Sample Problems in Pre-Stressed Concrete PDF [PDF]

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PS1-Balitayo-Eejay-D. A rectangular beam of cross section 350 mm deep and 250 mm wide is prestressed by means of 15 wires of 5 mm diameter located 65 mm from the bottom of the beam and 3 wires of diameter of 5 mm, 25 mm from the top. Assuming the prestress in the steel is 840 MPa and the density of concrete is 24KN/m 3 a. Calculate the stresses at the extreme fibers of the mid-span section then the beam is supporting its own weight over a span of 6m.

y=

(15 𝑥 65 ) + (3 𝑥 25) 18

y = 108.33 mm Eccentricity = c – y e = 175 – 108.33 e = 66.67 mm Prestessing Force P = 840 (18)(19.7) P = 3x105 N Area of Cross-Section A = 350(250) A = 8.75 x104 mm2 Moment of Area

I=

250(350)3 12

=89.32X107 mm2

Section Modulus S=

89.32𝑥107 175

= 5.10x106 mm3

Self – weight Beam weight = 24(0.25)(0.35) = 2.1 KN/m Mweight =

𝑤𝑙2 8

=

2.1(6)2 8

Due to Fweight =

= 9.45 KN.m

𝑀𝑐 9.45𝑥106 (175) 250(350)3 𝐼

= 1.85 MPa

12

Due to Prestress f=

𝑃 𝐴

=

3𝑥105 8.75𝑥104

= 3.43MPa

Bending Stress due to Prestress f=

𝑃𝑒𝑐 𝐼

=

3𝑥105 (66.67)(175) 250(350)3 12

= 3.91 MPa

ftop = -3.43 + 3.91 – 1.85 = -1.41 (compression) fbot = -3.43 – 3.91 + 1.85 = 2.33 (tension)

b. Construct the stress diagram for question A

-3.43

-3.43

+ 3.91

-1.85

- 3.91

+1.85

-1.41

+2.33

PS-Abelardo-Dan-Angelo-L. An 8m long cantilever T-beam shown supports a uniformly distributed load of 8kN/m including its own weight throughout its length. To prevent excessive deflection of the beam, it is pre-stressed with 16mm diameter strands causing a pre-stress force of 650kN.

a. Calculate the Moment of Inertia of Section. b. Determine the resulting stress at the top fiber of the beam at the fixed end if the center of gravity of the strands is 75mm above the neutral axis of the section. c. Determine the eccentricity of the pre-stress force at the fixed end such that the resulting stress at the top fiber of the beam at the fixed end is zero. a.

Calculate the Moment of Inertia of Section. I = I + Ad2A2 = (500)(100) = 50000mm2

30000mm2

A1 = (300)(100) = 1 100 I = (300)(1003 ) + 30000( + 500 − 362.5)2 12

1

+ 12

2

(100)(5003 ) + 50000(362.5 − 250)2

ATӮ = ƩAӮ I = 27.541 x 108 mm4 80000Ӯ = 30000(550) + 50000(250) Ӯ = 362.5mm from bottom fiber b. Resulting stress at the top fiber of the beam at the fixed end if the center of gravity of the strands is 75mm above the neutral axis of the section. M = WL(L/2) M = 8(8)(4) M = 256kNm Due to bending cause by load of 8kn/m ftop = ftop =

𝑀𝑐

Due to Pre-stressing and Moment 𝑃

𝐼 256(600−362.5)(106 ) 27.541 x 108

ftop = 22.076 Mpa

ftop = − 𝐴 − ftop = −

𝑀𝑐

𝐼 650(1000) 80000

−

650(.75)(600−362.5)(1000) 27.541 x 108

ftop = -8.167 Mpa

Stress at top fiber of the beam ftop = 22.076 – 8.167 ftop = 13.91 Mpa (tension) c. The eccentricity of the pre-stress force at the fixed end such that the resulting stress at the top fiber of the beam at the fixed end is zero. M = 256knm ftop due to moment Due to Pre-stressing force ftop = ftop =

