy)} ~ 1.05DL+1.275LL+1.275(Wx or Wy)

Design the structural elements separately by considering their integrity and construction

feasibility of that design.


Page | 9


Design Criteria

U.S.D Method

ƒc' = Cylindrical strength of concrete

ƒy = Yield strength of reinforcement

Vc =Allowable shear force without web

reinforcement = 2 Φ b′

c wd

V = Allowable shear force with web

reinforcement = 8 Φ b′

c wd

V =Allowable peripherial shear force in slab

and footing without web reinforcement

=4 Φ b′

c wd

Strength reduction factors :

# Flexure, without axial load = 0.90

# Axial compression and axial compression

with flexure:

Members with spiral Reinforcement = 0.75

Other reinforcement = 0.70

# Shear and torsion = 0.85

# Bearing on concrete = 0.75

W.S.D Method

ƒc' = Cylindrical strength of concrete

ƒc =0.45 ƒc'

ƒy = Yield strength of reinforcement

Ec =w1.5 ×33 c'

n = Es

Er=

29x106

14515 x33 fc ′

k =n

(n+r)

j = 1- k/3

R =1

2fckj

vc =Allowable shear stress without web

reinforcement =1.1 c ′

v= Allowable shear stress without web

reinforcement =5 c ′

Vc =Allowable peripherial shear stress in

slab and footing without web reinforcement

=2 c ′


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Design of Stair

Figure 4: Typical stair

Live Load= 100 psf = 0.1 ksf

Dead load Calculation:

Rises & Steps= 1

2∗

6

12∗

10

12∗3.5∗9∗150

1000 = 0.98 k

Waist= 7.52+4.52∗

6

12∗3.5∗150

1000 =

8.75∗6

12∗3.5∗150

1000 =2.3 k

Total Dead Load=Landing slab + (Rises & Steps+ Waist)

= 6

12∗150

1000 +

0.98+2.3

3.5∗8.75 =0.2 ksf

Floor Finish= 25 psf = 0.025 ksf

Total load, W= (0.1*1.6) + [1.2*(0.2 +0.025)] =0.43ksf


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𝐌+ =𝐖𝐋𝟐

𝟏𝟒=

0.43 ∗ 2∗3.5+8.75 ²

14=7.6 k-ft/ft

𝐌− =𝐖𝐋𝟐

𝟗=

0.43 ∗15.752

9=11.85 k-ft/ft

d= (t-1) = (6-1) = 5"

fy= 60000 psi

f'c= 3000 psi

𝐩𝐦𝐚𝐱 = 𝟎.𝟖𝟓 ∗ 𝛃𝟏 ∗𝐟′𝐜

𝐟𝐲∗

𝟎.𝟎𝟎𝟑

𝟎.𝟎𝟎𝟑+ є𝐭= 0.85 ∗ 0.85 ∗

3000

60000∗

0.003

0.003 +0.004= 0.015

Mu = ф ∗ pmax ∗ fy ∗ b ∗ d2 ∗ 1 − 0.59 ∗pmax ∗ fy

f′c

d2 = 11.85∗12

0.9∗0.015∗60∗12∗ 1−0.59∗0.015∗60

3 =

142 .2

8=17.8

d = 4.21"< provided, 5" (ok)

Table 1: Minimum ratios of temperature and shrinkage reinforcement in slabs based on gross

concrete area. (Ref: ACI Code, Design of Concrete Structure, 13th edition, Chap-13, P-417)

𝐀𝐬𝐦𝐢𝐧 = 𝟎.𝟎𝟎𝟏𝟖 ∗ 𝐛 ∗ 𝐭 = 0.0018 ∗ 12 ∗ 6 = 0.129in.2

+𝐀𝐬 = 𝐌∗𝟏𝟐

ф∗𝐟𝐲∗ 𝐝−𝐚

𝟐

= 7.6∗12

0.9∗60∗ 5−0.7

2 = 0.37in.2(controlled)

𝐚 = 𝐀𝐬∗𝐟𝐲

.𝟖𝟓∗𝐟′𝐜∗𝐛=

0.37∗60

.85∗3∗12= 0.7(ok)

Now,0.16∗12

0.37= 5.2"; use Ø12mm@5" c/c alt ckd

Again,


Page | 12


−𝐀𝐬 = 𝐌∗𝟏𝟐


𝟐

= 11.85∗12

0.9∗60∗ 5− 1

2 =0.58in.2(controlled)


.𝟖𝟓∗𝐟′𝐜∗𝐛=

0.5∗60

.85∗3∗12= 1.1(ok)

The distance between two cranked rod is 10".

So, Required reinforcement = 0.58- 0.16∗12

10=0.39in.2

The extra negative reinforcement required, 0.39

0.16*

10

12= 2. So, use 2-Ø12mmas extra top.

For shrinkage, Asmin = 0.0018 ∗ 12 ∗ 6 = 0.129in.2

Now, 0.121∗12

0.129= 11.23"; useØ10mm@10" c/c

Stair Beam

Assume beam size, 10"x12"

d= (t-1) = (12-2.5) = 9.5"

So, self weight = .83*1*150= 0.12k/ft

Load on Stair beam = 0.5∗ 15.75∗3.5∗0.2 ∗2 +0.5∗ 15.75∗3.5∗0.1 ∗2

8 + 0.42*10*0.12 + 0.12 = 2.7 k/ft

𝐌+ =𝐖𝐋𝟐

𝟏𝟒=

2.7 ∗8²

14=12.3 k-ft


𝟏𝟔=

2.7 ∗82

16=10.8 k-ft

𝐩𝐦𝐚𝐱 = 𝟎.𝟖𝟓 ∗ 𝛃𝟏 ∗𝐟′𝐜

𝐟𝐲∗

𝟎.𝟎𝟎𝟑

𝟎.𝟎𝟎𝟑+ є𝐭= 0.85 ∗ 0.85 ∗

3000

60000∗

0.003

0.003 +0.004= 0.015


f′c

d2 = 12.3∗12

0.9∗0.015∗60∗12∗ 1−0.59∗0.015∗60

3 = 18.45

d = 4.3" < provided, 5" (ok)

