Pre Stressed Concrete Solved Problems

Prestressed Concrete Practice Problems

1. A Prescon cable, 18.00 m long is to be tensioned from one end to an initial prestressed of 1040 MPa immediately after transfer. Assume that there is no slack in the cable, that the shrinkage of concrete is 0.0002 at the time of transfer, and that the average compression in concrete is 5.50 MPa along the length of tendon. E c = 26.2 GPa; Es = 200 GPa. Compute the length of shims required, neglecting any elastic shortening of the shims and any friction along the tendon. Ans: 100.98mm

Fig. 1

length of shims elongation of steel

shortening of concrete

end of beam after transfer

end of beam before transfer

18.00 m

Solution: Elasttic elongation of steel: ∆s =

f s L 1040 (18 x10 3 ) = = 93 .6mm Es 200 x10 3

Shortening of concrete due to shrinkage:

∆c

shrinkage

= 0.0002(18x10 3 ) = 3.6mm

Elastic shortening of concrete: δD =

ML2 (18.093)(10 2 ) x1012 = = 3.61 mm 8 EI 8(27.5 x103 )(2.278x109 )

Length of shims required:

∆T = ∆ s

elastic

+ ∆c

+ ∆c

shrinkage

elastic

= 93.6 + 3.6 + 3.78 = 100.98mm

2. A pretensioned member has a section shown 200mmx300mm. It is concentrically prestressed with 516mm2 of high tensile steel wire which is anchored to the bulkheads of a unit stress of 1040 MPa. Assuming n = 6, compute the stresses in the concrete and steel immediately after transfer. Ans: fc = 8.575 MPa; fy = 988.55 MPa Qi

Qi

300

200

Fig. 2

Solution: Exact Method fc =

Qo 516 x1040 = = 8.575 MPa Ac + ( n − 1) As ( 200 x300 ) + ( 6 − 1) 516

nf c = (6)8.575 = 51 .45 MPa

1

G. P. Ancog

Prestressed Concrete Practice Problems

Stress in steel after transfer f s = f so − nf c = 1040 − 51 .45 = 988 .55 MPa

Approximate Method The loss of prestress in steel due to elastic shortening of concrete is approximated by: fs = n

Qo 516 x1040 = (6) = 53 .664 MPa Ag 200 x300

Stress in steel after loss f s = f so − nf c = 1040 − 53 .664 = 986 .335 MPa

Stress in concrete is: fc =

net stress of steel x As Q 986 .335 x516 = net = = 8.482 MPa Ag Ag ( 200 x300 )

Approximations introduced: 1. using gross area instead of net area 2. using initial stress in steel instead of the reduced stress 3. A pretensioned member has a section 200mmx300mm. It is eccentrically prestressed with 516mm2 of high tensile steel wire which is anchored to the bulkheads at a unit stress of 1040 MPa. The c.g.s. is 100mm above the bottom fiber. Assuming n = 6, compute the stresses in the concrete immediately after transfer. Ans: fT = 0.00 MPa; fB = +16.918 MPa

initial cgc

cgc e

300

200

Fig. 3

cT final cgc

cB

cgs 100

Beam Section

ys

Transformed section

Solution Exact Method ( n −1) As = (6 −1)( 516 ) = 2580 mm 2 Ag = 200 x300 = 60000 mm 2 e = (300 / 2) −100 = 50 mm

Summing up moment at initial cgc: AT y o = A1 y1 + A2 y 2 yo =

A1 y1 + A2 y 2 (200 x300 )( 0) + 2580 (50 ) = = 2.06 mm AT 60000 + 2580

c B = (300 / 2 − y o ) = 147 .94 mm cT = (300 / 2 + y o ) = 152 .06 mm e = c B −100 = 47 .94 mm

Compute transformed section moment of inertia:

2

G. P. Ancog

Prestressed Concrete Practice Problems

1 bh 3 + Ag ( y o ) 2 + (n − 1) As y s 12 200 x300 3 = + 60000 ( 2.06 ) 2 + 2580 ( 47 .94 ) 2 = 4.562 x10 8 mm 4 12

IT =

Fiber stresses: f =

Qi Q ey 516 x1040 (516 x1040 )( 47 .94 ) y ± i = ± AT IT 60000 + 2580 4.562 x10 8