𝑀𝑐 𝐼 256(600−362.5)(106 ) 27.541 x 108

ftop = 22.076 Mpa

M = 650(103)e 𝑃

ftop = − 𝐴 − ftop = −

𝑀𝑐

𝐼 650(1000) 80000

−

650(𝑒)(600−362.5)(1000)

ftop = -8.125 – 0.056e ftop = 8.125 + 0.056e Assume Tension is equal to Compression to produce zero stress

27.541 x 108

22.076 = 8.125 + .056e .056e = 13.95 e = 248.89 mm

PS-Aguilar-Mary-Jane-S.C. A 300 mm x 400 mm concrete beam has a span of 6 m. a posttension force of 640 kN was applied at a point 70 mm above the bottom of the beam. Assume concrete wont crack in tension. f’c = 20.7 MPa. Unit weight of concrete is 23.5 kN/m3. a. Compute the deflection due to pre stressing force of 240 kN. b. Compute the net deflection of the beam immediately after transfer. c. Compute the safe uniform live load that maybe imposed on the beam so that there will be a net deflection upward of 5 mm. Solution: a. Deflection due to pre stressing force of 240 kN e = 200-70 e = 130 mm M = Pe M = 640(0.13) M = 83.2 kN.m ᵟ1 = I=

𝑀𝐿2

8𝐸𝐼 300(400)2 12

I = 1600x106 E = 4700√f’c E = 4700√20.7 E = 21384 MPa ᵟ1 =

𝑀𝐿2 8𝐸𝐼 83.2𝑥106(6000)2

ᵟ1 = 8(21384)(1600)106 ᵟ1 = 10.94 mm (upward) b. Net deflection of the beam immediately after transfer. W = 0.3(0.4)(23.5)(6) W = 2.82 kN/m 5𝑊𝐿2

ᵟ2 = 384𝐸𝐼 5(2820)(6)(6000)2

ᵟ2 = 384(21384)(1600)106 ᵟ2 = 1.39 mm Net deflection ᵟ = ᵟ1 - ᵟ2 ᵟ = 10.94 - 1.39 ᵟ = 9.55 mm (upward) c. Safe uniform live load that maybe imposed on the beam so that there will be a net deflection upward of 5 mm. 9.55 - ᵟ4 = 5 mm ᵟ4 = 4.55 mm 5𝑊𝐿2

ᵟ4 = 384𝐸𝐼

5𝑊(1000)(6)(6000)2

4.55 = 384(21384)(1600)106

W = 9.23 kN/m

PS-Andres-Charlene-D. A beam with width b = 300 mm and depth d = 600 mm is to be prestressed. Considering a 15% prestress loss, compute the value of initial prestressing force P and eccentricity e. A) If the compressive stress is 21 MPa. B) If the compressive stress at the bottom fiber is 12 MPa and a tensile stress at the fiber is 2 MPa. C) If he compressive stress at the top fiber is 16 MPa and zero at the bottom fiber.

Solution: A) b = 300 mm

15% loss

d = 600 mm

Pἱ =? e =? P = 0.855Pἱ --- equation 1 e=0

F=

𝑃

P

𝐴

-21 = −

P

𝑃 300(600)

P = 3780 KN Substitute P to equation 1 3780 = 0.85 Pἱ Pἱ = 4447.06 KN, e = 0 (ANSWER)

B) Top: 2=

−𝑃 300(600)

6𝑃𝑒

+ 300(600)2

Bottom: -12 =

, P = 900 KN e = 140

−𝑃

900 = 0.85 Pἱ Pἱ = 1059 KN (ANSWER)

6𝑃𝑒

-300(600) 300(600)2

C) Top: -16 =

−𝑃

6𝑃𝑒

300(600) 300(600)2

Bottom: 0=

, P = 1440 KN e = 100

−𝑃 300(600)

6𝑃𝑒

+ 300(600)2

1440 = 0.85 Pἱ = 1694 KN (ANSWER)

PS-Bayle-Georgia-Venice-V. A beam with width b = 250 mm and depth d = 450 mm is pre stressed by an initial force of 600 kN. Total loss of pre stress at service loads is 15%. 1. Calculate the resulting final compressive stress if the pre stressing force is applied at the centroid of the beam section. 2. Calculate the final compressive stress if the pre stressing force is applied at an eccentricity of 100 mm below the centroid of the beam section. 3. Calculate the eccentricity at which the pre stressing force can be applied so that the resulting tensile stress at the top fiber of the beam is zero. Solution: 1. Final compressive stress if the pre stressing force is applied at the centroid of the beam section.

