de si gn of str uctures - weebly

DESIGN OF

STRUCTURES

Engr. Faizan Tahir

RETAINING WALLS

Retaining walls are used to hold back masses of earth or other

loose material where conditions make it impossible to let those

masses assume their natural slopes.

Function of retaining wall

Such conditions occur when the width of an excavation, cut,

or embankment is restricted by conditions of ownership, use of

the structure, or economy. For example, in railway or highway

construction the width of the right of way is fixed, and the cut or

embankment must be contained within that width.

Similarly, the basement walls of the buildings must be located

within the property and must retain the soil surrounding the

base.

Free standing retaining walls, as distinct from those that form

parts of structures, such as basement walls, are of various types.

Types of retaining walls

The gravity retaining wall retains the

earth entirely by its own weight and

generally contains no reinforcement. It

is used up to 10 to 12 ft height.

The reinforced concrete cantilever

retaining wall consists of the vertical arm

that retains the earth and is held in position

by a footing or base slab. In this case, the

weight of the fill on top of the heel, in

addition to the weight of the wall,

contributes to the stability of the structure.

Since the arm represents a vertical

cantilever, its required thickness increase

rapidly, with increasing height. It is used in

the range of 10 to 25 ft height.

In the counterfort wall the

stem and base slab are tied

together by counterforts

which are transverse walls

spaced at intervals and act as

tension ties to support the

stem wall. Counterforts are of

half or larger heights.

Counterfort walls are

economical for heights over

25 ft.


Property rights or other restrictions sometimes make it necessary

to place the wall at the forward edge of the base slab, i.e. to omit

the toe. Whenever it is possible, toe extensions of one-third to

one-fourth of the width of the base provide a more economical

solution.

A buttress wall is similar to a

counterfort wall except that

the transverse support walls

are located on the side of the

stem opposite to the retained

material and act as

compression struts. Buttress,

as compression elements,

are more efficient than the

tension counterforts and are

economical in the same

height range.


A counterfort is more widely used than a buttress because the

counterfort is hidden beneath the retained material, whereas the

buttress occupies what may otherwise be usable space in front of

the wall.

This is an free standing wall category. A wall type bridge abutment

acts similarly to a cantilever retaining wall except that the bridge

deck provides an additional horizontal restraint at the top of the

stem. Thus this abutment is designed as a beam fixed at the

bottom and simply supported or partially restrained at the top.


w is unit weight of the soil

Earth Pressure

C0 is a constant known as the coefficient of earth pressure at

rest According to Rankin, the coefficient for active and passive

earth pressure are

22

22

CosCosCos

CosCosCosCosCa

22

22

CosCosCos

CosCosCosCosCp

For the case of horizontal surface =0

sin1

sin1Cah

sin1

sin1Cph

For liquid P=wwh, ww is the

unit weight of water.

Soil retaining structure Ph=Cowh

Earth pressure for common condition of loading

2whC2

1P

3

hy

ah

CosCFor

Cawh2

1p

3

hy

a

2

Earth pressure for common condition of loading

)h2h(whC2

1P

)h2h(3

hh3hy

ah

2

1. Individual parts should be strong enough to resist the applied

forces

Stability Requirement

2. The

wall as

a whole

should

be

stable

against

(i)

Settlem

ent (ii)

Sliding

(iii)

Overtur

ning


It is necessary to ensure that the pressure under the footing does

not exceed the “permissible bearing pressure” for the particular

soil.

By the formula

Settlement

I

MC

A

Nq

minmax

If , compression will act throughout the section

3a

v21

2v

22v

1

2v

2vvv

2,1

3

v

2

v

v3

v

2

v

v2,1

3

vv

2,1

Rqq

2awhen

2a6R

q&a64R

q

a63RaR6R3R

q

12

4

aR2RR

12

2

aR

4

RR

q

12

2Ra

2Rq

Unit dimension in the direction perpendicular to the paper.

Stability Requirement Settlement

0q&R2

q

11Rq

2R

612

1R

2R

6

2312

1Rq

12

2R

32Rq

2v

1

v2,1

v

v

v

v2,1

vv

2,1

33

3


a3

R2q

a3q2

1R

v

v


Sliding

F = Rv


5.1P

F

h

Overturning

2moment gOverturnin

moment gStabilizin

1. Lateral earth pressure will be considered to be live loads and

a factor of 1.7 applied.

