Eng. Ahmed Ali
Design fundamentals of R.C structures
Faculty of Eng., Helwan university
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
2 (kg/cm )
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
cu
2 (kg/cm )
Ɣc
if (qumax < qu) increase dimensions if (qumax > qu) ok 5- calculate the (qcu):
cu
qcu = 0.75
Ɣ
c
2 (kg/cm )
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
Design fundamentals of R.C structures
Faculty of Eng., Helwan university
Page No: 2
Calculate (qus):
cu
qus = qu-
if use stirrups: qus=
st( y Ć&#x201D;s)
(kg/cm2)
where: Ast= n * Ab
Ab = area of one branch of stirrups b= beam width n= number of branches (if bË&#x201A;40cm â&#x20AC;Ś. n=2)(if bâ&#x2030;Ľ40cmâ&#x20AC;Śn=4) S= spacing between stirrups (20 â&#x2030;Ľ S â&#x2030;Ľ 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 Ć&#x201D;s)
where: Ast= n * Ab
sin
Ab = area of one bent bar b= beam width n= number of bent bars in one raw. d= depth of beams Îą= angle of bent bars
Eng. Ahmed Ali
Design fundamentals of R.C structures
Faculty of Eng., Helwan university
Page No: 3
using 2 raw or more from bent bars qus= where: Ast= n * Ab
st( y Ć&#x201D;s)
(sin + cos )
Ab = area of one bent bar b= beam width n= number of bent bars in one raw. S= spacing between bent bars Îą= 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
Design fundamentals of R.C structures
Faculty of Eng., Helwan university
Page No: 4
Sheet NO.(1) Problem No.(2). For the shown beams It is required to: Calculate the Max. Load (Wmax) according to given data (b=12 cm, t=40cm, ts=12cm)
From flexure (Moment) capacity capacity. + 12 " 72 () + 16 ∗ 12 + 12 " 204 ()
B= min. Of B = 72 cm
∗
Assume (a<ts) 0..67 ∗
,cu ,y ∗ - ∗ . " /s ∗ Ɣc Ɣs
250 3600 0.67 ∗ ∗ 72 ∗ . " (3 ∗ 2) ∗ 1 1.5 1.15 a = 2.43 cm
< ts
ok
Eng. Ahmed Ali
Design fundamentals of R.C structures
Faculty of Eng., Helwan university
Page No: 5
1u ,y ∗ 8 ∗ 3 1u " (3 ∗ 2) ∗ 3600 ∗ 0.826 ∗ 35 " 624456 ((>?. ()) 1u " 6.24 (:. )) 2 ; 2 ∗ 3 1u " " 6.24 " 8 8 2m " 5.55 (:/))<<.(1)
a = 2.43 cm < 0.1 d
at this case ( j =0.826)
/s "
From shear capacity. ( 3 + " 0.2 + 0.175 " 0.375 375 ) 2 2 4"
4"
1.5 5u 1.125
6∗7
5ɸ8/m
case(1)
case(2)
may be used as a min.stirrups
may be calculated to resist shear
qu= qcu
qu= qus+ 0.5*qcu
Eng. Ahmed Ali
Design fundamentals of R.C structures
Faculty of Eng., Helwan university
Page No: 6
@u ≤ @cu
(Case 1)
5u ,cu " 0.75 ∗ C B∗3 Ɣc
@u "
u
@u "
4"
∗ .
.
2"
∗D 7
(Case 2)
" 0.75 ∗
.
∗ 4.066 " 5.42 :
"
∗ .E
<..
5u " 4.066 :
<..
4 " 5.42 :
" 3.61 :/)′ <<2s1 " 3.61 :/)′<<<<(2)
@u " @us + 0.5 ∗ @cu
,y 2 ∗ /b ∗ ( ) 2 ∗ 0.5 ∗ (3600) Ɣs 1.15 " 8.69 >?/()2 @us " " H∗B 20 ∗ 12 250 0.75 ∗
1.5 @u " 8.69 + " 13.53 >?/()2 2 u @u " ∗ " 13.53 >?/()2 <..
4"
.
.
2"
∗D 7
∗ 5.685 " 7.52 :
"
∗J.
<..
5u " 5.685 : 4 " 7.52 :
" 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).