Project of Principles of Structure Design -------Reinforced Concrete Beam
Name：Wang Pengzhi
Class ：2011210701
NO. ：08
SCHOOL OF HIGHWAY COLLEGE
CHANG’AN UNIVERSITY
DECEMBER 7, 2013
Project
Known: the member is a concrete T beam (simple support beam), the standard span is L b=13m, calculating span L=12.6m. The reinforcement distribution placed and ction dimensions of normal ction flexural capacity are shown in the diagram. Main steel bar is HRB335 (4ϕ32+4ϕ16), A s=4021mm2, erection steel bar is HRB335 (2ϕ22), an eight-layer steel skeleton is welded. Concrete is
C30, grade Ⅰenvironment, λ0=1.1. For support ction, shearing calculated value V0 =γ0Vd,0 = 342kN,bending moment calculated value M0 =γ0Md,0 = 0；For mid-span ction, shearing calculated value Vl/2 =γ0Vd, l/2 =71.0kN, bending moment calculated value Ml/2 =γ0Md, l/2 =
Solution:
(1)Check of the ctional dimensions
To meet the construction requirements, the bottommost layer of steel bar must be through the bearing ction.
And the effective depth of bearing ction is (t the thickness of cover c=38mm)
ℎ0=ℎ−(35+35.8
2
)=946mm
(0.51×10−3)√f cu,k bℎ0=0.51×10−3×√25×180×948=475.65kN>γ0V d,0(
=342kN)
So, the ctional dimensions meet the construction requirements.
(2)Check that if necessary to equip the web reinforcement
Mid-span ction:(0.5×10−3)f td bℎ0=(0.5×10−3)×1.39×180×915=114.76kN Bearing ction:(0.5×10−3)f td bℎ0=(0.5×10−3)×1.39×180×944=118.34kN So, γ0V d,l2⁄<(0.5×10−3)f td bℎ0<γ0V d,0
The stirrup must be ud with some zone of mid-span. As for other zones, the construction requirements are suit.
(3) Distribution of the envelope diagram of shear force
From the envelope diagram of shear force and V l 2⁄= γ0V d,l 2⁄, V 0= γ0V 0 and V x = γ0V d,x =(0.5×10−3)f td bℎ0=114.76kN . The distance between V x ction and mid-span ction can be get as
l 1=L 2×V x −V l 2⁄V 0−V l 2⁄=6300×114.51−71342−71=1017mm
It ’s justifiable to place stirrup reinforcement in the zone of 1023mm length.
And calculated shearing force of the ction with a distance ℎ2⁄ to the support center can be obtained proportionally from the envelope diagram of shear force.
V ′=
LV 0−ℎ(V 0−V d,l 2⁄)L =12600×342−1000×(342−71)
12600
=320.49kN
Hence, V cs is responsible for the 60% load and V sb is responsible for the rest load. V sb =0.4V ′=128.2 kN .
Besides, the ssion length of bent-up reinforcement is 2980 mm.
(4) Stirrup reinforcement design
The ϕ8mm stirrup is taken for reinforcement design. Its ction area is A sv =nA sv1=2×50.3=100.6mm 2.
The longitude reinforcement ratio p and effective depth ℎ0 of diagonal ction can be the average of the support ’s ction value and mid-span ction value. Support ction: p l 2⁄=2.44<2.5 ,ℎ0=917mm Mid-span ction:p 0=0.94,ℎ0=946mm So, the average value is p =
p l 2⁄+p 0
2
=1.69,ℎ0=
919+948
2
=931.68mm
Hence, the spacing of stirrup S v is
342k N
71k N
114.76k N
500
2980
1017
128.2k N
192.3k N
Figure 1 Allocation diagram of shearing force
S v=α12α32×(0.56×10−6)×(2+0.6p)√cu,k A sv f sv bℎ02
(V)
=12×1.12×(0.56×10−6)×(2+0.6×1.69)×√30×100.6×195×180×929.682
(320.49)2
=334mm
And the design value is eventually determined as 300mm with concerning the requirements of construction.