Asmin= 𝟐𝟎𝟎

𝐟𝐲bd = 0.32 in

2

+𝐀𝐬 = 𝐌∗𝟏𝟐


𝟐

= 12.3∗12

0.9∗60∗ 9.5−0.7

2 = 0.3 in.2


Page | 13



.𝟖𝟓∗𝐟′𝐜∗𝐛=

0.3∗60

.85∗3∗12= 0.7(ok)

So, use 2-Ø16mm st at top and bottom.

Figure 5: Reinforcement Details of stair.


Page | 14


Design of OWR

Figure 6: Roof top water reservoir

Given Data:

f'c= 3000 psi

fy= 60000 psi

Size of a water Reservoir:

6th

floor building of 2 units& 5 members in each unit.

[Water consuming 210 per capita per day]


Page | 15


Total members= 6*2*5= 60 persons.

Total water consuming= 60*210= 12600 litters for a full day.

= 12600

1000m3= 12.6*3.283 = 445ft3

Inner length & width of Reservoir is,

Length=14.5 ft

width=7.5 ft

so, Height=445

7.5∗14.5=4.09 ft+1 ft=5.09 ft~ 6 ft ; [where, free Board= 1 ft]

Height= 6 ft

Figure 7: Pressure distribution on reservoir wall

For Vertical Reinforcement:

Let wall thickness = 5"

so, Effective depth, d= 5-1 = 4"

𝐩𝐦𝐚𝐱 = 𝟎.𝟖𝟓 ∗ 𝛃𝟏 ∗𝐟′𝐜

𝐟𝐲∗

𝟎.𝟎𝟎𝟑

𝟎.𝟎𝟎𝟑+є𝐭= 0.85 ∗ 0.85 ∗

3000

60000∗

0.003

0.003+0.004= 0.015


f′c

d2 = 2.25∗12

0.9∗0.015∗60∗12∗ 1−0.59∗0.015∗60

3 =

27

8=3.8


Page | 16


d = 1.84"< provided, 4" (ok)

𝐀𝐬𝐦𝐢𝐧 = 𝟎.𝟎𝟎𝟏𝟖 ∗ 𝐛 ∗ 𝐭 = 0.0018 ∗ 12 ∗ 5 = 0.12 in2/ft

[Let, a=0.25]

𝐀𝐬 = 𝐌∗𝟏𝟐


𝟐

=2.25∗12

.90∗50∗ 4−.25

2

= 0.13 in2/ft(controlled)


.𝟖𝟓∗𝐟′ 𝐜∗𝐛=

0.13∗60

.85∗3∗12= 0.26(ok)

use, ф10mm@8in c/c.

For Horizontal Reinforcement:

𝐅𝐨𝐫𝐜𝐞 = 𝛄 ∗ 𝐡 ∗(𝟏𝟒.𝟓

𝟐+

𝟏𝟒.𝟓

𝟐) = 62.5 * 6* (

14.5

2+

14.5

2) = 5438 lb

Again,

force

stress=

5438

fy =

5438

60000 =0.09 in2/ft

Use ф10 @ 8in c/c

Slab Load Distribution:

Bottom slab:

Table 2: Minimum thickness of nonprestressed one-way slabs. (Ref: ACI Code, Design of

Concrete Structure, 13th edition, Chap-13, P-416)

Thickness=7.5

20∗ 12 = 4.5 in


Page | 17


Self weight of slab =𝟏.𝟐 ∗1∗

4.5

12∗150

1000 = 0.067ksf

5wA lA4

384 EI=

5wB lB4

384 EI

wA lA4 = wB lB

4

wA = wB lB

lA

4

Wa =16* Wb

Wa +Wb = 0.067 ksf

Wb = .004k/ft

Wa = .064k/ft

Floor Finish= 25 psf = 0.025 ksf

Total load, W= (0.0625*1.6) + [1.2*(0.064 +0.025)] =0.21ksf

Moment for short Direction

𝐌+ =𝐖𝐋𝟐

𝟏𝟒=

0.21∗7.5²

14= 0.84 k-ft/ft


𝟐𝟒=

0.21∗7.52

9= 0.5 k-ft/ft

𝐩𝐦𝐚𝐱 = 𝟎.𝟖𝟓 ∗ 𝛃𝟏 ∗𝐟′𝐜

𝐟𝐲∗

𝟎.𝟎𝟎𝟑

𝟎.𝟎𝟎𝟑+ є𝐭= 0.85 ∗ 0.85 ∗

3000

60000∗

0.003

0.003 +0.004= 0.015


f′c

d2 = 0.84∗12

0.9∗0.015∗60∗12∗ 1−0.59∗0.015∗60

3 =

10

8=1.25

d = 1.11" < provided, 3.5" (ok)

𝐀𝐬𝐦𝐢𝐧 = 𝟎.𝟎𝟎𝟐 ∗ 𝐛 ∗ 𝐭 = 0.002 ∗ 12 ∗ 4.5 = 0.11in2/ft


Page | 18


+𝐀𝐬 = 𝐌∗𝟏𝟐


𝟐

= 0.84∗12

0.9∗60∗ 3.5−0.1

2 = 0.05in.2


.𝟖𝟓∗𝐟′ 𝐜∗𝐛=

0.05∗60

.85∗3∗12= 0.1(ok)

So, all the reinforcement will be controlled by Asmin.

