### Example 7: design of inverted T-beam

Select reinforcing bars for the beam shown if Mu= 250 ft-k, fy= 60,000 psi, and fc'= 4000 psi. (Hint: Assume that the distance from the c.g. of the tensile steel to the c.g. of the compression block equals 0.9 times the effective depth, d, of the beam.) After a steel area is computed, check the assumed distance and revise the steel area if necessary. Is εt ≥0.005?

Figure 1

Mu=ФMn

Mu=ФAsfy*(d-a/2)

d=24in

a will be equal to 4.80in if we assumed the distance between the center of compression and tensile force=0.9*d=21.6in

a=4.8

assuming Ф=0.9

Mu=250k-ft=3,000,0000lb-in

3,000,0000=0.9*60,000*As*(24-4.8/2)

3,000,0000=54,000As*(24-3.52As/2)

3,000,0000=1,296,000As-95,040As^2

47,520As^2-1,296,000As+3,000,000=0

As=2.95 in2

checking if the value of a is correct

a=As*fy/(0.85*fc'*b)

a=2.95*60,000/(0.85*4000*5)

a=10.40 in

so our assumption is wrong, we will calculate steel area again assuming a=10.4in

for a=10.40in, compression stress will be distributed over the first 6in of a beam with a width of 5in and another 4.40in of a beam with 15in width, for this case, we will have two blocks of compression stress as shown in figure 3

the force in first bloack=0.85*fc'*5*6=102,000lb

the force in second block=0.85*fc'*15*(10.4-6)=224,400lb

total compression force=326,400lb

locating the center of compressive force, taking the top of the beam as a reference line

x=(102,000*3+224,000*8.2)/326,400=6.575in

the moment arm between tensile and compressive force=24-6.56=17.44in

Mu=09*As*fy*moment arm

3,000,0000=0.9*60,000*As*17.425

As=3.18in2

checking a

As*fy=0.85*fc'*6*5+0.85*fc'*(a-6)*15

a=7.74in

so our assumption is wrong, we will calculate steel area again assuming a=7.74in

for a=10.40in, compression stress will be distributed over the first 6in of a beam with a width of 5in and another 1.74in of a beam with 15in width, for this case, we will have two blocks of compression stress as shown in figure 3

the force in first bloack=0.85*fc'*5*6=102,000lb

the force in second block=0.85*fc'*15*(7.74-6)=88,740lb

total compression force=190,740lb

x=(102,000*3+88740*6.87)/190,740=4.80in

the moment arm between tensile and compressive force=24-4.8=19.2in

Mu=09*As*fy*moment arm

3,000,0000=0.9*60,000*As*19.2

As=2.89in2

checking a

As*fy=0.85*fc'*6*5+0.85*fc'*(a-6)*15

a=7.4in very close so ok

(εt+0.003)/(d)=0.003/c

C=a/β1

β1=0.85 for fc'=4000 psi or less

a=7.40in

C=a/β1=7.40/0.85=8.70in

εt=0.003*(24)/8.7-0.003=0.00>0.0053 so section is tension control Ф=0.9 is ok

Figure 2

Figure 3

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