= 8.575 ± 0.056393 y

Top fiber stress:

f T = 8.575 − 0.056393 (152 .06 ) = 0.00 MPa

Bottom fiber stress:

f B = 8.575 + 0.056393 (147 .94 ) = 16 .918 MPa

Approximate Method Loss of prestress: f sL =

nQ i 6(516 x1040 ) = = 53 .664 MPa Ag 60000

Net prestress: f sn = f si − f sL = 1040 − 53 .664 = 986 .336 MPa Qnet = f sn As = 986 .336 (516 x10 −3 = 508 .949 kN

Fiber stresses: f =

Qnet Q ey ± net Ag Ic

508 .949 x10 3 508 .949 x10 3 (50 ) y ± 60000 200 (300 ) 3 12 = 8.48248 ± 0.0565498 y =

Top fiber stress:

f T = 8.48248 − 0.0565498 (150 ) = 0.00 MPa

Bottom fiber stress:

f B = 8.48248 + 0.0565498 (150 ) = 16 .964 MPa

Approximation introduce: 1. using approximate values of reduced prestressed 2. using the gross area of concrete 4. A post-tensioned beam has a mid span cross-section with a duct of 50mm x 75mm to house the wires. It is pretensioned with 516mm2 of steel to an initial stress of 1040 MPa. Immediately after transfer, the stress is reduced by 5% owing to anchorage loss and elastic shortening of concrete. Compute the stresses in the concrete at transfer. Ans: fT = 4.829 MPa, fB = +23.913 MPa 200

300

Fig. 4 yo 50x75 cgs

75

Beam Section

cgs

75

Solution Method 1: Using net section of concrete Ac = Ag − Aduct = 200 x300 − 50 x75 = 56250 mm 2

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G. P. Ancog

Prestressed Concrete Practice Problems

Locate the cg of net section: yo =

Aduct (75 ) (50 x 75 )( 75 ) = = 5.00 mm Anet 56250

y s = 75 + y o = 75 + 5 = 80 mm cT = 150 − y o = 150 − 5 = 145 mm c B = 150 + y o = 150 + 5 = 155 mm

Compute the moment of inertia of net section: bh 3 b ' h' 3 + bh ( y o ) 2 − − b' h' (80 ) 2 12 12 200 x300 3 50 x 75 3 = + 60000 (5) 2 − − 3750 (80 ) 2 = 4.527 x10 8 mm 4 12 12

I =

Total prestress in steel: Q = η( As f s ) = 95 %( 516 x1040 ) x10 −3 = 509 .808 kN Fiber stresses: f =

Q (Qe ) y 509 .808 x10 3 509 .808 x10 3 (80 ) ± = ± y Ac I 56250 4.257 x10 8

= 9.063 ± 0.095806 y

Top fiber stress:

f T = 9.063 − 0.095806 (145 ) = −4.828 MPa

Bottom fiber stress:

f B = 9.063 + 0.095806 (155 ) = 23 .913 MPa

Method 2: Using gross section of concrete Q Qec 509 .808 x10 3 509 .808 x10 3 (75 )(150 ) ± = ± 1 Ag I 200 x300 ( 200 x300 3 ) 12 = 8.4968 ±12 .7452

f =

Top fiber stress:

f T = − 4.2484 MPa

Bottom fiber stress:

f B = 21 .242 MPa

If eccentricity does not occur along one of the principal axes of the section, it is necessary to further resolved the moment into two components along the two principal axes. f =

Q Qe x y Qe y x ± ± A Ix Iy

5. A post-tensioned bonded concrete beam has a prestress of 1560 kN in the steel immediately after prestressing which eventually reduces to 1330 kN. The beam carries two live loads of 45 kN each in addition to its own weight of 4.40 kN/m. Compute the extreme fiber stresses at mid-span: a) under the initial condition with full prestress and no live load b) under final condition after all the losses have taken place and with full live load. Ans: Initial condition: fT = 2.234 MPa, fB = 15.10 MPa; Final condition: fT = 13.803 MPa, fB = 0.975 MPa 45kN 4.50m

