Uncracked section

fc=My/I

Ft=Mcr*c/I Steel Area:

fs=Mcr*yt*n/I (n-1)A s= in2

Mcr=fr*Ig/yt

Cracked Section

n =Es/Ec

K=-nρs+√ ¿s+(nρs)2¿ Es =29000

T=Asfs Es=57500√ fc '

C=fcb(kd)/2 ρs =As/bd

Moment Arm=jd c =kd

M=Tjd=cjd

Ms=Tjd=Asfsjd stress block depth

Mc=cjd=fcb(kd)/2jd=Rbd2 J =1-k/3

R =1/2*f c kj

Alternative

f s =MYt*n/I

I/fc =Mc/I f c =Ms/Mc*fc

Design of Unreinforced Concrete section

Allowable tensile stress of concrete, ft’ =6M/bh2

Hreq=√ 6 M /bft '

Design of cracked Elastic section

K=n/n+r n =Es/Ec

T=Asfs r =fs/fc =Allowable stress ratio

C=fcb(kd)/2 fs =0.40 fy

Mc=Rbd2 Fc =0.45 fc
dreq=√ M /Rb J =1-k/3

Ms=Tjd =Asfsjd

As=M/jd

Design of Singly Reinforced beams (WSD) (without Sw)

1. Mmay=wL2/8 or from structural

Analysis calculate all small values
2. Assume b= in n=
3. hreq=√ (Mmax ¿¿ Rb)¿ f c=
4. hreq=dreq+cover f s=
5. As=Mmax/fsall*jd j=
K=
r=

R=

Design of Singly reinforced beam(wsd) (with Sw)

1. Assume 1.calculate all small values

b= in n=
h= in f c=
2. Calculate self hlt from assumed value f s=
Sw=b*h*0.15/144= k/ft r=
3. Calculate total load on beam. K=
W=SW+Wo j=
4. Mmax=wL2/8 r=
5. dreq=√ Mmax/ Rb R=
6. hreq=dreq+covers
7. follow steps 2 to 6 untill our dreq and

hreq value matches with our last assumed value.

8. As=Mmax/fs*jd

Doubly Reinforced Beam (Wsd) (Basics)

dreq>dpro,so Applied moment > Mc
Applied moment is divided into two parts
M1=Mc=Rbd2=As1(fs jd)
As1=M/fsjd [this will resist M1 moment]
M2=M-Mc
M2 will be resisted by Additional tensile steel and compression stell (A s’)
M2=As2fs (d-d’)=As’fs’(d-d’)

d’
A s’

A s2
d

As2=M2/fs(d-d’)
Total tensile reinforcement, As= As1+As2
Compression Reinforcement, As’=M2/fs (d-d’)
Original, fs’=fs(k-d’/d)/(1-k)
ACI recommends fs’=2fs (k-d’/d)/(1-k)=fs

Design of Doubly Reinforced Beam(wsd)

1.calculate Beam self weight from
given Beam Dimensions. 1.Calculate all small values
SW=((b*h)/144)*0.15 n=
2.Total w=SW+Wo fc=
3.Calculate,Mmax=(wL2)/8 f s=
4.Assume, d r=
One layer = 2”(cv) k=
Two layer = 4” (cv) j=
For two layer, d=h-4”=16” R=
5.Calculate,M1=Rbd2
6.Calculate,M2=Mmax-M1 ***Based on layer ,we have to select d’
7.Calculate,As=As1+As2 & then we have to calculate A s & As’

8.Calculate,fs’
9.As’=M2/fs’(d-d’)
 (parabolic stress block)

Analysis of singly reinforced beam(USD)

1st Method

1. Calculate Mn 1.calculate α,β,ρb,Asb

Mn=Asfy(d-ßc)=….k’ 2.calculate ρs < ρb
3.calculate c=A sfy/αfc’b*

2nd Method

1.calculate,M n 1.calculate α,ß,ρ b,Asb

Mn=αfc’bc(d-ßc)=---k’ ρb=(αfc’/fy){87/(87+fy)

2.calculate,fs A sb=ρb*b*d

fs=87(d-c)/c 2.calculate ρ s=As/bd

3.m=Es*ε u/αfc’

4.mρ s=

5.c=d[-(mρ s/2)+√ {(mρ s /2) 2+mρs}]

Rectangular stress block

Analysis singly reinforced Beam (USD)

1st method

1.calculate Mn 1.calculate Calculate α,ß, ρ b,Asb

Mn=Asfy(d-a/2) 2.calculate ρ s

Or, 3.calculate,a

Mn=ρsfy(1-0.59ρsfy/fc’)/bd2 a=A sfy/0.85fc’b

2nd method

1.calculate Mn 1.calculate α,ß,ρ b,Asb

Mn=0.85fc’ab(d-a/2) 2.calculate ρ s

3.calculate m,mρ s
 Calculate c & a

C=d[-(mρ s/2)=√ ( m2ρ s ) +mρ }]

2
s

A=(α/0.85)*c

Over load & resistance factor

Mu<ΦMn Φ=1.4(DL)

Vu<ΦVn =1.7(LL & W)

Pu<ΦPn =1.87(EQ)

=0.90(S)

=0.85(S)

=0.70(A)

=0.75(A)

Example

Given,Mn=177.8k’ Mu=0.90*Mn=160k’