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8/11/2019 Chapter 3 Example 1
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Dept. of Civil & Urban Eng., iOTec-HU. Hawassa University
___________________________________________________________________________
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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Dept. of Civil & Urban Eng., iOTec-HU. Hawassa University
___________________________________________________________________________
Advanced Structural Design (CEng 5721 ) Chapter 3 Examples 2
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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Dept. of Civil & Urban Eng., iOTec-HU. Hawassa University
___________________________________________________________________________
Advanced Structural Design (CEng 5721 ) Chapter 3 Examples 3
= 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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Dept. of Civil & Urban Eng., iOTec-HU. Hawassa University
___________________________________________________________________________
Advanced Structural Design (CEng 5721 ) Chapter 3 Examples 4
= 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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Dept. of Civil & Urban Eng., iOTec-HU. Hawassa University
___________________________________________________________________________
Advanced Structural Design (CEng 5721 ) Chapter 3 Examples 5
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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Dept. of Civil & Urban Eng., iOTec-HU. Hawassa University
___________________________________________________________________________
Advanced Structural Design (CEng 5721 ) Chapter 3 Examples 6
∴ = + = 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