45kN 3.00m

4.50m 300

4

600

Fig. 5

G. P. Ancog 175

Mid span section

Prestressed Concrete Practice Problems

Solution To be theoretically exact, net concrete section should be used up to the time of grouting, after which the transformed section should be considered. Section Properties: Ag = bh = 300 x 600 = 180000 mm 2 Ig =

1 1 bh 3 = (300 x 600 3 ) = 5.4 x10 9 mm 4 12 12

Initial condition M =

f =

wL 2 4.4 x12 2 = = 79 .2 kN − m 8 8

Qo Q ey My 1560 x10 3 1560 x10 3 (125 )( 300 ) 79 .2 x10 6 (300 ) ± o ± = ± ± Ag Ig Ig 180000 5.4 x10 9 5.4 x10 9

= 8.667 ±10 .833 ± 4.40

Top fiber stress: f T = 8.667 −10 .833 + 4.4 = 2.234 MPa

Bottom fiber stress: f B = 8.667 + 10 .833 − 4.4 = 15 .10 MPa

Final condition Live load moment at mid-span: M L = Pa = 45 ( 4.5) = 202 .5 kN − m

Dead load moment at mid-span: MD =

wL 2 4.4(12 2 ) = = 79 .2 kN − m 8 8

Total moment: MT = 79.20 + 202.5 = 281.7 kN-m Stresses: f =

Q Qey M T y ± ± Ag Ig Ig

1330 x10 3 1330 x10 3 (125 )( 300 ) 281 .7 x10 6 (300 ) ± ± 180 x10 3 5.4 x10 9 5.4 x10 9 = 7.389 ± 9.236 ±15 .65

f =

5

G. P. Ancog

Prestressed Concrete Practice Problems

Top fiber stress: f T = 7.389 − 9.236 + 15 .65 = 13 .803 MPa

Bottom fiber stress: f B = 7.298 + 9.236 −15 .65 = 0.975 MPa

Note: For pre-tensioned beam, steel is always bonded to the concrete before any external moment is applied. Values of A, y and I should be computed on the basis of transformed section. For approximation, gross area of concrete can be used in calculation. For post-tensioned and bonded beams, for any load applied after the bonding has taken place, transformed section should be used. For post-tensioned unbonded beams, the net concrete section is the proper one for all stress calculation.

45 kN 4.50 m

1.50 m

13.8202 MPa β 

C α

T 0.976 MPa

C = T, M = Cα = Tα M M 281 .7 x10 6 = = = 211 .8 mm C T 1330 x10 3 β = α − e = 211 .8 −125 = 86 .8 mm

α=

Stresses: C Cβy ± A I 1330 x10 3 1330 x10 3 (86 .8)( 300 ) = ± 180 x10 3 5.4 x10 9 = 7.389 ± 6.413

f =

Top fiber stress: f T = 7.389 + 6.413 = 13 .802 MPa

Bottom fiber stress: f B = 7.389 − 6.413 = 0.976 MPa

Computation of average strain for unbonded beams:

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G. P. Ancog

Prestressed Concrete Practice Problems

ε=

f My = E Ec I

∆=∫

My dx Ec I

εave =

1 L

My dx cI

∫E

Average stress in steel is: MyE s n My f s = E s ε ave = ∫ dx = ∫ dx LE c I L I 6. A post-tensioned simple beam on a span of 12 m carries a superimposed load of 11.00 kN/m in addition to its own weight of 4.40 kN/m. The initial prestress in the steel is 950 MPa, reducing to 830 MPa after deducting all loses and assuming no bending of the beam. The parabolic cable has an area of 1612.9mm2, n = 6. Compute the stresses in the steel at mid-span assuming: a) the steel is bonded by grouting b) the steel is unbonded and entirely free to slip. Ans: Bonded condition: fs = 845.258 MPa, Unbonded condition: fs = 838.137 MPa wT = 11.00+4.40=15.40 kN/m









  

 

Section properties: A = bh = 300x600 = 180000.00 mm2 I = bh3/12 = 300(600)3/12 = 5.4x109 mm4 c = h/2 = 600/2 = 300 mm