Use ф10mm @ 8 in c/c ckd and 1-ф10mm as extratop.

Top Slab:

For top slab there is no water load and some life load which is negligible. As the bottom slab is

controlled by 4.5" thickness and Asmin, so will be the top slab.

Figure 8: Reinforcement details of top slab overhead water reservoir


Page | 19


Beam Design:

Here, Load from Bottom Slab= 0.21ksf

Beam Thickness, t=12 in

Effective Depth, d=(12-2.5)=9.5 in

Self weight = .83*1*150= 0.12k/ft

Figure 9: Load distribution of slab

Moment for

Triangular portion=

1

2∗ 7.5∗3.5 ∗0.21

7.5+ 0.12 + .42 ∗ 6 ∗ 0.15 = 0.86k/ft

Trapezoidal portion=

1

2∗ 14.5+7.5 ∗3.75∗0.21

14.5+ 0.12 + .42 ∗ 6 ∗ 0.15 =1.1k/ft


Page | 20


Table 3: Moment and shear values using ACI coefficients. (Ref: ACI Code, Design of Concrete

Structure, 13th edition, Chap-12, P-395)


Page | 21


Figure 10: Moment coefficients for beam.

𝐌+ =𝐖𝐋𝟐

𝟏𝟒=

1.1∗14.5²

14=16.5 k-ft


𝟏𝟔=

1.1∗14.5²

16=14.5 k-ft


f′c

𝐝𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝=1.44in< 9.5 in Provided. (Ok)

+As = M∗12

ф∗fy∗ d−a

2

=16.5∗12

0.9∗60∗ d−1

2

= 0.41in2

Where, a = As∗fy

.85∗f′c ∗b=0.95in (Ok)

Now, Asmin=200

fy∗ bd =

200∗10∗9.5

60000= 0.32 in

2(Ok)

Use 2-ф16mmst at top and bottom.


Page | 22


Figure 11: Reinforcement details of roof top water reservoir beam


Page | 23


Figure 12: Reinforcement details of roof top water reservoir

Lateral Load Calculation

Earthquake Load Calculation:

Here,

Seismic Zone-coefficient, Z=0.15 [Dhaka]

Structural Importance Coefficient, I=1

[Residential]

Response Modification Coefficient, R=9

Now, Numerical Co-efficient, C=1.25∗S

T23

= 2.21

Figure 13: Plan of the building

47.875 ft

40.575 ft


Page | 24


s2=1.2

T=Ct*(hn)3/4

=0.073*(50

3.28)

3/4=0.565˂0.75;

∴ C =1.25 ∗ 12

(0.56)23

= 2.21

W= DL* Area* Storied= 175*47.875*40.575*5*1

1000=1699.7 k~1700 k

V =Z ∗ I ∗ C ∗ W

R=

0.15 ∗ 1 ∗ 2.21 ∗ 1700

9= 62.6 kip

Wi= DL*Area*1

1000= 175*47.875*40.575*

1

1000= = 339.9 k~ 440 k

ΣWi ∗ hi= 440*(10+20+30+40+50) =60000

Here, Ft=0 as, T<0.75

Load on each floor, Fx= V−Ft ∗W i∗hx

ΣW i∗h i

Fx=(62.6−0)∗440∗hx

60000= 0.417 ∗ hx

Table 4: Equivalent earth quake forces at different levels.

Floor hx Force, Fx=0.417 * hx

Ground Floor 10 ft 4.17

1st 20 ft 8.34

2nd

30 ft 12.51

3rd

40 ft 16.68

4th

50 ft 20.85+ Ft = 20.85

Wind Load Calculation:


Page | 25


(a) (b)

Figure 14: a) Plan b) Elevation of the building

Here,

Gust Co-efficient, CG = 1.43

Cc = 47.2 ∗ 10−6

B= 46.43’

L= 39.88’

Heght, h= 50’

Now,

Important Co − efficient, CI = 1.00; for Office Building.

Combined height & 𝑒𝑥𝑝𝑜𝑠𝑢𝑟𝑒 𝐶𝑜 − 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, Cz = Table 6.2.15

Wind Velocity= 210 km

hr (Dhaka)

qz = CC ∗ CI ∗ Cz ∗ Vb2= 2.08 ∗ Cz

Table 5: Overall pressure coefficients, Cp for rectangular building with flat roof. (Ref: BNBC

1993)

B= 47.875

ft

40.575 ft

5@10’= 50 ft


Page | 26


Here,

L

B= 0.85;

h

B= 1.04;

∴ Cp = 1.49

Pz = CG ∗ Cp ∗ qz = 4.43 *Cz

Table 6: Equivalent wind forces at different floor levels.

Height (m) Cz qz pz = (

kn

m2)

Fz = Pz *A (KN) Fz (KN) F (K)

3.048 0.368 0.76544 1.63 1.63*3.0478*4.33 21.51 1.80

6.096 0.415 0.8632 1.84 1.84*3.0478*4.33 24.28 5.42

9.144 0.498 1.03584 2.21 2.21*3.0478*4.33 29.17 6.51

12.192 0.57 1.1856 2.53 2.53*3.0478*4.33 33.39 7.45

15.24 0.63 1.3104 2.79 2.79*3.0478

2*4.33 18.41 4.109

Design of Slab


Page | 27


Figure 15: Typical floor plan

Thickness, t = 𝑙𝑜𝑛𝑔 𝑙𝑒𝑛𝑔𝑡 𝑕(0.8+

𝑓𝑦

200000 )

36+9𝛽

Considering the largest two panels of 22'-10"x13'-2" and 22'-10"x11'-6". As the longest

dimension among both of them are same the thickness will depend on β.