Web reinforcement ratio ρsv(take S V=300mm) is
ρsv=A Sv
b S V
=
100.6
180×300
=0.1863%>ρmin=0.18%
Besides that
S v≤1
2
ℎ=500mm and 400 mm
Obviously, the design meets the requirements.
In conclusion, the stirrup spacing is 100mm of the zone from the support center to the mid-span direction at the distance 1000mm. as for the rest zone; the stirrup spacing is 300mm.
(5)Bent-up reinforcement design
As the erection steel bar (HRB335) of weld reinforcement skeleton is ϕ22and the distance between C.G. of compression rebar and outer fiber of concrete is a s′=56mm.
Now, it’s planned to bent up the steel bar N1~N3. And a sheet of the calculating values is listed here.
The detailed calculation process is in the following:
The perpendicular distance between upper bent-up point and lower bent-up point △ℎ1=1000−(36+35.8×1.5)+(43+25.1+35.8×0.5)=824mm
As the bent-up angle is 45°,the distance to the center of support for first row x1=824mm
For the cond row of steel bar,
△ℎ2=1000−(36+35.8×2+18.4×0.5)+(43+25.1+18.4×0.5)=806mm The distance to the center of support for first row x2=824+806=1630mm As for the cond row’s calculated shearing value of distribution V sb2
2980+500−824
2980=
V sb2 128.20
V sb2 =114.26 kN Notation: the designed length of bent-up ssion is 2980mm.
Required area of bent-up bars A sb2=1333.33(V sb2)
f sd sin45°=1333.33(114.26)
280×sin45°
=769mm2
The distance from interction point of bent-up bar and axis to the center of support x c′=1630−(1000
2
−(36+35.8×2)−0.5×18.4)=1246.8mm
The calculation mode of other arrow is the same with the cond arrow. And the results are placed in the sheet.
Since the main steel bars have been t, the corresponding calculation of normal ction carrying capacity can be done and the results are in the following sheet.
fulcrum
ction
Figure 2Envelope diagram for bending moment and resistance bending moment moment
•
And the preliminary design of bent-up steel bar is shown in the diagram as well as the diagram of resistance bending moment and the envelope of bending moment. Plot parallel lines by reprenting carrying capacity of flexural member M ui and substitute each M ui into the interction of parallels and envelope of bending moment which is named as i ,j,…,q(theatrical points ). Substitute each M ui into
M d,x=M d,l2⁄(1−4x2 L
)
and got distance from i ,j,…,q mid-span ction x i.
Check that if initial position of bent-up steel bar is qualified for the requirements by each point.
(a)For first arrow steel bar (2N3)
The abscissa of fully utilized point k is x=2702mm, and the abscissa of bent-up point 1 is x1=6300−824=5476, so the point 1 is left to the point l and x1−x= 5476−2702=2774mm>ℎ02⁄(=9462⁄=473mm)
The abscissa of non-utilized point l is x=4794mm, while the abscissa of the interction 1′of 2N3 bar and axis is x1′(=6300−414=5886mm)>x(= 4794mm).
Hence, the position of bent-up point 1 is qualified.
(b)For cond arrow steel bar (2N2)
The abscissa of fully utilized point j is x=1885mm, and the abscissa of bent-up point 2 is x2=6300−1630=4670, so the point 2 is left to the point j and x2−x= 4670−1885=2785mm>ℎ02⁄(=9282⁄=464mm)
The abscissa of non-utilized point k is x=2702mm, while the abscissa of the interction 2′of 2N2 bar and axis is x2′(=6300−1247=5053mm)>x(= 2702mm).
Hence, the position of bent-up point 2 is qualified.
(c)For third arrow steel bar (2N1)
The abscissa of fully utilized point I is x=0mm, and the abscissa of bent-up point 3