𝑃 𝐴

𝜕= =

600(0.85)(1000) 250(450)

𝑀𝐴 = −4.53 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

2. Final compression stress if the pre stressing force is applied at an eccentricity of 100 mm below the centroid of the beam section.

𝑓𝑡 =

𝑃 𝑃𝑒 𝐶 + 𝐴 𝐼

𝑓𝑡 =

𝑃 6𝑃𝑒 + 𝑏𝑑 𝑏𝑑 2

𝑓𝑡 =

600000(0.85) 6(600000)(0.85)(100) − 250(450) 250(450)2

𝑓𝑡 = −10.58 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

3. Eccentricity at which the pre stressing force can be applied so that the resulting tensile stress at the top fiber of beam is zero.

𝑓𝑡 = 0=

𝑃 6𝑃𝑒 + 𝑏𝑑 𝑏𝑑 2 −600000 6(600000)𝑒 + 250(450) 250(450)2

600000 6(600000)𝑒 = 250(450) 250(450)2 𝑒 = 75𝑚𝑚

PS-Binuya-Anna-E. A pre stressed concrete simple beam having a cross section of 300mmx600mm

is

subjected

to a

pre

stressing

force

in

steel

which

eventually reduce to 1200KN due to losses acting on a parabolic tendons. The beam carries a total uniform distributed load of 20 KN/m including beam’s weight. Using three methods calculate the total stresses at mid span section of the beam. Given: B D P W E E L

= = = = @ @ = 

300mm 600mm 1200KN 20 KN/m mid span = 50 mm end section = 0 mm 10 mm Stress calculation

𝑃

𝐹𝑡𝑜𝑝 = − 𝐴 +

6𝑃𝑒 𝑏𝑑 2

6𝑀

− 𝑏𝑑2 (20)(102 )

𝐹𝑡𝑜𝑝 =

6( )(106 ) 1200 (103) (6)1200 (103 )50 8 − 300 𝑥 600 + 300 𝑥 6002 − 300 𝑥 6002

𝐹𝑡𝑜𝑝 = -17.222 MPa 𝑃

𝐹𝑏𝑜𝑡 = − 𝐴 − 𝐹𝑏𝑜𝑡 = −

6𝑃𝑒 𝑏𝑑 2

6𝑀

+ 𝑏𝑑2

1200 (103) 300 𝑥 600

−

(6)1200 (103 )50 300 𝑥 6002

+

𝐹𝑏𝑜𝑡 = 3.889 MPa



Internal Resisting Couple Method

(𝑤)(𝐿2 ) 𝑀= 8 (20)(102 ) 𝑀= 8

(20)(102 ) )(106 ) 8 300 𝑥 6002

6(

𝑀 = 250 𝐾𝑁. 𝑚 𝑀 = 𝑃ℎ 250(106 ) = 1200 (103 )ℎ ℎ = 208.333 𝑚𝑚 208.333 = 50 + 𝑒′ 𝑒′ = 158.333 mm 𝑃

𝐹𝑡𝑜𝑝 = − 𝐴 + 𝐹𝑡𝑜𝑝 = −

6𝑃𝑒′ 𝑏𝑑 2

1200 (103) (6)1200 (103 )158.333 + 300 𝑥 600 300 𝑥 6002

𝐹𝑡𝑜𝑝 = -17.222 MPa

𝑃

6𝑃𝑒′ 𝑏𝑑 2 1200 (103)