So, β = 22.83

13.17= 1.73

Thickness, t =5.63 in. 5.5 in.

Load:


Page | 28


Self weight of slab =

Floor finish= 30 psf

Partition wall= 40 psf

Live Load = 40 psf

Total, W = (174+64) = 231psf

f’c=3000 psi

fy= 60000 psi

m = = 0.58 ~ 0.6 and case 4

m = = 0.5and case 9

Table 7: Moment coefficients for two-way slabs. (Ref : BNBC 1993)

Conditions Case 4 Case 9

- CA 0.089 0.088

-CB 0.011 0.003

+CA(DL) 0.053 0.038

+CB(DL) 0.007 0.002

+CA(LL) 0.067 0.067

+CB(LL) 0.009 0.004

For, case 4

Short distance A, +M= { =

short distance A, -M = { *W* =

For, case 9

Short distance A, +M= { =

short distance A, -M = { *W* =


Page | 29


So, in short direction –M = and +M=

= 0.132

= 0.11

(ok)

Now, lt ckd

Again,

= 0.18 (controlled)

(ok)

The distance between two cranked rod is 16".

So, Required reinforcement = 0.18 - =0.09

The extra negative reinforcement required, * =1. So use 1- as extra top.

By observing the moment coefficients it can be said that, all the reinforcement in long direction

will be controlled by Asmin.

So, the reinforcement will be lt ckd and 1- as extra top.


Page | 30


Figure 16: Reinforcement Details of Slab


Page | 31


Design of Beam Only for Gravity Load

Figure 17: Beam layout

Load on slab, W = 231psf

and,

f’c=3000 psi

fy= 60000 psi

= = 0.016

1

3

1

2

4

1

E C A


Page | 32


=

Beam in-between A and B grid on grid 2

Trapezoidal panel:

= = *(22.92 +11.42)*5.75 = 98.73

= = *(18.29 +4.08)*7.105 = 79.4

Load on bam:

Let, Depth, h = 18''

Self weight= * = 0.187 kip/ft *1.2 = 0.22 kip/ft

Load from Slab = + = 1+1 = 2 kip/ft

Partition wall on beam = 0.42* 9* 120 = 0.45 * 1.2= 0.54 k/ft

Total load = 0.022 + 2 +0.54 = 2.57 kip/ft


Page | 33


Figure 18: Moment coefficient for slab and beam.

At grid 2-A joint

- M = = = 54 kip-ft

At grid 2-B joint

- M = = = 86 kip-ft

At mid span

+ M = = = 62 kip-ft

As we always provide reinforcement at both top and bottom of a beam section, a designer could

consider the beam always as double reinforcement for economy.

Assume fs' = fy, 2-Ø16mm at compression zone and two later of reinforcement,

so d - d´= 18 - 2.5 – 2 = 13.5'',


Page | 34


ØMn = ØAs' fs' (d - d') = 0.9*(2*0.31)*60*(13.5-2.5)/12 = 31 k-ft

So Mu(concrete) = 86 – 31 = 55 k-ft

=

d = 9.6'' < 13.5" (Ok)

-As(at 2-B joint) = = 1 in²

a = 2.4 (ok)

ƿmin = 200/fy = .003

ƿ = = = 0.012> ƿmin and < 0.016 (ƿmax). So ok

Use 2-Ø20mm or 1-Ø20mm and 2-Ø16mm as extra top.

-As(at 2-A joint) = = 0.4 in²

a = 1 (ok)

Use 1-Ø20mm as extra top.

At mid span

Assuming stress block depth (a) equal to flange thickness hf and two layer of reinforcement.

d = 18-2.5-2=13.5".

+As = = 0.64 in²

a = 1.5 < 5.5, so rectangular beam.

+As = = 0.55 in²

a = 1.2 (ok)

use 2-Ø16mm as extra bottom.


Page | 35


If it was a T beam then,

Effective flange width

i) One fourth of the beam span = = 5.73' ~ 69"

ii) Stem width plus 8 times the slab thickness on both side = bw + (8*2)hf = 10" + 16*5.5" = 98"

iii) Steam width plus a flange overhang not greater than half of the clear distance to the next

beam = 10" + + = 155"

so, effective flange width, b = 69".

Asf =

ØMn1 = Ø*Asf*fy*(d - hf/2)

ØMn2 = Mu - ØMn1

As - Asf = ; a =

As = Asf + (As – Asf)

N.A. c = > hf (T beam checked)

Shear Design:

Vu =0 .5WL=18.3k

Ф*Va = 2*Ф b*d = 2*0.85 10*19.5 = 18.16 kip

Use Ø10mm as shear reinforcement.

Smax = = = 29 in

Smax = = = 6.5 in (Govern)

Smax = 24"

S = = = 833 in


Page | 36


So provide Ø10mm @ 6.5" c/c although the beam.

Beam in-between B and D grid on grid 2

Load on bam:

For UDL,

Depth, h = 18'', as it is the continuation of previous beam.

Self weight= * = 0.187 kip/ft *1.2 = 0.22 kip/ft

Load from Slab = + = 1 + 3.38 = 4.38 kip/ft

Partition wall on beam = 0.42* 9* 120 = 0.45 * 1.2= 0.54 k/ft

Total load = 0.022 + 4.38 +0.54 = 4.94 kip/ft

At grid 2-B and 2-D joint

- M = = = 38 kip-ft

At mid span

+ M = = = 26 kip-ft

For point load

= = *11.5*5.75 = 33.06

Point load from Slab = = 3.8 kip


Page | 37


M = = = 4.4 kip-ft

So finally

- M =38 + 4.4= 42.4 kip-ft

+ M= 26 + 4.4= 30.4kip-ft

By observing the values of

moment the negative moments at

supports will be controlled by the

continuation of adjacent beam. For

positive moment no extra

reinforcement other than the

continues reinforcement bars.

