Chapter 3 Example 1

8/11/2019 Chapter 3 Example 1

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Dept. of Civil & Urban Eng., iOTec-HU. Hawassa University

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Advanced Structural Design (CEng 5721 ) Chapter 3 Examples 1

Example 1: A transfer girder with a clear span

of 6m is to support a column having a factored

axial load of 5000KN and an additional uniform

factored load including self weight of

Wd=268.45KN/m. Using the empirical method

of Deep beam design, design the beam for

flexure and shear. Use C-25 for concrete and S-

360 for steel class I works.

Solution:

Check for deep beam

≤ / = 6 + 0.3 + 0.3 = 6.61.15 ∗ = 1.15 ∗ 6 = 6.9 = 6.6

= 6.64

= 1.65 < 2

∴

Design for flexure

Material design strength

=0.85 =

0.85 ∗ 0.8 ∗ 25

1.5= 11.33/2 =

=360

1.15= 313.04/2

Max span moment

=

2

8

+

4

=268.45

∗6.62

8

+5000

∗6.6

4

= 9711.71

Max shear at support

=

2+

2

=268.45 ∗ 6.6

2+

5000

2= 885.88 + 2500 = 3385.89

% =2500

3385.89∗ 100 = 73.84 > 50

∴ Main reinforcement

Lever arm: 1 ≤ / ≤ 2 ∴ = 0.2( + 2) = 0.2(6.6 + 2 ∗ 4) = 2.92 = 2920

As =M

sdf yd ∗ z =

9711.71

∗106

313.04 ∗ 2920 = 10624.612

Provide 24 24mm diameter bars, As,provided = 10857.34mm2

This reinforcement should be placed within the lower y depth of the deep beam where:

= 0.25 − 0.05 = 0.25 ∗ 4000 − 0.05 ∗ 6600 = 670




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Assume cover=50mm, stirrup=12mm, main =24mm

∴ = 50 + 12 + 24/2 = 74

c/c distance b/n first row bars to last row bars =670-74-24/2=584mm

/ = 584/7 = 83.43

row d from bottom

1st row d1 74

2nd row d2 157.43

3rd row d3 240.86

4th row d4 324.29

5th row d5 407.72

6th row d6 491.15

7th row d7 574.58

8th row d8 658.01

the centroid of the reinforcements d’ from the bottom is

’ =61 + 62 + 23 + 24 + 25 + 26 + 27 + 28

=6 ∗ 74 + 2(157.43 + 240.86 + 324.29 + 407.72 + 491.15 + 574.58 + 658.01)

24= 256.34

Effective depth d=h-d’=4000-256.34=3743.66

=0.6 =

0.6

313.04 ∗ 400 ∗ 3743.66 = 2870.172 ≤ = 10857.34mm2 … .ok!

Design for shear

B/c more than 50% of the shear at the support is

caused by the concentrated load, the load is a

principal load. Thus the critical shear is calculated

at av/2, where

≤ / = 3300

1.15 ∗ / = 1.15 ∗ 3000 = 3450

∴ = 3300

Vmax=3385.89

Vsd=2943

2500

2500

Vmax=3385.89

av/2=1.65

3.3

SFD

y = 6 7 0 m m

d

d’




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= 2500 +3.3 − 1.65

3.3∗ 3385.89 − 2500 = 2943

Limiting shear capacity

= 0.25

= 0.25

∗11.33

∗400

∗3743.66

∗10−3 = 4241.57

Shear carried by concrete

= ∗ 0.25 12

=2 =

2 ∗ 3743.66

3300= 2.27, 1 = 1 + 5 0 = 1 + 5 0

= 1 + 5 0 ∗ 10624.61

400 ∗ 3743.66= 1.35

2 = 1.6 − = 1.6 − 3.743 = −2.14 ≥ 1 − − − not ok ‼ take 2 = 1

=

0.21

2

3

=

0.21

∗20

23

1.5

= 1.03

= ∗ 0.25 12 = 2.27∗ 0.25 ∗ 1.03 ∗ 1.35 ∗ 1 ∗ 400 ∗ 3743.66 ∗ 10−3 = 1181.66

= − = + = 2943 − 1181.66 = 1761.34

Assume 1000KN is carried by vertical stirrups and the remaining 761.34KN is carried by