Mo

M

x Parabolic moment diagram x y

yo

Parabolic y diagram

Solution 1: Moment at mid-span: Mo =

wL 2 15 .4(12 ) 2 = = 277 .2kN − m 8 8

Moment at mid-span due to prestress: M s = Qe = (1612 .9 x830 )(125 ) x10 −6 = 167 .34 kN − m

Net moment at mid-span: MN = 277.2 – 167.34 = 109.86 kN-m Stress in concrete at the level of steel due to bending: 7

G. P. Ancog

Prestressed Concrete Practice Problems

Using I for gross section My 109 .86 x10 6 (125 ) fc = = = 2.543 MPa I 5.4 x10 9

The stress in steel is increased by: f s = nf c = 6( 2.543 ) = 15 .258 MPa

Resultant stress in steel: fsf = 830 + 15.258 = 845.258 MPa Solution 2: If the cable is unbonded and free to slip. n My dx L∫ I   x 2  M = M o 1 −  L 2         fs =

  x  y = y o 1 −  L 2     

n fs = LI

2

    2

  x 2  ∫−L 2 M o y o 1 −  L 2   dx   L 2

L/2

nM o y o  2 x3 x5  = + x −  LI 3 ( L 2 ) 2 5( L 2 ) 2  −L 2  =

8 nM o y o 8 (15 .258 ) = 8.137 MPa = 15 I 15

Resultant stress in steel: fsf = 830 + 8.137 = 838.137 MPa Ultimate strength analysis:

.85fc’

ε

a/2

a = β 1c

C

c

= 0.0034

c d

z T

ε

s

C = .85 f c ' ba = T = As f s ' a=

As f s ' C = .85 f c ' b .85 f c ' b

z = d −a 2 M = As f s ' ( d − a 2 )

7. A rectangular section 300mm x 600mm deep is prestressed with 937.5 mm 2 of steel wires for an initial stress of 1040 MPa. The cgs of the wires is 100mm above the bottom fiber. For the tendons, fs’ = 1650 MPa, fc’ = 34.4 MPa. Determine the ultimate resisting moment. Ans: Mu = 637.05 kN-m 8

G. P. Ancog

Prestressed Concrete Practice Problems

Solution Total tension of steel at rupture T = 937 .5(1650 ) x10 −3 =1546 .875 kN C =T .85 f c ' ba = T a=

T 1546 .875 x10 3 = =176 .34 mm .85 f c ' b .85 (34 .4)( 300 )

Ultimate moment M u = As f su ( d − a 2 )

= 1546 .875 ( 500 −176 .34 2 ) x10 −3 = 637 .05 kN − m

8. A post-tensioned bonded beam with a transfer prestress of Ft = 1560 kN is being wrongly picked up at its mid-span point. Compute the critical fiber stresses. If the top fiber cracks and the concrete is assume to take no tension, compute the bottom fiber stresses. If the beam is picked up suddenly so that an impact factor of 100% is considered compute the maximum stresses. Ans: Case 1: fT = -6.566 MPa, fB = 23.90 MPa; Case 2: fB = 27.905 MPa; Case 3: fB = 35.39 MPa  











Solution Section properties: A = 300 x 600 = 180 x10 3 mm 2 1 I = (300 )( 600 ) 3 = 5.4 x10 9 mm 4 12 600 c= = 300 mm 2

External moment at pick-up point M =−

wL 2 4.4(6) 2 =− = −79 .2 kN − m 2 2

a) Fiber stress at mid-span Q Qey My ± ± A I I 3 1560 x10 1560 x10 3 (125 )300 79 .2 x10 6 (300 ) = ± ± 180 x10 3 5.4 x10 9 5.4 x10 9 = 8.667 ± 10 .833 ± 4.4 f T = 8.667 −10 .833 − 4.4 = −6.566 MPa f =

f B = 8.667 +10 .833 + 4.4 = 23 .9 MPa

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G. P. Ancog

Prestressed Concrete Practice Problems

b) If the fiber cracks and concrete is assume to take no tension. e = 50.77 Q C

372.69

124.23

Triangular Stress Block

Locate center of pressure, C: M = Qe e=

M 79 .2 x10 6 = = 50 .77 mm Q 1560 x10 3

From bottom: 175-50.77 = 124.23 mm Assuming a triangular stress block, height y: y = 3(124 .23 ) = 372 .69 mm 1 T = C = f c by 2 2T 2(1560 x10 3 ) fc = = = 27 .905 MPa by 300 (372 .69 )