Figure 19: Moment co-efficient for single span beam

Figure 20: Reinforcement detail of beam


Page | 38


Design of Column

Figure 21: Interaction diagram for compression plus biaxial bending a) uniaxial bending about

Y axis; b) uniaxial bending about X axis; c) biaxial bending about diagonal axis; d) interaction

surface. (Ref: ACI Code, Design of Concrete Structure, 13th

edition, Chap-8, P-274)


Page | 39


Figure 22: Interaction diagram for nominal column strength in combined bending and axial load.

(Ref: ACI Code, Design of Concrete Structure, 13th edition, Chap-8, P-260)

Considering,

fy= 60000 psi

f'c=4000 psi

For a column,

P = 554 K

Mx = 85 K-ft

My = 120K-ft

For, tied column, due to accidental eccentricity strength reduction factor α = 0.8 and


Page | 40


Based on importance strength reduction factor ɸ = 0.65, (ACI Code, Design of Concrete


let, ρg = 2%

Now, ɸPn = αɸ[0.85 f’c*Ag + ρg *Ag*fy]

554 = 0.65*0.8[0.85*4*Ag+0.02*Ag*60]

Ag = 232 in2

Let, 18"x15"

For My or dimension parallal to X axis,

ɤ = dx/Dx = (18-2.5*2)/18 = 0.72~ 0.7

Eccentricity ex = My/p = 120/554 = 0.21' = 2.6"

ex/h = 2.6/18 = 0.14

From graph, Kɳ = 0.79

Py = 853 k

For My or dimension parallal to X axis,

ɤ = dy/Dy = 0.67 0.6

ey = 85/554 = 0.15' = 1.8"

ey/h = 1.8/15 = 0.12

From graph, Kɳ = 0.85

Px = 918

For Po,Kɳ = (1.1+1.12)/2 = 1.11


Page | 41


Po = 1200

+ -

= +

ɸ Pn = 0.65*700 k =455k < 554 k (not ok)

As the difference between capacity and load is 20% so, size increment will be required. But

neglecting it continue the design.

So, column size 18"x15"and As = 0.02*18*15 = 5.4 in2. Use 12- .

Figure 23: Minimum spacing between reinforcement bars


Page | 42


The distance between reinforcement bars must be such to allow the largest expected concrete

size gravel to pass between them. In order to have properly anchored reinforcement, it is

mandatory for rebars to be surrounded by concrete.

The minimum spacing between two reinforcement bars should be at least equal to the maximum

coarse aggregate dimension plus a margin of 5 mm.

Tie bar

bars are used.

Longitudinal Spacing

16 dbof main bar = 16*20/25.4 = 12"

48 dbof tie bar = 48*10/25.4 = 18"

Least dimension = 15"

So, spacing at top and bottom 12/2 = 6" c/c and at middle span 12" c/c.

Figure 24: Failure mechanism of a column

A column with 10% fewer rebars has around 10% lower capacity strength. However, if we

remove even a single intermediate stirrup, the capacity strength of that same column will be

lowered even by 50%. This happens because the stirrup’s removal doubles the buckling length of

the rebars previously enclosed by it.


Page | 43


Cross sectional Spacing

# the reinforcement at a distance greater than 6" from the outer most bar should be under a lateral

tie and

# Alternate bar should be under lateral tie.

Figure 25: Tie arrangement of rectangular column ((Ref: ACI Code, Design of Concrete



Page | 44


Figure 26: Standard bar hook for tie and stirrup. (Ref: ACI Code, Design of Concrete Structure,

13th edition, Chap-5, P-177)

Figure 27: Typical column detail


Page | 45


References

Bangladesh National Building Code (BNBC), 2006.

Concrete Technology by Nevill.

Design of Concrete Structure by Nilson (13th edition).

Design of RCC Members by WSD and USD Methods, Public Works Department (PWD),

1997.

Treasure of RCC Designs by Sushil Kumar (16th edition).

www.buildinghow.com

http://www.buildinghow.com/


Page | 46


Part II

Preliminary Design of the Superstructure of a Balanced Cantilever

Bridge for Gravity loading


Page | 47


LECTURE SCHEDULE

Lecture 1

Introduction to Bridge Engineering

About Balanced Cantilever Bridge

Design of Deck Slab

Lecture 2

Design of Railing, Post and Sidewalk

Design of Interior Girder

(Dead load Calculation, Shear force diagram, Bending Moment Diagram for dead load, Influence line for shear & moment at different sections)

Lecture 3

Design of Interior Girder, Exterior Girder

(Shear force diagram, Bending Moment Diagram for live load including truck load and Lane load at different sections, Corresponding Impact shear & moment, Design of reinforcement for shear & moment)

Lecture 4

Design of Cross Girder/ Diaphram and Articulation


Page | 48


BRIDGE DESIGN SUBMISSION GUIDELINE

The Design Report shall explain the details of the design process. It shall include the

following items:

• Design Specification, Standards followed in Analysis & Design

• Loads and Load Combinations

• Design of Slab

• Design of Railing, Post and Sidewalk

• Design of Interior Girder

• Design of Exterior Girder

• Design of Diaphrams or Cross Girders

• Design of Articulation

[Note: Appropriate hand sketches showing the details of reinforcements must

accompany all design calculations.]