Horizontal stirrups. i.e Shear carried by vertical stirrups = 1000 and Shear carried by

horizontal stirrups = 761.34. Assume 12 is used for both vertical and horizontal stirrups

=

= 2

∗ ∗ 122

4= 226.19

2

= −

2 ≤

= −

2 =

226.19 ∗ 313.04 ∗ 3300 − 3743.662

1000 ∗ 103

= 101.12

≤

=226.19 ∗ 313.04 ∗ 3743.66

1000

∗103

= 265.08 ≤ 4

=3743.66

4= 935.91

Provide 12 c/c 100mm for vertical stirrups.

= 3

2 − ≤




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= 3

2 − =

226.19 ∗ 313.04 ∗ 32

∗ 3743.66 − 3300761.34 ∗ 103

= 215.35

≤

=

226.19∗313.04∗3743.66

761.34

∗103 = 348.17

≤3

=3743.66

3= 1247.89

Provide 12 c/c 200mm for horizontal stirrups.

Example 2: Design a corbel to support a factored ultimate load of 400KN at a distance of 360mm

from the face of a column 300x300mm in cross section. Proportion the various dimensions of the

corbel and find the amount of reinforcement required to resist the load without failure. Use C-

30 for concrete and S-400 for steel class I works.

Solution:

Proportions corbel 0.4d< av <d

= 0.8 = 0.8

=360

0.8= 450

= 50 = + = 450 + 50 = 500 ∴ = 500, = 450 = 300( )

Design material strength

=0.85 =

0.85 ∗ 0.8 ∗ 30

1.5= 13.6/2 =

=400

1.15= 347.83/2

Main Rebar As 24 24mm

Av 2mm c/c 100mm

Ah 2mm c/c 200mm

y= 0.25h-0.05l=670mm




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Check for diagonal compression failure (limiting shear capacity)

= 0.25 = 0.25 ∗ 13.6 ∗ 300 ∗ 450 ∗ 10−3 = 459 > = 400−−−−!

∴

Main reinforcement

= 2 ∗ (1 − 1 − 2

22 Determine = 0

= = = − − − −1

= 0 ∗ = ∗

=

−

2

=

∗ = ∗

( − 2) = ∗ =

1

∗ ∗ ( −

2 ) = ∗

( −

2

) = −

sin (2

)

=

= − sin (2)

=

450 − 400 ∗ 103

300 ∗ 13.6 ∗ sin (2)

360= 1.25 − 0.2723

sin (2)

= 1.25 − 0.2723

sin (2) = 44.360

∴ = 2 ∗ (1 − 1 − 2

22

= 300 ∗ 450 ∗244.36 ∗13.6

347.83 (1 − 1 −2

∗400

∗103

∗360

13.6 ∗ 300 ∗ 4502244.36 = 1176.482

Horizontal design force ≥ 0.2 = 0.2 ∗ 400 = 80

= =

80 ∗ 103

347.83= 2302 ≥ 0.25 = 0.25 ∗ 1176.48 = 294.122




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∴ = + = 294.12 + 1176.48 = 1470.62 ≥ , = 0.004 = 6002 ‼

Provide 520 diameter bars at the top, , = 1570.82

Shear reinforcement

≥ 0.5 − = 0.5 ∗ 1570.8 − 294.12 = 637.952 ≥ , = 0.4 = 0.4 ∗ 1570.95 = 628.382 −−−−!

2

3 =

2

3∗ 450 = 300

Assume 10 diameter bars is used as stirrup

= 2 ∗ ∗ 102

4= 157.082

= =

637.95

157.08≈ 4.

Provide 4 10 closed stirrup bars

/ / = 3004

= 75

Minimum width of bearing

To resist bearing failure, the bearing stress developed should be less than 0.4fcu ≤ 0.4 = 0.4 ∗ 30 = 12

= ≤ 12

≥ 12 =

400 ∗ 103

12 ∗ 300= 111.11

Provide a bearing with width of w=115mm

= 5∅20

= ∅10 / 75

b

w

F vd

Chapter 3 Example 1

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