Architectural Engineering

Pre Stress Double Tee Design Calculation

Pre Stress Double Tee Design Calculation

Today I have a topic about double Tee design calculation |Pre Stress Double Tee Design Calculation.

A pre-tensioned merely supported 10LGT24 double T-beam while not topping, it’s subjected to an identical superimposed gravity dead-load intensity Wsp and burden W₁.

W=420lb/ft                                  w=Wsd+Wι

fρ¡=189000 psi                            fρ¡≈0.70fpu


Determine the intense fiber stresses at the midspan because of two

1.1- The initial full prestress and no external gravity load

1.2- the ultimate service load conditions once prestress losses have taken place

Allowable stress 

fc = 6000 psi     concrete cylindrical strength

fo=270000 psi

fpy= 220000 psi

fpe= 150000 psi

ft = 12√f’c foot = 930 psi most allowable lastingness at the time of initial stress in conc.

fci = 4800 psi   conc. compressive strength at the time of initial prestress

fci= 2880 psi      fci= zero.6f’ci most allowable stress in concrete at initial prestress

fc = 2700 psi fc=0.45f’c most allowable compressive stress in concrete at service.

n = ten nos      7- wire strand tendons with a 108-Di strand pattern area unit wont to prestress beam

d=12.70 millimetre military intelligence of tendons

d = 0.50 in

Section Properties

L=64 ft

Ay = 1623.7 sq.inch                    y = two.00 neutral axis

A=689sq.inch                             At=240 tension space

Ac =449sq.inch                           ec=14.77 inch

lc=22469 inch4                           ee=7.77inch

r²=5004sq.inch   = lc/ac              Sb=126inch³

Cb=17.77inch                              St=3607inch³

Ct=6.23inch                                Wd=359lb/ft

122489 inf

Initial Conditions at Prestressing

A = 1.53 sq.inch

Pi=289170 pound

Pe = 229500 pound

Maximum moment at mid-span

Md= 2205696 inch-lb

M= wl²/8

ft = Pi/Ac(1+ec/r²)+Md/Sb foot=  Pi/Ac(1+ec/r²)-Mt/St


Fb=Pi/Ac(1+ec/r²)+Md/Sb         fb= Pi/Ac(1+ec/r²)-Mt/Sb


fb= -2277 psi

fb 2277 psi    <  2880psi allowed.

2. Final condition at Service Load, Basic methodology.

Midspan moment because of superimposed dead and loading has

M= 2580480 inch-lb Msd+Ml=Wl²/8

Mt= 4786176 inch-lb

f’ = -898 psi ft=pe/Ac(1-ec/r²)-Mr/St

f’=-898psi < fc=0.45f’c 2700psi

fb= 594psi T fb=-pe/Ac(1+ecb/r²)+Mt/Sb

fb=594psi < ft=12√f’c 930psi

3. Final service load condition by the line-of-thrust, C-Line methodology.


Pe =229500 pound

Mt= 4786176 inch-lb

a = 20.85 inch        a=Pe/Mt

e’=6.08inch            e’=a-ec

f’=-898psi              ft= pe/Ac(1-e’ct/r²)

fb=592       T          fb=-pe/Ac(1-e’ct/r²)


4. Final service load condition when losses victimization the Load-Balancing methodology

At middle spen

Pt= Pe= 229500 pound

a=ec= 14.77 inch

For balance load

W₂ = 552 lb/ft

Thus, if the whole gravity load would are 552 lb/ft solely the axial load P’/A would act as if the beam had a parabolically drooped sinew with no eccentricity at the supports. this can be a result of the gravity load being balanced by the sinew at the mid-span. therefore

Total load w/c beam = 779 lb/ft Wd + (WED + WL)

Unbalanced load, Wub = 227 lb/ft

unbalanced moment Mub= 1394688 inch-lb.

f= -898 psi C ft=Pt/Ac-Mub/St

fb = 592 psi T fb= PtAc+Mub/Sb

fb=592 psi < 930 psi enable


Learn More

Bar Bending Schedule For Pipe Sleeper

Difference B/w Flexible and Rigid Pavements


Thanks for Reading Article Get more Information and share it with others.




Raja Numan

Hi, My name is Engr. Raja Numan author of Engineering Information Hub and I am a Civil Engineer by Profession and I've specialized in the field of Quantity Surveying, Land Surveying as QC Engineer in national and multinational companies of Pakistan & Saudi Arabia.

Related Articles

Leave a Reply

Your email address will not be published.

Back to top button
error: Content is protected !!