Table 8: Design Data For Students

Parameters For Odd Roll No. For Even Roll No.

fc/ 3000 psi 3500 psi

fy 40 ksi 60 ksi

Lane Width 10 ft 13 ft

No. of lanes 3 2

No. of Longitudinal girders 5 4

Live load HS 15-44 HS 20-44

Total Span Length = Student Number + 120 ft

Wearing Surface + Protective Coating = 10 psf + 15 psf = 25 psf

Width of Sidewalk = 3.5 ft


Page | 49


INTRODUCTION TO BRIDGE ENGINEERING

What is a Bridge?

• A Bridge is a structure providing passage over an obstacle without closing the

way beneath.

• The required passage may be for a road, a railway, pedestrians, a canal or a

pipeline.

Requirements of an Ideal Bridge

• Economical

• Serves the intended functions with safety and convenience

• Aesthetic elegant look

Selection of Bridge Site

• A straight reach of the river

• Steady river flow without serious whirls and cross currents

• A narrow channel with firm banks

• Suitable high banks above high flood level on each side

• Rock or other hard strata close to the river bed level

• Absence of sharp curves in the approaches

• Avoidance of excessive underwater construction

• Avoidance of expensive river training work

• Proximity to a direct alignment of the connected road


Page | 50


Choice of a type of a Bridge

• Channel Section

• Sub-soil condition

• Grades and Alignment

• Hydraulic Data

• Weather

• Navigation requirements

• Economic and Strategic considerations

• Labour availability

• Materials of Construction available

• Period of Construction

• Type of loading

• Erection Facilities

Types of Bridge

• Slab Bridge

• Deck-girder Bridge

• Balanced- Cantilever Bridge

• Suspension Bridge

• Cable-stayed Bridge

• Arch Bridge


Page | 51


Figure 28: Deck-girder Bridge – Niteroi Bridge, Rio De Janeiro, Brazil

Figure 29: Arch Bridge - Sydney Harbour Bridge, Australia


Page | 52


Figure 30: Cable-stayed Bridge – Rion Antirion Bridge, Greece

Figure 31: Suspension Bridge – Akashi Kaikyo Bridge, Japan


Page | 53


Different Parts of a Bridge

Foundation: The portion below the bed level of a river.

Substructure: The parts below the bearings level and above the foundation.

Superstructure: Components above the level of bearings.

Components above the level of bearings

Figure 32: Different parts of a Bridge

Components of a Bridge

• Deck Slab

• Girder

• Diaphram or Cross Girder

• Bearings for the decking

• Abutment, Wingwall

• Pier, Viaduct

• Foundation (i.e.Pile)

• Handrail, Curb/ Sidewalk

• Approach to the Bridge (to connect the bridge proper to the roads on either side)

Superstructure

Substructure

Foundation


Page | 54


Softwares for Bridge Design

• SAP 2000

• CSiBridge

• ADAPT ABI 2012

• Structural Bridge Design

• CRSI (Slab Bridge Designer)

• ANSYS Civil FEM Bridge

Components of a Balanced Cantilever Bridge

Figure 32: Longitudinal Profile showing Different components


Page | 55


BRIDGE TERMINOLOGY

Abutment

• The end supports of the superstructure of a bridge.

• Supports the bridge deck at the ends.

• Retains the approach road embankment.

Wing walls

• The walls constructed on both sides of the abutments.

• Anchor the bridge to its approach road.

• Support the embankments of approach road.

• Protect the embankments from the wave action of running water.

Figure 33: Transverse section

Curb/ Sidewalk

• Raised portion of a roadway slab on both sides.

• Provided to check the vehicle to fall out the bridge.

• Width of 60cm & Height of 22.5 cm are adopted.

• Roadside slope is kept as 1 in 8 upto 20cm & top portion is curved.


Page | 56


Footpath

• The passage where only pedestrians are allowed to walk.

• Width may be taken as 1.5 to 2.2 metre.

Handrail

• Protective measures adopted to prevent the falling to river of the bridge users.

Pier

• Intermediate supports of the superstructure of a bridge.

• Transfer load from the superstructure to the sub-soil through the foundation.

• Obstruct the flow of water on the upstream.

• Facilitate a long bridge to be converted into segments.

Figure 34: Afflux

Afflux

• The rise in water level of the river near bridge due to obstruction created by

obstruction of piers.

• Afflux = Difference of levels of downstream and upstream water surface of

bridge.

•


Page | 57


Freeboard

• The difference between the high flood level and the level of the crown of the road

at its lowest point.

Approaches/ Embankments

• The structures that carry the road or railway track upto the bridge.

Approach Slab

• The slab provided to join the approach road with the bridge. One end rests on the

backfill of the abutment and extends into the approach at least by 3.5m.

Backfill

• Materials used to fill the space at the back of the bridge. They are the broken

stone, gravel, sand etc. and should be clean.

Figure 35: Total span, total clear span, span and clear span

Total Span & Total Clear Span


Page | 58


• The centre to centre distance between the end supports of a bridge is termed as

total span. Clear distance between the end supports is termed as total clear

span.

Span & Clear Span

The centre to centre distance between any two adjacent supports is termed as

span. Clear distance between any two adjacent supports is termed as clear span.

Headroom

• The distance between the highest point of the vehicle using that bridge and the

lowest point of any protruding member of the bridge.

•

High Flood Level (HFL)

• The highest water level ever recorded during a flood in a river or stream.

Low Flood Level (LFL)

• The lowest water level in a river or stream during dry weather

Mean or Ordinary Flood Level (MFL)

• The flood level that normally occurs every year. •

Loads on Bridge

• Dead load

• Live load (i.e. Vehicles and Pedestrians)

• Dynamic or Impact effect of live load

• Wind loading

• Seismic Forces

• Buoyancy

• Water current forces


Page | 59


• Thermal Forces

• Erection Forces

• Earth Pressure

• Centrifugal Forces (for curved deck)

• Longitudinal Forces (for stopping vehicle)

• Ice loading

•

AASHTO Live Load (Truck load)

Figure 36: Truck loading as per AASHTO 2002


Page | 60


ABOUT BALANCED CANTILEVER BRIDGE

What is a Balanced Cantilever Bridge?