c) 100% impact factor M T = M +100 % M = 2 M = 2(79 .2) = 158 .4 MPa MT 158 .4 x10 6 = = 101 .538 mm Q 1560 x10 3 From bottom : 175 −101 .54 = 73 .46 mm Assu min g a triangular stress block : y = 3(73 .46 ) = 220 .38 mm

e=

2T 2(1560 x10 3 ) = = 47 .19 MPa by 300 ( 220 .38 ) Assu min g a rec tan gular stress block : fc =

y = 2(73 .46 ) = 146 .92 mm T = C = f c by fc =

T 1560 x10 3 = = 35 .39 MPa by 300 (146 .92 )

9. Determine the total dead and live uniform load moment that can be carried by the beam with a simple span of 12m:1. for zero tensile stress in the bottom fibers. 2. for cracking in the bottom fibers at a modulus of rupture of 4.13 MPa and assuming concrete to take up tension up to that value. Ans: Case 1: wT = 16.21 kN/m; Case 2: wT = 20.34 kN/m 13.853



 

  



  

4.13

18.534

      





      0 291.78 kN-m

10

4.13 74.34 kN-m

4.13 366.12 kN-m G. P. Ancog

Prestressed Concrete Practice Problems

Solution Section properties: A = bh = 300 (600 ) = 180 x10 3 mm 2 1 1 I = bh 3 = (300 )( 600 ) 3 = 5.4 x10 9 mm 4 12 12 h 600 c= = = 300 mm 2 2

Prestress Q: Q = As f s = 1562 .5(830 ) x10 −3 = 1296 .8 kN

1. Moment for zero tensile stress at the bottom: Q Qey My + − =0 A I I 1296 .8 x10 3 1296 .8 x10 3 (125 )( 300 ) Mx 10 6 (300 ) 0= + − 180 x10 3 5.4 x10 9 5.4 x10 9 M = 291 .78 kN − m 8M 8( 291 .78 ) w= 2 = = 16 .21 kN / m L 12 2 fB =

Top fiber stress: Q Qey My − + A I I 3 1296 .8 x10 1296 .8 x10 3 (125 )( 300 ) 281 .78 x10 6 (300 ) = − + 180 x10 3 5.4 x10 9 5.4 x10 9 = 13 .853 MPa

fT =

2. For cracking in the bottom fibers. Additional moment carried by the section up to the beginning of crack. ∆M =

f 'I 4.13 (5.4 x10 9 ) = x10 −6 = 74 .34 kN − m c 300

Total moment capacity: 

M T = M 1 + ∆M = 291 .78 + 74 .34 = 366 .12 kN − m     w=

   

8M 8(366 .12 ) = = 20 .34 kM / m L2 12 2

 







10. A concrete beam of 10m simple span is post-tensioned with a 750mm 2 of high tensile steel to an initial prestress of 965 MPa immediately after prestressing. Compute the initial deflection at the mid-span due to-3 prestress Q = 965(750)x10 = 723.75 kNand the beam’s own weight assuming Ec = 27.5 GPa. Estimate the deflection after 3 mos. Assuming creep coefficient of cc = 723.75x25 1.8 and an due effective prestress of 830 MPa at that time. If the beam carry a 45 kN Moment concentrated to prestressload applied at mid-span when the beam is 3 mos. old after prestressing, 723.75x150 what is the deflection at mis-span? Ans: After 3 mos. δ = 0.5579mm upward; When 45 kN is added after 3 mos. δ = 14.407 mm downward. wL2/8

Moment due to beam weight

Moment due to load P

G. P. Ancog

11 PL/4

Prestressed Concrete Practice Problems

Solution Section properties: A = bh = 300 x 450 = 135 x10 3 mm 2 I =

1 1 bh 3 = (300 )( 450 ) 3 = 2.278 x10 9 mm 4 12 12

The parabolic tendon with 150mm mid-ordinate is replaced by a uniform load acting along the beam. wP L2 8 8Qh 8(723 .75 )(150 ) x10 −3 wP = 2 = = 8.685 kN / m L 10 2