𝐹𝑏𝑜𝑡 = − 𝐴 − 𝐹𝑏𝑜𝑡 = −

300 𝑥 600

−

(6)1200 (103 )158.333

𝐹𝑏𝑜𝑡 = 3.889 MPa 

Load Balancing

𝑤 ′ = 𝑤 − 𝑤𝑝 8𝑃𝑒 𝑤𝑝 = 2 𝐿 𝑤𝑝 =

8(1200) (0.05) (10)2

𝑤𝑝 = 4.80

𝐾𝑁 𝑚

300 𝑥 6002

𝑤 ′ = 20 − 4.80 = 15.20 KN/m 𝑃

6𝑀

𝐹𝑡𝑜𝑝 = − 𝐴 − 𝑏𝑑2 (15.20)(102)

𝐹𝑡𝑜𝑝 =

6( )(106) 1200 (103) 8 − 300 𝑥 600 − 300 𝑥 6002

𝐹𝑡𝑜𝑝 = -17.222 MPa 𝑃

6𝑀

𝐹𝑏𝑜𝑡 = − 𝐴 + 𝑏𝑑2 𝐹𝑏𝑜𝑡 = −

1200 (103) 300 𝑥 600

+

𝐹𝑏𝑜𝑡 = 3.889 MPa

(15.20)(102 ) )(106) 8 300 𝑥 6002

6(

PS-BulananHectorM. A pre-stressed concrete beam, 250 mm wide by 600 mm deep, supports a uniformly distributed dead load of 8.5 kN/m in addition to its weight throughout its 10 meter span. It is pre-tensioned with steel tendons placed 100 mm from the bottom of the beam and subjected to an initial prestress force of 1880 kN. The beam is used to support a live load of 25 kN/m. Use weight of concrete 24 kN/𝑚3 . 1. Determine the resulting temporary stress at the bottom fiber of the beam at midspan immediately after the transfer of prestress. 2. Determine the stress at the bottom fiber of the beam at midspan after allowance for 25% losses at final service loads. 3. Determine the stress at the top of the beam at the supports after allowance for 25% losses at final service loads. Solution: 1.

L=10 m 𝑃

𝑓𝑏𝑜𝑡 = – 𝐴 −

600 200

6𝑃𝑒 𝑏𝑑2

1880(103 ) 6(1880)(103 )(200) 𝑓𝑏𝑜𝑡 = – − 250(600) 250(6002 ) 𝑓𝑏𝑜𝑡 = – 37.6 𝑀𝑃𝑎

P

100 250

2.

𝑤 = 25 + 8.5 + 0.25(0.60)(24) 𝑘𝑁 𝑤 = 37.1 𝑚 𝑀= 𝑀=

𝑤𝐿2 8 37.1(10)2 8

𝑀 = 463.75 𝑘𝑁. 𝑚 𝑃

6𝑃𝑒

𝐴

𝑏𝑑2

𝑓𝑏𝑜𝑡 = – 0.75 − 0.75 𝑓𝑏𝑜𝑡 = – 0.75

1880(103) 250(600)

+

− 0.75

6𝑀 𝑏𝑑2 6(1880)(103)(200) 250(6002 )

𝑓𝑏𝑜𝑡 = 2.72 𝑀𝑃𝑎

3. 𝑃

6𝑃𝑒

𝐴

𝑏𝑑2

𝑓𝑡𝑜𝑝 = – 0.75 − 0.75 𝑓𝑡𝑜𝑝 = – 0.75

1880(103 ) 250(600)

𝑓𝑡𝑜𝑝 = 9.4 𝑀𝑃𝑎

+ 0.75

6(1880)(103)(200) 250(6002 )

+

6(463.75)(106 ) 250(600)2

PS-CONSTANTINO-AILENE-P. 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.