• A cantilever bridge is a bridge built using cantilevers, structures that project

horizontally into space, supported on only one end.

• The suspended span is designed as a simply supported span with supports at

the articulations.

• A simple cantilever span is formed by two cantilever arms extending from

opposite sides of an obstacle to be crossed.

Why Balanced Cantilever Bridge?


Page | 61


Figure 37: A bridge having simply supported span

Figure 38: A bridge having continuous span


Page | 62


Figure 39: A bridge having intermediate hinges

Advantages of Balanced Cantilever Bridge

• Being a Determinate Structure.

• The problem of large stress due to differential support settlement is eliminated

due to the internal hinges.

• The design section becomes economic.

• Less concrete, steel are required for cantilever design.

Disadvantages of Balanced Cantilever Bridge

• Requires a little more skill on the part of the designer.

• Requires more elaborate detailing of the reinforcements.

• Articulations are very congested with steel and anchorages.


Page | 63


Example- World’s largest Cantilever Bridge

Figure 40: Quebec bridge, CANADA

Details of Quebec Bridge, CANADA

• Total length: 987 m (3,239 ft)

• Width: 29 m (94 ft) wide

• Longest span : 549 m (1,800 ft)

• Opened: December 3, 1919

• Carries: 3 lanes of roadway

1 rail line

1 pedestrian walkway

• Crosses: St. Lawrence River


Page | 64


Example - Bangladesh China Friendship Bridge

Figure 41: Bangladesh China Friendship Bridge or Mukterpur Bridge, Bangladesh

(Source: Googlemap)

Details of Bangladesh China Friendship Bridge

• Bridge Type : Pre-stressed concrete box girder

• Length : 151 m (over river Dhaleswari on Dhaka-Munshigonj road)

• Width : 10 m (carriage way - 7.5 m & sidewalk - 2x1.25 m)

• No. of Lanes : 2 Lanes

• No. of Span: 37 nos.

• No. of Abutment: 2 nos.

• No. of Piers: 38 nos.

• Type of Foundation : Pile foundation


Page | 65


Figure 42: All spans of Bangladesh China Friendship Bridge

Cantilever Span


Page | 66


Figure 43: Articulation/ Halving joint

Figure 44: A back view showing diaphragm/cross girder and longitudinal girder


Page | 67


DESIGN OF DIFFERENT COMPONENTS

Longitudinal Profile

Figure 45: Longitudinal profile a three spanned balanced cantilever bridge

• Total Span, L g = 2 ft

• Middle Span, LM

Assumed Relations:

• End Span, LE L

M = 1.25 L

E

• Suspended Span, LS

Lc = 0.6 L

s

• Cantilever Span, LC

H > LS/14


Page | 68


Figure 46: Transverse section

• S = [(Clear Road width – Number of Girder × Girder width) /(Number of Girder -1)] – 2 × ½ × (Haunch width)

Design Specification & Method

AASHTO 2002

(American Association of State Highway & Transportation Officials)

USD – Ultimate Strength Design Criteria

Width of Longitudinal Girder,


Page | 69


DESIGN OF DECK SLAB

Figure 47: Reinforcement in Deck Slab

Design Steps

Assume Slab thickness, t

Determine self-weight of slab = (150 × t /12) psf

Consider wearing surface

Determine total dead load in psf , wDL

(Self weight of slab + load of wearing surface

Determine Dead load Moment , MDL

= + (1/9 )wDL

s2


Page | 70


Determine Live load Moment , MLL

= [{(s +2) /32} P15 or 20

] × 0.8

[Here 0.8 is continuity factor as the slab is continuous over > 3 supports]

Determine Impact Moment, MI = I × (M

LL). Here Impact Factor, I = 50/(L +125)

Determine Total Moment, M = MDL

+ MLL

+ MI

Check, dreq.

= [Mu / {Ø ρ

max b f

y (1- 0.59 ρ

max fy / f

c' )}]

1/2

where Mu = Ø ρ b d

2

fy (1- 0.59 ρf

y / f

c' )

Determine main steel area, As (req)

= Mu / [Ø f

y {d - (a/2)} where a = A

s f

y / 0.85 f

c' b.

Check, As (req)

> As(min)

Determine the distribution steel, Pdistr

= 2.2/ < 0.67


Page | 71


DESIGN OF RAILING

Figure 48: Railing and Post

• Each railing shall be designed for 50 lb/ft uniformly distributed live load acting

simultaneously in both vertical and horizontal direction.

• Opening between rails < 6 inch for portion 27 in. vertically from walkway surface.

• Opening between rails < 8 inch for portion above 27 in. from walkway surface.

Design Steps:

• Assume, 5in. X 5in. Railing

• Consider Live load on each railing = 50lb/ft

• Determine Dead load per unit length

• Determine total load wT per unit length

• Determine Maximum Moment = (1/10) wT l2

• Determine steel Area As .


Page | 72


DESIGN OF RAIL POST

Figure 49: Rail Post

• Minimum Height of Rail Post = 42 inch.

• Rail posts shall be designed for only a lateral load of P = 50 L acting at the top of

the post where L= Spacing of the posts.


Page | 73


DESIGN OF CURB/SIDEWALK

Figure 50: Cross-section of Curb or Sidewalk

• Determine P1, P2 , P3, P4.

• Determine bending moment M & F at critical section

• Determine steel area, As1 due to M

• Determine As2 = F / (Ø fy )due to F

• Place 50% of As2 at top & remaining 50% at bottom.


