Video from my You tube channel

Subscribe to My You tube channel

Example 3: design of rectangular beam

design rectangular sections for the beams, loads, and ρ values shown. Beam weights are not included
in the loads shown.  Assume concrete weighs 150 lb/ft3 , fy= 60,000 psi, and fc'= 4000 psi, unless given otherwise.

Figure 1








first, we will calculate the ultimate load, we have point live load and distributed dead load, therefore will take each load separately

ultimate live load=1.6*PL=1.6*20=32k

assuming the weight of beam=0.50k/ft

dead load=2+0.5=2.5 k/ft

ultimate dead load=1.2*2.5=3.00k/ft

Mu from moment diagram in figure 1=657.5k-ft=7,890,000lb-in

                                                                               Figure 1



Mu=Ф*Mn=Ф*Asfy(d-a/2)    

Mu/(b*d^2)=Ф*ρ*fy(1-fy*ρ/1.7*fc')    

ρ=0.18*4,000/60,000=0.012

bd^2=Mu/(Ф*ρ*fy(1-fy*ρ/1.7*fc')   assuming Ф=0.90

Mu=7,890,000lb-in

bd^2=7,890,000/(0.9*.012*60,000*(1-60,000*0.012/1.7*4000))=13,617.88

taking bd=    16x29.5
          

using 16*33 as beam dimension (bxh)

checking beam weight=16*33*150/144=0.55k/ft    not ok

then calculating ultimate dead load again

dead load=2+0.55=2.55 k/ft

ultimate dead load=1.2*2.55=3.06k/ft

Mu=664.25 k-ft=7,971,000lb-in

bd^2=7,971,000/(0.9*.012*60,000*(1-60,000*0.012/1.7*4000))=13,757.61

bd=    16x29.5 is ok 

As=ρ*b*d=0.012*16*29.5=5.664in2

 using 4#11=4*1.56=6.24 in2   ok


checking for Ф=0.9 or no

(εt+0.003)/(d)=0.003/c

C=a/β1

β1=0.85 for fc'=4000 psi or less

 a=Asfy/(0.85*fc'*b)

a=6.24*60,000/(0.85*4000*16)=6.88in

C=a/β1=6.88/0.85=8.1in

εt=0.003*(29.5)/8.1-0.003=0.0079>0.005 so section is tension control Ф=0.9  is ok

checking for ρmin

As,min=(3√fc'/fy)*bw*d and not less than 200bw*d/fy

As,min=3*√4000*16*29.5/60,000  and not less than 200*29.5*16/60,000

As,min=1.49 in2 and not less than 1.57in2 ok, our As is larger than the minimum 

checking Ф*Mn>Mu

Ф*Mn=0.9*6.24*60,000(d-a/2)


Ф*Mn=0.9*6.24*60,000(29.5-6.88/2)=8,781,177.6lb-in> Mu ok 





Comments

Popular posts from this blog

Field density test-sand cone method

Example 1: Design of one-way slab

Determinate and indeterminate structure

Zero force member for truss

Pile cap

Tributary area(Loading)

Prime coat

Flakiness Index and Elongation Index of Coarse Aggregates

Punching shear

Working piles (cast in situ piles)