Qh =

Moment due to eccentric load at the end of the beam M = Qe ' = 723 .75 ( 25 ) x10 −3 = 18 .093 kN − m

Dead load uniform load wD = γA = 23 .5(300 x 450 ) x10 −6 = 3.17 kN / m

Net uniform load: ∆w = wQ − wD = 8.685 − 3.17 = 5.515 kN / m

Upward deflection at mid-span due to net uniform load: δU =

5(∆w) L4 5(5.515 )(10 4 ) x10 12 = = 11 .462 mm 384 EI 384 (27 .5 x10 3 )( 2.278 x10 9 )

Downward deflection at mid-span due to end moment: 12

G. P. Ancog

Prestressed Concrete Practice Problems

ML 2 (18 .093 )(10 2 ) x10 12 = = 3.61 mm 8 EI 8(27 .5 x10 3 )( 2.278 x10 9 )

δD =

Initial deflection due to pretsress and beam weight: δ net = δU − δ D = 11 .462 − 3.61 = 7.852 mm , upward

Deflection due to prestress alone:

δP =

5(8.685 )(10 4 ) 18 .093 (10 2 )  5( wP ) L4 ML 2 10 12 − = − = 14 .44 mm  3 9 384 EI 8 EI 384 8   ( 27 .5 x10 )( 2.278 x10 )

Deflection due to dead load alone:

δ DL =

5wDL L4 5(3.17 )(10 4 ) x10 12 = = 6.59 mm 384 EI 384 (27 .5 x10 3 )( 2.278 x10 9 )

The initial deflection is modified by two factors: 1. loss of prestress 2. creep effect which tend to increase deflection Deflection after 3 months:  f  830 δ f = δ P  s  + δ DL ( c c ) = 14 .44 − 6.59(1.8) = 0.5579 mm , upward 965  f so 

Deflection due to applied concentrated load of 45kN: δ LL =

PL 3 45 (10 3 ) x10 12 = = 14 .965 mm , downward 48 EI 48 (27 .5 x10 3 )( 2.278 x10 9 )

The resultant deflection: δ R = δ LL + δ f = 14 .965 −.5579 = 14 .407 mm , downward

11.A double cantilever beam is to be designed so that its prestress will exactly balance the total uniform load of 23.5 kN/m on the beam. Design the beam using the least amount of prestressed assuming that the cgs must have a concrete protection of 75 mm. If a concentrated load P = 65 kN is applied at the mid-span, compute the maximum top and bottom fiber stresses. Ans: F = 1410 kN; fT = 14.934 MPa, fB = -2.40 MPa 

       





 





13

G. P. Ancog

Prestressed Concrete Practice Problems

Solution Section properties: A = bh = 300 x 750 = 225 x10 3 mm 2 I =

1 1 bh 3 = (300 )( 750 ) 3 = 1.0546875 x10 10 mm 4 12 12

In order to balance the load on the cantilever, the cgs at the tip must coincide with the cgc with a horizontal tangent. To use the least amount of pretsress, the eccentricity over the support should be a maximum. Assume a gross cover of 75mm, emax = 750/2-75 = 300 mm. 65 kN w = 23.5 kN/m 300

The prestress required: 6.00 m

Qe =

h

15.00 m

e

750

e

6.00 m

wL 2 wL 2 23 .5(6 2 ) ;Q= = = 1410 kN 2 2e 2(300 x10 −3 )

In order to balance the load at the mid-span, using the same prestress Q, the sag of the parabola must be: Qh =

wL 2 wL 2 23 .5(15 2 ) ;h= = x10 −3 = 468 .75 mm 8 8Q 8(1410 )

The result will be a concordant cable and under the action of the uniform load and prestress, the beam will have no deflection any where and will only have a uniform compressive stress. fc =

Q 1410 x10 3 = = 6.267 MPa A 225 x10 3

Due to concentrated load P: M =

PL 65 (15 ) = = 243 .75 kN − m 4 4

The extreme fiber stresses: Mc 243 .75 x10 6 (375 ) = = 8.667 MPa I 1.0546875 x10 10 Q Mc fT = + = 6.267 + 8.667 = 14 .934 MPa A I Q Mc fB = − = 6.267 − 8.667 = −2.4 MPa A I f =