Page | 74


DESIGN OF INTERIOR GIRDER

Figure 51: Longitudinal and transverse profile of Interior Girder


Page | 75


Figure 52: Dead Loads and different sections of Interior Girder


Page | 76


Table 9: Determining depth of interior girder at different sections

Section Distance from

Centreline (1loc.) (ft)

Depth of Girder, y

(inch)

Depth of Variable

part (inch)

1

2

3

4

5

6

7

8

9

10

11

12

13

14


Page | 77


Dead load Analysis of Interior Girder

• Determine Dead load coming from Self weight of slab & wearing surface.

• Determine self weight of Longitudinal Girder from constant & variable part.

• Determine self weight of cross girder/diaphram.

Dead Load from Variable part of Girder

Figure 53: Variable part of Interior Girder


Page | 78


Table 10: Calculation of necessary data for Variable Part of Interior Girder

Section

wij (ft)

yi

(in.)

yj

(in.)

Load

from

Varying

part (lb)

C.G. from

left

Location

of load

from CL

(1loc.) (ft)

1-2

2-3

3-4

4-5

5-6

6-7

7-8

8-9

9-10

10-11

11-12

12-13

13-14


Page | 79


Table 11: Determining concentrated load of cross girder/diaphram

Load of Diaphram Depth of Cross

girder (in.)

Width of Girder, bd

(inch)

Load (lb)

P1

P2

P3

P4

P5

P6

Figure 54: Typical SFD and BMD of interior Girder due to dead loading


Page | 80


Live Load Analysis of Interior Girder

Influence Line (IL)

• IL is a digram showing the variation in shear, moment, reaction, stress in a

structure due to a unit load moving across the structure.

• Miller Breslay’s Principle

“The ordinates of IL for any stress element (such as axial force, shear force,

bending moment or reaction) of any structure are proportional to those of the deflection

curve which is obtained by removing the restrain corresponding to that element from

structure & introducing in its place, a corresponding deformation into the primary

structure which remains.”

Figure 55: Influence line for shear and Moment at 2-section


Page | 81


Figure 56: Influence line for shear and Moment at 6&10-section

Figure 57: The dimension SG for live load multiplier


Page | 82


Figure 58: Forward and Reverse truck loading

Figure 59: Truck direction for Maximum positive and negative Shear


Page | 83


Figure 60: Lane loading as per AASHTO 2002

Figure 61: Lane load for Maximum positive and negative Moment at 10-section


Page | 84


Figure 62: Lane load for Maximum positive and negative Shear at 10-section


Page | 85


Explanation of Loaded length (L) in Impact Formula

Figure 63: The length L for determining Impact Factor I


Page | 86


Table 12: Design shear of Interior Girder at different sections


Page | 87


Table 13: Design Moment of Interior Girder at different sections

Note: Use Dead load with its sign to combine with both positive and negative live load

moments


Page | 88


Flexural Reinforcement Design of Interior Girder

• Determine Effective width beff. for Interior Girder.

• Choose minimum of Span/4, Centre to centre distance of Girder,

(16 x Slab thickness) + Girder Width for beff .

• Consider the Design Moments for each sections.

• Determine Steel Area As for maximum design moment.

• Bar Cut-off will be done where required.

Table 14: Required flexural reinforcement of Interior Girder


Page | 89


Shear Reinforcement Design of Interior Girder

• Determine Shear Strength of Concrete,

Ø Vc = Ø 2 bw d

• Determine required spacing of shear reinforcement,

sreq. = Ø Av fy d / (Vu - Ø Vc)

• Also determine Maximum spacing for the check,

smax. = Av fy / (50 bw ) or, d/2 or, 24 inch (which one is the smallest)

Table 15: Required Shear reinforcement of Interior Girder

Section

Design Shear

(kip)

Ø Vc (kip)

Av

Srequired

(in.)

Sprovided

(in.)

1 2 3 4 5 6 7 8 9

10 11 12 13 14


Page | 90


DESIGN OF EXTERIOR GIRDER

Figure 64: Loads considered for Exterior Girder


Page | 91


Live Load Analysis of Exterior Girder

Figure 65: The dimension d & SG for live load multiplier


Page | 92


DESIGN OF CROSS GIRDER/ DIAPHRAM

Figure 66: Cross-section of Cross Girder or Diaphram


Page | 93


ARTICULATION

• The connection between the suspended span and the edge of the cantilever is

called „Articulation‟.

• The bearings at articulations can be in the form of sliding plates, roller-rocker

arrangement or elastomeric pads.

Figure 67: Possible Cracks near Articulation


Page | 94


Figure 68: Clearance requirement around bearing pad

Figure 69: Minimum Edge distance and expansion gap at articulation


Page | 95


Figure 70: Widening of Girder section for providing enlarged section near articulation


Page | 96


Design of Articulation

Figure 71: Bearing and shear friction criteria considered for minimum width of bA


Page | 97


Figure 72: Elevation and plan view of Articulation showing reinforcement

Design Steps

Determine flexural steel area As1 based on moment MA .

Determine steel area As2 based on FA.

Determine steel area Ash based on direct tension of VA.

Determine required spacing s.


Page | 98


DETAILING OF ARTICULATION

Figure 73: Detailing of reinforcement


Page | 99


REFERENCES

Lecture Note on CE 412, Prepared by Dr. Khan Mahmud Amanat,

Professor, Dept. of Civil Engg., BUET.

Lecture Note on CE 412, Prepared by Mr. Ruhul Amin, Assistant

Professor, Dept. of Civil Engg., BUET.

AASHTO Code 2002.

Nilson A.H., Darwin D., Dolan C.W. (2003), Design of Concrete

Structures, 13th edition, Mc. Graw Hill.

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