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Example 5: allowable uniform load can beam support using transformed area method

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Using transformed area, what allowable uniform load can this beam support in addition to its own
weight for a 28-ft simple span? Concrete weight = 150 lb/ft3 ,fs= 24,000 psi, and fc= 1800 psi.






Figure 1             


Transform area=As*n
As=1.27*5=6.35in4
Transform area=6.35*8=50.8in2

Location of neutral axis by equalling moment of area around neutral axis as shown in figure 2(assuming neutral axis below the void)

X(20)(X/2)-8*8*(x/2-4)=50.8*(29-X)
10X^2-32X+256=1473.2-50.8X
10X^2+18.8X-1217.2=0
X=10.3, assumpitation is not valid

Assuming X in void area as shown in figure 3


X(20)(X/2)-8*(X-4)*(x-4)/2=50.8*(29-X)
10X^2-4(X^2-4X-4X+16)-1473.2+50.8X=0
10X^2-4X^2+32X-64-1473.2+50.8X=0
6X^2+82.8X-1409.2=0
X=9.91in

now we will calculate moment of inertia using parallel axis theorem 

I=20*(9.91^3)/12+20*9.91*4.955^2-((8*5.91^3)/12+5.91*8*2.955^2)+50.8*19.09^2
I=24450.76in4=1.18ft4




Figure 2

Figure 3

the moment for extreme compression fiber for y=9.91in
M=fcI/y
M=1800*24450.76/9.91
=4441106.76lb-in
=370.09k-ft
from the moment we can calculate the load 
M=WL^2/8
W=M*8/L^2
W=370.09*8/(28^2)=3.776K

the moment at the center of reinforcing rebar for y=19.09in

fs=n*M*y/I
M=fs*I/(n*y)
M=24000*24450.76/(19.09*8)
=3842445.259lb-in
=320.20k-ft

M=WL^2/8
W=M*8/L^2
W=320.2*8/(28^2)=3.27K

we will select the lowest load from above W=3.27K(including beam own weight)
W without beam own weight=3.27-0.15*(32*20-8*8)/144=2.67k/ft









Comments

  1. struturalAugust 22, 2020 at 11:19 PM

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  2. AnonymousAugust 25, 2020 at 9:27 AM

    you have error : x not equal x = 9.91 , your -64-1473.2 = -1537.2 , you can also easily calc moment of inertia as : bh^3/3+bh^/3+As*D^2 , Saudi Eng bypass from here. good luck mohammed.

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