Basis of Structural Design

2. In general, the reactive pressure of the soil under the structure

at the service load stage will be taken equal to 1.7 times the soil

pressure found for service load conditions in the stability

analysis.

3. For cantilever retaining walls, the calculated dead load of the

toe slab, which causes moments acting in the opposite sense to

those produced by the upward soil reaction, will be multiplied by

a factor of 0.9.

4. For the heel slab, the required moment capacity will be based

on the dead load of the heel slab itself, plus the earth directly

above it, both multiplied by 1.4.

5. Surcharge, if resent, will be treated as live load with load

factor of 1.7.

6. The upward pressure of the soil under the heel slab will be

taken equal to zero, recognizing that for severe over load stage

a non linear pressure distribution will probably be obtained, with

most of the reaction concentrated near the toe.

Basis of Structural Design

Drainage

Reduction in bearing capacity.

Hydrostatic pressure.

Ice pressure

Drainage can be provided in various ways

i. Weep holes, 6 to 8 in. 5 to 10 ft horizontally spaced. 1 ft3

stone at the bottom weep holes to facilitate drainage and to

prevent clogging.

ii. Longitudinal drains embedded in crushed stone or gravel,

along rear face of the wall.

iii. Continuous back drain consisting of a layer of gravel or

crushed stone covering the entire rear face of the wall with

discharge at the ends.

Problem A gravity retaining wall consisting of plain concrete w=144

lb/ft3 is shown in fig. The bank of supported earth is

assumed to weigh 110 lb/ft3 ø=300 and to have a coefficient

of friction against sliding on soil of 0.5.

Solution

333.030sin1

30sin1

sin1

sin1Ca

330sin1

30sin1

sin1

sin1Cp

lb36.2637)12)(110)(333.0(2

1whC

2

1P 22

aah

lb.ft 44.10549

3

1236.2637momentgOverturnin

Restoring moment:

Moment about toe

Component Weights

Force Moment arm Moment

W1 (6)(1)(144) 864 x 3 2592 lb.ft

W2 (1)(11)(144) 1584 x 1 1584 lb.ft

W3 (4)(144)(11) 8976 lb.ft

W4 (4)(11)(110) 10083.33 lb.ft

W5 (0.5)(11)(110) 605 x 5.75 3478.75 lb.ft

Rv = 8641 lb M=26714.08 lb.ft

2

1

43

15.13168

Solution

43

25.12420

2

1

Safety factor against overturning = O.K 253.244.10549

08.26714

Distance of resultant from the toe

The max. soil pressure will be )87.1)(3(

)8641)(2(

a3

R2q v

q = 3080.57 lb/ft2

Solution

ft87.18641

44.1054908.26714a

Sliding

Assuming that soil above footing toe has eroded and thus

the passive pressure is only due to soil depth equal to

footing thickness.

Solution

lb 516)1)(110)(3(2

1whC

2

1P 22

pph

Friction force between footing concrete and soil.

F.O.S. against sliding = O.K

lb 5.4320)8641)(5.0(RF v 5.17.1

36.2637

1655.4320

Estimating size of cantilever retaining wall

The base of footing should be below frost penetration about 3’ or 4’.

Height of Wall

Stem is thickest at its base. They have thickness in the range

of 8 to 12% of overall height of the retaining wall. The minimum

thickness at the top is 10” but preferably 12”.

Stem Thickness.

Preferably, total thickness of base fall between 7 and 10% of

the overall wall height. Minimum thickness is at least 10” to 12” used.

Base Thickness

Estimating size of cantilever retaining wall

For preliminary estimates, the base length can be taken

about 40 to 60% of the overall wall height.

Base Length

Another method refer to fig. W is assumed to be equal to

weight of the material within area abcd.

Take moments about toe and solve for x.

Problem Design a cantilever retaining wall to support a bank of earth of

16 ft height above the final level of earth at the toe of the wall.

The backfill is to be level, but a building is to be built on the fill.

Assume that an 8’ surcharge will approximate the lateral

earth pressure effect.

Weight of retained material = 130 lb/ft3

Angle of internal friction = 35o

Coefficient of friction b/w

concrete and soil = 0.4

fc’=3000 psi

fy=40,000 psi

Maximum soil

pressure

= 5 k/ft2

Solution

Allowing 4’ for frost penetration to the bottom of footing in

front of the wall, the total height becomes.

Height of Wall

h = 16 + 4 = 20 ft.

At this stage, it may be assumed 7 to 10% of the overall

height h.