SOLUTION: Section Properties: 𝐴 = 𝑏ℎ = 300(600) = 180000 𝑚𝑚2 𝐼=

𝑏ℎ3 (300)(600)3 = = 5.4𝑥109 𝑚𝑚4 12 12

(a) Initial Condition: 𝑀=

𝑤𝐿2 4.4(12)2 = = 79.2 𝑘𝑁. 𝑚 8 8

𝑓=−

𝑃 𝑃𝑒𝑐 𝑀𝑐 ± ± 𝐴 𝐼 𝐼

𝑓=−

1560𝑥103 1560𝑥103 (125)(300) 79.2𝑥106 (300) ± ± 180000 5.4𝑥109 5.4𝑥109

𝑓 = −8.667 ± 10.833 ± 4.40 𝑓𝑡 = −8.667 − 10.833 + 4.40 = −15.10 𝑀𝑃𝑎 𝑓𝑏 = −8.667 + 10.833 − 4.40 = −2.434 𝑀𝑃𝑎

(b) Final Condition: Moment at mid-span

𝑀𝐿𝐿 = 𝑃𝑎 = 45(4.5) = 202.5 𝑘𝑁. 𝑚 𝑀𝐷𝐿 =

𝑤𝐿2 4.4(12)2 = = 79.2 𝑘𝑁. 𝑚 8 8

𝑀𝑇 = 79.20 + 202.5 = 281.7 𝑘𝑁. 𝑚 Stresses: 𝑓=−

𝑃 𝑃𝑒𝑐 𝑀𝑇 𝑐 ± ± 𝐴 𝐼 𝐼

𝑓=−

1330𝑥103 1330𝑥103 (125)(300) 281.7𝑥106 ± ± 180000 5.4𝑥10^9 5.4𝑥10^9

𝑓 = −7.389 ± 9.236 ± 15.65 𝑓𝑡 = −7.389 − 9.236 + 15.65 = −0.975 𝑀𝑃𝑎 𝑓𝑏 = −7.298 + 9.236 − 15.65 = −13.803 MPa

PS-De-Leon-Cherrie-s. Problem: The flooring of a warehouse is made up of double tee joists (DT). The joists are simply supported on a span of 7.5m and are pre tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75mm above the bottom fiber, loss of stress service load is 18%. Load imposed on the joists are: Dead load = 2.3 KPa Live load = 6.0 KPa Properties of DT: A = 200,000 𝑚𝑚2 I = 1880x106 𝑚𝑚4 a = 2.4 m Yt = 88 mm Yb = 267 mm

a = 2.4 m Yt = 88 mm N.A. 192 mm Yb = 267 mm 75 mm

a. Compute the stress at the bottom fibers of the DT at mid-span due to the initial prestressing force alone. b. Compute the resulting stress at the bottom fibers of the DT at mid-span due to the service loads and pre stress force. c. What additional super imposed load can DT carry such that the resulting stress at the bottom fiber at mid-span is zero.

Solution: a.) 𝑃 𝑃𝑒𝐶 − 𝐴 𝐼 745(1000)(2) 745(1000)(2)(192)(267) 𝑓𝑏 = − − 200,000 1880𝑥106 𝑓𝑏 = −7.45 − 40.63 𝒇𝒃 = −𝟒𝟖. 𝟎𝟖 𝑴𝑷𝒂 𝑓𝑏 = −

b.) 𝑤 = 2.3(2.4) + 6(2.4) 𝑘𝑁 𝑤 = 19.92 𝑚 𝑀=

𝑤𝐿2 19.92 (7.5)2 = = 140.06 𝑘𝑁. 𝑚 8 8

𝑃 = 0.82(745)(2) = 1221.8 𝑘𝑁

𝑃 𝑃𝑒𝐶 𝑀𝐶 − + 𝐴 𝐼 𝐼 1221.8(1000) 1221800(192)(267) 140.06𝑥106 (267) 𝑓𝑏 = − − + 200,000 1880𝑥106 18880𝑥106 𝑓𝑏 = −6.11 − 33.32 + 19.89 𝒇𝒃 = −𝟏𝟗. 𝟓𝟒 𝑴𝑷𝒂 𝑓𝑏 = −

