2nd civil shear

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



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



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


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


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


) according to given data

Eng. Ahmed Ali


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�


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)


case(2)

may be calculated to resist shear

= qus+ 0.5*qcu

(>?. ())

Eng. Ahmed Ali



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).

2nd civil shear

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