Thickness of Base

Assume a uniform thickness t = 2’ ( 10% of h )

h = 20’ h’ = 8’

Base Length

ft148.8

82203

820320

h2h3

hh3hy

lb2.11707

3620120271.05.0

822020120sin1

sin1

2

1

)h2h(whC2

1P

22

ah

Solution

Moments about point a

W = (120)(x)(20+8) = 3360 x lb

yP)2

x)(W(

)148.8)(2.11707()2

x)(x3360(

x = 7.54 ft

So base length = 1.5 x x = 11.31 ft

Use 11 ft 4” with x = 7’-8” and 3’-8’ toe length

Solution

Stem Thickness Prior computing stability factors, a more accurate

knowledge of the concrete dimensions is necessary.

Solution

The thickness of the base of the stem is selected with the

regard for bending and shear requirements.

P for 18’ height and h’ = 8’

lb12.9951

)8218(18120271.02

1

)h2h(whC2

1P ah

ft412.7

)8218(3

)8)(18)(3()18(

)h2h(3

hh3hy

22

'

'

Mu = (1.7) Py = (1.7) (9951.12) (7.412)

= 125388.09 lb.ft

Solution

02784.075.0

03712.0

000,4087000

87000

000,40

)3000)(85.0)(85.0(

f87000

87000

f

f)85.0(

bmax

yy

cb

1

For adequate deflection control, choose

686.15

300085.0

000,40

f85.0

fm

)m2

11(fRthen

c

y

yn

01392.0

2

1max

Solution

67.16d

65.3370

4969.0

1209.125388

R

M

R

MquiredRebdquiredRe

496

)686.1501392.02

11)(000,40)(01392.0(

m2

11fR

n

n

n

n2

yn

Total thickness = 16.7 + 0.5 + 3 = 20.26”

Try 21” thickness of base of stem and select 12” for top of the wall

Shear at d d used now = 17.5” = 1.458’ At 18’ – 1.458’ = 16.542’ from top

.requiredisorcementinfreshearnoSo,VVSince

lb7.19553

5.17123000285.0

bdf2V

lb96.17879

P7.1V

lb92.8752

82542.16542.16120271.02

1

)h2h(whC2

1P

uu

cu

u

ah

Solution

F.O.S Against Overturning

Component Force Arm Moment

W1 (5.92)(18)(120)=12787.2 107156.74

W2 (0.75)(18)(150)=1012.5 3.67+1+0.25=4.92 4981.50

W3 (18)(1)(150)=2700 3.67+0.5=4.17 11259

W4 (11.33)(2)(150)=3399.00 19255.34

W5 (18)(0.75)(120)=810 3.67+1+0.5=5.17 4187.7

W6 (6.67)(8)(120)=6403.2 51257.62

Total 27111.9 198097.9

)2

1(

)2

1(

38.82

92.52167.3 "

665.52

33.11

005.82

67.6167.3

Solution

P = 11707.2 lb y= 8.148 ft Overturning Moment = 11707.2 x 8.148 = 95390.27 lb.ft

F.O.S. against overturning O.K 2 077.227.95390

9.198097 Location of Resultant & Footing Soil Pressure

Distance of the resultant from the front edge is

7883.39.27111

27.953909.198097a

)righting( load Total

moment gOverturninmoment Rightinga

Middle third = 3.7778 ft, So resultant is within the middle third.

Solution

7883.3633.11433.11

9.27111q

)a64(R

q

21

v21

2

22ft/lb 74.14)2a6(

Rq v

q1= 4771.12 lb/ft2 < 5 k/ft2

So O.K against bearing pressure.

Solution

Passive earth pressure against 2’ height of footing aphCwh

2

1 2

Solution

.requirediskeySo5.1964.02.11707

44276.10844.S.O.F

lb82.442

sin1

sin1)2)(120(

2

1

F.O.S. against sliding

force causing sliding = P = 11707.2 lb

Frictional resistance = R

= (0.4) (27111.9)

= 10844.76 lb

The front of key is 4” in front of back face of the stem. This

will permit anchoring the stem reinforcement in the key..

ft/lb58.2635ordinateTotal

ft/lb84.2620x

33.11

38.4756

243.6

x

Frictional resistance between soil to soil = R

lb17.13191

)087.5(2

58.263512.4771)(tan

Solution

Frictional resistance between heel concrete to soil = R

lb19.3309

)243.6(2

74.1458.2635)4.0(

lbh4.22169.3h1202

1

Cwh2

1pressureearthPassive

22

ph

2

Solution

F.O.S. against sliding = 1.5

ft19.2h

2.11707

h4.22119.330917.131915.1

2

So use key of height = 2’-3” = 2.25’

