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2nd civil shear
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Eng. Ahmed Ali
Faculty of Eng., Helwan university
Design fundamentals of R.C structures
Page No: 1
Steps for check shear in RC beams.
1- Draw shear force diagram.
2- Calculate (Qu) at critical section at distance (��+ ��) from center line of column.
3- Calculate (qu) :
qu = �u�� (kg/cm
2)
where:
(Qu) shear force at critical section.
(b) the width of the beam.
(d) depth of the beam.
4- calculate the (qumax)
qumax = 2.2�cuƔc (kg/cm2
)
if (qumax < qu) increase dimensions
if (qumax > qu) ok
5- calculate the (qcu):
qcu = 0.75�cuƔc (kg/cm2
)
if (qcu ≥ qu) safe use min. stirrups (5 ɸ8/m)
use stirrups or bent bars.
if (qcu < qu) or
use stirrups only.
Eng. Ahmed Ali
Faculty of Eng., Helwan university
Design fundamentals of R.C structures
Page No: 2
Calculate (qus):
qus = qu-�cu� (kg/cm2)
if use stirrups:
qus=
�st(�y Ɣs� )��
where:
Ast= n * Ab
Ab = area of one branch of stirrups
b= beam width
n= number of branches (if b˂40cm …. n=2)(if b≥40cm…n=4)
S= spacing between stirrups (20 ≥ S ≥ 10 cm)
Assume Ab then get s
No. of stirrups in meter = ����
if use bent bars:
using one raw from bent bars
qus=
�st(�y Ɣs� )�� sin �
where:
Ast= n * Ab
Ab = area of one bent bar
b= beam width
n= number of bent bars in one raw.
d= depth of beams
α= angle of bent bars
Eng. Ahmed Ali
Faculty of Eng., Helwan university
Design fundamentals of R.C structures
Page No: 3
using 2 raw or more from bent bars
qus=
�st(�y Ɣs� )�� (sin � + cos �)
where:
Ast= n * Ab
Ab = area of one bent bar
b= beam width
n= number of bent bars in one raw.
S= spacing between bent bars
α= angle of bent bars
6- Draw shear force diagram to scale
-calculate (Qcu= qcu b d)
The calculated stirrups is provided in the distance y
The min. stirrups is provided in the distance x
Eng. Ahmed Ali
Faculty of Eng., Helwan university
Problem No.(2).
For the shown beams It is required to:
Calculate the Max. Load (W
(b=12 cm, t=40cm, ts=12cm)
From flexure (Moment) capacity
B= min. Of � �∗���� + 12 "16 ∗ 12 + 12 "B = 72 cm
Assume (a<ts)
0.0.67 ∗ 2501
a
Design fundamentals of R.C structures
Page No: 4
SSShhheeeeeettt NNNOOO...(((111)))
For the shown beams It is required to:
Calculate the Max. Load (Wmax) according to given data
=12cm)
From flexure (Moment) capacity.
" 72()" 204()+
.67 ∗ ,cuƔc ∗ - ∗ . " /s ∗ ,yƔs 2501.5 ∗ 72 ∗ . " (3 ∗ 2) ∗ 36001.15
a = 2.43 cm < ts ok
Design fundamentals of R.C structures
) according to given data
Eng. Ahmed Ali
Faculty of Eng., Helwan university
a = 2.43 cm < 0.1 d at this case ( j =0.826)
1u " (3 ∗ 2) ∗ 36001u2
From shear capacity.
(2 + 32 " 0.2 + 0.175 " 0.3754 " 1.51.1255u4 " 6∗7�
Design fundamentals of R.C structures
Page No: 5
5ɸ8/m
case(1)
may be used as a min.stirrups
qu= qcu
may be calculated to resist shear
qu= q
2.43 cm < 0.1 d at this case ( j =0.826)
/s " 1u,y ∗ 8 ∗ 3
) 3600 ∗ 0.826 ∗ 35 " 624456(1u " 6.24(:.)) u " 2;�8 " 6.24 " 2 ∗ 3�8
2m " 5.55(:/))<<.(1)
shear capacity.
375)
Design fundamentals of R.C structures
case(2)
may be calculated to resist shear
= qus+ 0.5*qcu
(>?. ())
Eng. Ahmed Ali
Faculty of Eng., Helwan university
Design fundamentals of R.C structures
Page No: 6
(Case 1) @u ≤ @cu @u " 5uB ∗ 3 " 0.75 ∗ C,cuƔc
@u " �u��∗�� " 0.75 ∗ ����.� <.. 5u " 4.066: 4 " �.��.��� ∗ 4.066 " 5.42: <.. 4 " 5.42: 2 " �∗D7 " �∗�.E�� " 3.61:/)′ <<2s1 " 3.61:/)′<<<<(2)
(Case 2) @u " @us + 0.5 ∗ @cu @us " 2 ∗ /b ∗ (,yƔs)H ∗ B " 2 ∗ 0.5 ∗ (36001.15 )20 ∗ 12 " 8.69>?/()2
@u " 8.69 + 0.75 ∗ 2501.52 " 13.53>?/()2
@u " �u��∗�� " 13.53>?/()2 <.. 5u " 5.685: 4 " �.��.��� ∗ 5.685 " 7.52: <.. 4 " 7.52: 2 " �∗D7 " �∗J.��� " 5.05:/)′ <<2s2 " 5.05:/)′<<<<(3)
Choose the max. value of load from shear capacity because the two
cases are safe (Ws=5.05 t)
Choose the min. value of load from shear and moment capacity because
the beam will failure first in the small value (W=5.05 t).
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