12. A hollow member is reinforced with 4 wires of 62.5 mm 2 each pretensioned fsi = 1030 MPa. If fc’ = fci = 34.4 MPa, n = 7, determine the stresses when the wires are cut between members. Determine the moment that can be carried at a maximum tension of 0.5√(fc’) and a maximum of fc = 0.45fc’. If 240 MPa of the prestressed is lost (in addition

14

G. P. Ancog

Prestressed Concrete Practice Problems

to the elastic deformation) determine this limiting moment. Ans: when the wires are cut, fs = 936.98 MPa; Limiting moment, MT = 9.665 kN-m 

200

    

100

200

(n-1)As = (7-1)(62.5)





100







= 375 mm2

open

Transformed Section

Solution Transformed section: AT = 200 x 200 − 100 x100 + 4( n − 1) As = 31 .5 x10 3 mm 2 IT =

[

]

1 200 4 − 100 4 + 4( n − 1) As (70 2 ) = 1.3235 x10 8 mm 4 12

Initial prestressing force, Qi before transfer:

Qi = Ast f si = ( 4 x 62 .5)(1030 ) x10 −3 = 257 .5 kN fc =

Qi 257 .5 x10 3 = = 8.175 MPa AT 31 .5 x10 3

∆f s = nf c = 7(8.175 ) = 57 .225 MPa

Net stresses right after transfer (loss due to elastic shortening): f c = 8.175 MPa f so = f si − nf c = 1030 − 57 .225 = 972 .775 MPa

Allowable concrete stresses: f c = 0.45 f c ' = 0.45 (34 .4) =15 .48 MPa f t = 0.5

f c ' = 0.5 34 .4 = 2.93 MPa

Total moment : MT = MD+ML Additional concrete stress on top:

∆ f t c = 15.48 − 8.175 = 7.305 MPa compression Additional concrete stress on botton: ∆f b = 2.93 + 8.175 = 11 .105 MPa tension

Total moment that can be carried: f =

∆f I MT c 7.305 (1.323 x10 8 ) ;MT = t = x10 −6 = 9.665 kN − m I c 100

Concrete stress on top reach full allowable limit: 15

G. P. Ancog

Prestressed Concrete Practice Problems

f T = f c = 15 .48 MPa compression

Concrete stress at the bottom: f B = 8.175 − ∆f t = 8.175 − 7.305 = 0.87 MPa compression

8.175 MPa

15.48 MPa

=

-2.93 MPa

Allowable value of stress

7.305 MPa

8.175 MPa

-11.105 MPa

Initial concrete stress

Additional concrete stress

concrete stress at the level of steel

Net stress in steel:

 70  f sn = f so ± nf cs = 972 .775 ± 7 7.305 = 972 .775 ± 35.7945 .1135  100 

Top steel: f snT = 972 .775 − 35 .7945 = 936 .98 MPa

Bottom steel: f snB = 972 .775 + 35 .7945 = 1008 .5695 MPa

After 240 MPa of prestress is lost (in addition to elastic deformation) Qi = f snet Ast = (1030 − 240 )( 4 x 62 .5) x10 −3 = 197 .5 kN fc =

Qi  197 .5  = 8.175   = 6.27 MPa AT  257 .5 

f se = f snet − nf c = (1030 − 240 ) − 7(6.27 ) = 746 .11 MPa

Additional concrete stress on top:

∆ f t c = 15.48 − 6.27 = 9.21 MPa compression Additional concrete stress on botton: ∆f b = 2.93 + 6.27 = 9.2 MPa tension

Total moment that can be carried: f =

∆f I MTc 9.2(1.323 x10 8 ) ;MT = t = x10 −6 = 12 .17 kN − m I c 100

16

G. P. Ancog

Prestressed Concrete Practice Problems

6.27 MPa

15.48 MPa

=

-2.93 MPa

Allowable value of stress

9.21 MPa

6.27 MPa

-9.20 MPa

Initial concrete stress

Additional concrete stress

Therefore the limiting moment: M T = [ 9.665 , 12 .17 ] min = 9.665 kN − m

17

G. P. Ancog