Design of Heel Cantilever Wu = (1.7) (120) (8) + (1.4) [18 x 120 + 2 x 150 ]

= 5076 lb/ft

ft.lb76.88947

92.576.52

1

2

WM

2

u

2

Vu = Factored shear a joint of stem and heel

Solution

When the support reaction introduces compression into

the end region, then critical shear is at a distance d

from face of support. However, the support is not

producing compression, therefore, critical shear is at

joint of stem and heel.

Design of Heel Cantilever

Vu = Factored shear a joint of stem

and heel

= (5.92) (5076)

= 30049.92 lb

Solution

u

cc

V11.24023

5.21123000285.0

bdf2V

So depth is required to be increased.

98.26)12)(3000)(2)(85.0(

92.30049

bf2

Vd

c

u Therefore heel thickness 30”, d = 27.5”

Now Wu=(1.7)(120)(8)+(1.4)[(17.5)(120)+(2.5)(150)] = 5097 lb/ft

.ft.lb 75.89315)92.5)(5097(2

1M 2

u

005.0f

200

00337.0f

mR211

m

1

psi 23.131)5.27)(12)(9.0(

1289315

bd

MRquiredRe

y

min

y

n

un 22

Solution Design of Heel Cantilever

As = min bd = 1.8 in2

Use # 8 @ 5” c/c (As = 1.88 in2 )

Dev. Length required = 23” top bars so 23 × 1.3 = 29.9”

Available = 5.92’-3” = 68.04” O.K

Design of Toe Slab

33.11

66.7

38.4756

x x = 3215.7

3215.7 + 14.74 = 3230.44

(6801.33)lb/ft

2

12.477144.3230)7.1(Wu

Wu = 3801.33 lb/ft – 270 = 6531.33

Solution

Overload factor = 0.9

{ d=24”-3.5”=20.5”

=1.71 ft

Self load=(0.9)(1×2×150)=270 lb/ft

3.1165.20129.0

1292.43984

bd

MRquiredRe

ft.lb 92.439842

)67.3)(33.6531(

2

WM

22u

n

uu

22

So min will control

As = (0.005)(12)(20.5) = 1.23 in2

Use # 8 @ c/c (As = 1.26 in2)

2

17

Available dev. Length = 3.76’ – 3” = 41.01” Required = 23”

At a distance d= 20.5” = 1.71’

3.67’ – 1.71’ = 1.96’

3.394874.1456.3933

56.3933x

33.11

37.9

38.4756

x

Solution

Earth pressure lb 55.14526)7.1)(96.1(2

12.47713.3948

Vu = 14526.55 – 270 x 1.96 = 13997.37 lb

u

cc

Vlb76.229055.20123000285.0

bd)f2(V

So no shear reinforcement is required

Reinforcement for stem

Solution

32.75.100

25.726

825.173

85.173)5.17(

h2h3

hh3hy

lb43.9532

825.175.17120271.02

1

)h2h(whC2

1P

22

ah

)in67.2A(c/c2

14@bars9#Use

in457.25.17120117.0bdA

0117.0

000,40

08.425686.15211

686.15

1

686.15300085.0

000,40

f85.0

fm

f

mR211

m

1

psi 08.425)5.17(129.0

121.117163

bd

MR

ft.lb1.11716323.743.95327.1M

2

s

2

s

c

y

y

n

un

u

22

Reinforcement for stem Solution

At 5’ from top

P=1707.3lb

y=2.302’ Mu=6681.34 lb.ft

=6.68 k.ft

At 10’ from top

P=4227.6 lb

y=4.36’ Mu=31334.97 lb.ft

=31.33 k.ft

At 15’ from top

P=7560.9 lb

y=6.29’ Mu=80848.7 lb.ft

=80.85 k.ft

Solution

ft.k16.117

in.lb2.1405957ft.lb1.117163baseatMu

With Full Reinforcement

C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb

T = Asfy = 2.67 x 40000 = 106800

a = 3.49 in

At top of wall d = 8.5”

ft.k11.54.in.lb6.649290

2

49.35.8)106800)(9.0(

2

adfA9.0M ysn

At base of stem d=17.5”

