Design Example 4

7/28/2019 Design Example 4

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1.054/1.541 Mechanics and Design of Concrete Structures Spring 2004Prof. Oral Buyukozturk Design Example Shear and Torsion

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1.054/1.541 Mechanics and Design of Concrete Structures (3-0-9)

Design Example

Shear and Torsion

Objective:To examine the adequacy of given cross section based on shear and torsion capacities.

Problem: At a section of a beam, the internal forces areVu =45 kips, Mu=300 kips-ft, andTu =120 kips-ft. The material strengths are fc =4 ksi and fy =60 ksi. Assume that thedistance from beam faces to the center of stirrups is 2 in and d is 21 in.

a

a

Massachusetts Institute of Technology

Figure 1. Plan view of overpass

24

16

Tu

MuVu

Figure 2. Cross section a-a


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Task: Answer or accomplish the following questions and tasks based on given assumption.

Will this given cross section (16 in wide and 24 in deep) be adequate for shear andtorsion requirements? If not, what width is required?

Assume that the depth is to be held at 24 in. Select the required torsion, shear, and bending reinforcement for the minimumrequired width (to nearest inch).

Summarize reinforcement on a sketch of the cross section.

[Design Procedures]

1. Check maximum torsion capacity (ACI)

For the given section: Section/(Equation)

( ) ( )2 2

16 24 6144x y = = in3

( )( )2

16 210.05469

6144w

T

b dC

x y= = =

in-1

( )' 2 32 2

0.8 0.8 4000 6144318.88 10

0.4 450.4 110.05469 120 12

c

c

u

T u

f x yT

V

C T

= = =

++

lbs-in

Since ( )max

4s cT T= 11.6.1 &

( ) ( ) ( )max 4 4.25u c s c c cTT T T T T = + = + = 0.85( for torsion) 11.6.2.2 =

( ) 3max

1355 10uT = lbs-in

However, lbs-in( ) 3actual

120 kips-ft 1440 10uT = =

( ) ( )actual maxu u

T T> Section is NOT adequate.

2. Selection of cross section dimensions

Let h =24 in =constant.

( ) ( )2

3

actual

0.8 4000 244.25 1440 10 lbs-in 4.25

1.02579u c

xT T < <

Assume to be about the same same dimension.TC

x>16.92 in

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Try x=17 in,

( ) ( )2 2

17 24 6936x y = = in3, =0.05147 inTC -1.

( ) 32

0.8 4000 6936341 10

0.4 4510.05147 120 12

cT

= =

+

lbs-in

( ) 3max

4.25 1449 10u cT T = = lbs-in > (O.K.)uT

Section 17 is OK. (Although predicting heavy reinforcement.)24

3. Selection of stirrups

3 31440 341 10 1353 10

0.85

us c

TT T

= = =

lbs-in

( )1 17 2 2 13x = = in

( )1 24 2 2 20y = = in

1

1

0.66 0.33 1.168Ty

x = + =

( ) ( ) ( )

3

31 1

1353 100.0743

1.168 13 20 60 10

t s

T y

A T

s x y f

= = =

in2/in

( ) ( )

min

25 17250.00708

60000t w

yv

A b

s f

= = =

in2/in (O.K .) 11.6.5.3

( ) ( )

( )

'

2 2

2 4000 17 21210657

120 121 2.5 0.051471 2.5

45

c wc

uT

u

f b dV

TC

V

= =

++

= lbs

3 345 10.66 10 42.28 100.85

us c

VV V

= = =

lbs

( )

32

3

42.28 100.0336 in /in

60 10 21

v s

y

A V

s f d

= = =

(11-15)

Since 2

stirrup

1 0.03360.0743 0.0911 in /in

2 2t vA AA

s s s

= + = + =

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1 1max

13 208.25

4 4

x ys

+ += = = in

( )( )'4 4 4000 17 21 90314c wf b d= = lbs > sV (Not applicable) 11.5.4.3

max 21 10.52 2ds = = = in 11.5.4.1

( ) ( ) ( )' 20.5 0.85 0.5 4000 6936 186 10cf x y = 3= lbs-in in;

(ii) s0.0071s in2.

Try #5 stirrups @ 3.5in, A =0.31 in2.

min 3.40.0911

A

s = = in (O.K.)

max 8.25s = in (O.K.)

( )min 0.0071 0.0071 3.5 0.0249A s= = = in2 (O.K.)

#5 stirrups @ 3.5in are O.K.

4. Selection of longitudinal reinforcement

( ) ( ) ( )1 1

1 12 2 2 0.0743 13 20 4.9t

l t

Ax y

A A x ys s

+

= = + = + = in2

'

,min

5 c cp yvtl

yl yl

h

f A fAA p

f s f= (11-24)

cpA is the area enclosed by outside perimeter of concrete cross section, in2, and hp is the

perimeter of centerline of outermost closed transverse torsional reinforcement, in.

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Assuming yv ylf f= ( yvf : yield strength of stirrups/ ylf : yield strength of longitudinal steels)

and substituting all known numbers ( 17 24 408cpA = = in2, ( )2 17 4 24 4 66hp = + = in)

provides


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( )balanced

0.00312.6

0.003 y

bC

= =

+in balanced balanced0.85 10.71a C= = in

( ) ( )( ) ( ) ( )

balanced

0.85 4 10.71 1710.32

60s s b

A A= = = in2

( )( )

10.320.0289

17 21b = =

max 0.75 0.0217b = = (O.K.)

As =3.47 in2 for flexure

5. Summary

Use

4 #10 @ the bottom (3.47 in2 for M, 1.61 in2 forT)

2 #6 @ intermediate level (1.50 in2 forT)

3 #6 @ the top (1.79 in2 forT, 0.46 in2 excess)

We have an excess of steel. In the worse case, we have a momentMu withoutTu. The 5.08in2

of steel at the bottom can all be considered for flexural tensile reinforcement purpose. In that

case,

( )( )

5.080.0142

17 21

= =


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Comments on the final design:

1. Different design configurations are possible, in general. Various combinations ofdifferent sizes of steel bars can achieve same reinforcement ratio. However, relevant

designs are made typically considering the convenience of construction and the spacing

between any two steel bars (the concrete between two steel bars will crash undesirably if

the spacing between them is not enough).

2. Considering the constructability, four corner positions are usually required to deploylongitudinal bars to fix stirrups.

3. It is preferable to use same size of steel bars on each cross section for economic reasonunless it is not possible to achieve the requirement of reinforcement.

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Design Example 4

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