c.) ; 𝑓𝑏 = 19.54 𝑀𝑃𝑎 𝑓𝑏 =

𝑀𝐶 𝐼

19.54 =

𝑀(267) 1880𝑥106

𝑀 = 137.58 𝑘𝑁. 𝑚 𝑀=

𝑤𝐿2 8

𝑤(7.5)2 8 𝒘 = 𝟏𝟗. 𝟓𝟕 𝒌𝑵/𝒎 137.58 =

PS-DIAZ-GERALDINE-DR. A concrete beam with cross-sectional area of 32 x 103 mm2 and the radius of gyration is 72 mm is prestressed by a parabolic cable carrying an effective stress of 1000 N/mm 2. The span of the beam is 8 m. the cable composed of 6 wires of 7 mm diameter has an eccentricity of 50 mm at the center and zero at the supports. Neglecting all losses, find the central deflection of the beam as follows: a) Self-weight + prestress b) Self-weight + prestress + live load of 2 kN/m

Given: A = 32 x 103 mm2 E = 38 kN/mm2 DC = 24 kN/mm3 I = 72 mm L = 8 m or 8000 mm e = 50 mm Solution: 2

I = Ai = 32 x 103 (72) = 166 x 106 mm4 Prestressing force P = (6)(38.5)(1000) = 231000 N or 231 kN Self- weight, g =

32 x 103 106

(24)

g = 0.77 kN/m g = 0.00077 kN/mm Downward deflection due to self- weight =

5𝑔𝐿4 384 𝐸𝐼

5(0.00077)(8000)4

= 384(38)(166 𝑋 106 ) = 6.5 mm

Upward deflection dur to prestressing force =

5𝑃𝑒𝐿4 48 𝐸𝐼 5(231)(50)(8000)4

=

48(38)(166 𝑋 106 )

= 12.2 mm

Downward deflection due to live load =

Downward deflection due to self− weight Self− weight

(2)

6.5

= 0.77 (2) = 16.9 mm

a) Self-weight + prestress = 12.2 – 6.5 = 5.7 mm (↑)

b) Self-weight + prestress + live load of 2 kN/m = 6.5 – 12.2 + 16.9 = 11.2 mm (↓)

PS-Duldulao-Nanette-G. A concrete beam with cross-sectional of 32 𝑥 103 𝑚𝑚2 & the radius of gyration is 𝑁 72 𝑚𝑚 is prestressed by a parabolic cable carrying an effective stresses1000 2 .The 𝑚𝑚

span of the beam is 8 𝑚. The cable, composed of 6 wires of 7𝑚𝑚 diameter, has an eccentricity of 50 𝑚𝑚 at the center & zero at the supports. Neglecting all losses, find the central deflection of the beam as follows: a. Self-weight + prestressed b. Self-weight + prestressed + live load of 2

𝑘𝑁 𝑚

.

Solution: Cross-sectional area of beam, 𝐴 = 32𝑥103 𝑚𝑚2 Modulus of elasticity, 𝐸 = 38

𝑘𝑁 𝑚𝑚2

Dead weight of concrete,𝐷𝐶 = 24𝑘𝑁/𝑚𝑚2 Radius of gyration, 𝑖 = 72𝑚𝑚 Span, 8𝑚 = 8000𝑚𝑚 Eccentricity, 𝑒 = 50𝑚𝑚 𝐼 = 𝐴𝑖 2 = (32𝑥103 𝑥722 ) = 166𝑥106 𝑚𝑚4 𝑃 = (6𝑥38.5𝑥1000) = 23100 𝑁 = 231 𝑘𝑁 32𝑥103 𝑘𝑁 0.00077𝑘𝑁 𝑔=( 𝑥24) = 0.77 = 6 10 𝑚 𝑚𝑚 Downward deflection due to self weight = (