Solution

ft.k2.126.in.lb6.1514370

2

49.35.17)106800)(9.0(Mn

With half Reinforcement

C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb

T = Asfy = 1.335 × 40,000 = 53400 for #9@9” c/c(As=1.33 in2)

a = 1.745 in


ft.k55.30.in.lb65.366577

2

745.15.8)53400)(9.0(Mn


Solution

ft.k59.66.in.lb65.799117

2

745.15.17)53400)(9.0(Mn

With one-fourth Reinforcement

C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb

T = Asfy = 0.67 × 40,000 = 26800 for #9@18” c/c(As=0.67 in2)

a = 0.88 in


ft.k2.16.in.lb2.194407

2

88.05.8)26800)(9.0(Mn


Solution

ft.k29.34.in.lb2.411487

2

88.05.17)26800)(9.0(Mn

The actual termination point is found by extending beyond the

intersection of capacity moment line with the factored moment

diagram a distance of either the effective depth d or 12 bar

diameters, whichever is greater.

Solution

Thus half bars should be cut at 4’-8” distance from bottom and

further half bars should be cut 8’-8” theoretically from bottom.

Solution

36.6y5.17

9x

125.17

9

y12

x

d at 4.6’ from bottom (y=12.9’)

= 8.5 + 6.63 = 15.13” 15”

85.4y5.17

9x

d at 8.6’ from bottom (y=8.9’)

= 8.5 + 4.58 = 13.08” 13”

Solution

bar used # 9 of diameter = 1.128”

12 db = 13.54” = 14”

Therefore half bars should be terminated actually at

4’-8”+15”=15’-11” from bottom

and further half bars should be terminated at

8’-8”+14” =9’10”

For tension bars to be terminated in the tension zone, one

of the following condition must be satisfied.

2. Continuing bars must provide at last twice the area

required for bending moment at the cut off point.

1. Vu at the cut-off point must not exceed two-thirds of the

shear strength øVn.

3. Excess shear reinforcement is provided.

at 12.9’ from top bd)f2(V cc

Solution

kips 68.9V3

2

kips 14.53

1000

113123000285.0V

.satisfiedis)1(Condition

kips31.10

1000

1h2hwhC

2

17.1V

kips 11.17V3

2

kips 16.76

1000

115123000285.0V

c

c

ahu

c

c

Solution

kips13.6

1000

1829.89.8120271.0

2

17.1

1000

1h2hwhC

2

17.1V ahu

Condition (1) satisfied so bars can be terminated.

The above condition are imposed as a check stress concentration.

Shear at bottom = Vu = 1.7 x 9.53243 = 16.21 kips

Solution

kips56.191000

15.17123000285.0Vc

Since øVc > Vu so no need of shear reinforcement.

The used should not be less than at any point. This

minimum limit, strictly speaking, does not apply to retaining

walls. However, because the integrity of retaining wall

depends absolutely on the vertical walls, it appear prudent to

use this limit un such cases.

yf

200

First termination point is 5’-11” from bottom

where d=14.46” As =1.335 in2

Second termination point is 9’-10” form bottom where

d=12.44” As=0.6675 in2

Solution

005.0f

2000077.0

y

= 0.0045 ≈ 0.005 Therefore the above condition is satisfied.

Another requirement is that maximum spacing of the primary

flexural reinforcement exceed neither 3 times the wall thickness

nor 18 in. These restrictions are satisfied as well.

For splices of deformed bars in tension, at sections where the

ratio of steel provided to steel required is less than 2 and where

no more than 50% of the steel is spliced, the ACI code requires a class-B splice of length 1.3 ld.

ld for # 9 bars =29”

Splice length = 1.3 × 29 = 37.7” or 3’-2” O.K

Solution

Temperature & shrinkage reinforcement

Total amount of horizontal bars (h is average thickness)

ft/in39.02

50.201212002.0bh002.0A 2

s Since front face is more exposed for temperature changes

therefore two third of this amount is placed in front face and one

third in rear face.

Solution

Accordingly # 4 @ 9 in. c/c As=0.26 in. ft/in26.0A3

2 2

s Use # 3 @ 10 in. c/c As=0.13 in2. ft/in 13.0A

3

1 2

s For vertical reinforcement on the front face, use any nominal

amount. Use # 3 @ 18 in. c/c

Since base is not subjected to extreme temperature changes,

therefore # 4@ 12” c/c just for spacers will be sufficient.

Solution

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