5gL4 5𝑥0.00077𝑥80004 )=( ) 384𝐸𝐼 384𝑥38𝑥166𝑥106

= 6.5 𝑚𝑚 Upward deflection due to prestressing force = (

5𝑃𝑒𝐿4 48𝐸𝐼

)=(

5𝑥231𝑥50𝑥80004 48𝑥38𝑥166𝑥106

)=

12.2𝑚𝑚 Downward deflection due to live load = (

6.5 0.77

x2) = 16.9mm

a. Deflection due to (self-weight + prestress) = (12.2-6.5) = 5.7mm (upward) b. Deflection due to (self-weight + prestress + live load) = 6.5-12.2+16.9) = 11.2 (downward) PS-Embien-Aljomar-C. A prestressed concrete beam, 100 𝑚𝑚 wide and 300 𝑚𝑚 deep, is prestressed by straight, wires carrying an initial force of 150 𝑘𝑁 at an eccentricity of 50 𝑚𝑚. The modulus of elasticity of stell and concrete are 210 and 35 𝑘𝑁/𝑚𝑚2 respectively. Estimate the percentage loss of stress in steel due to elastic deformation of concrete if the area of steel wires is 188 𝑚𝑚2 .

Solution: Force, 𝑃 = 150𝑘𝑁 Area of concrete section, 𝐴 = (100𝑥300) = 3𝑥104 𝑚𝑚 Area of the steel wire = 188 𝑚𝑚2 Section modulus, 𝐼 = 225𝑥106 𝑚𝑚4

𝐸

Modular ratio, 𝛼𝑒 = ( 𝑠 ) = 6 𝐸𝑐

𝑓𝑐 =

Initial stress in steel =

150𝑥103

Stress in concrete, 𝑓𝐶 =

188

3𝑥104

𝑁

= 800

150𝑥103

+

𝑃 𝑃 𝑃𝑒 +( + ) 𝐴 𝐴 𝑍𝑏

𝑚𝑚2

150𝑥103 𝑥50𝑥50 225𝑥106

= 6.66

𝑁 𝑚𝑚2

Loss of stress due to elastic deformation of concrete = 𝑎𝑐 𝑓𝑐 = (6𝑥6.66) = Percentage of loss of stress in steel =

40𝑥100 800

= 5%

40𝑁 𝑚𝑚2

PS-Fernando-Josiah-Abel-V. A Single-T prestressed concrete beam shown below is simply supported having a span of 10 m. It carries a superimposed live load of 15.08 kN/m in addition to the weight of beam. It is prestressed with 700 mm2 of steel to an initial stress of 1034 N/mm 2 located 400 mm from the topmost fiber of the beam section. Immediately after transfer, the stress is reduced by 12 %. Determine the stresses at L/4 from the support due to losses in prestress and final service loads. Use concrete weight equals to 24 kN/m3. Properties of Single-T section AT = 205,000 mm2 I NA = 4.78 mm x 10 9 mm4 y = 303.66 mm Solution: P’= 0.88 x 700 ( 1034 ) = 636.944 kN

C.G.

y

e = 303.66 – 100 = 203.66 mm Weight = 24 (

205000 10002

) = 4.92 kN/m

Total Weight = 4.92 + 15.08 = 20 kN/m

C

𝑥 =

𝐿 10 = = 2.5 𝑚 4 4

𝑅𝐴 = 𝑅𝐵 = 100 𝑘𝑁 𝑀𝑐 = 100 (2.5) − 20 (

2.52 ) 2

𝑀𝑐 = 187.5 𝑘𝑁. 𝑚 𝑃

𝑃𝑒𝑐

𝐴

𝐼

f top = − + f top = −

−

𝑀𝑐 𝐼

187.5 (106 )(196.34) 636.944 (103 ) 636.944 (103 )(203.66)(196.34) + − 205000 4.78 𝑥 109 4.78 𝑥 109

f top = - 5.48 MPa (compression)

f bottom = −

636.944 (103 ) 205000

−

636.944 (103 )(203.66)(303.66)

f bottom = 0.564 MPa (tension)

4.78 𝑥 109

+

187.5 (106 )(303.66) 4.78 𝑥 109

PS-GUILANGROMEO-CESAR-J. ]A prestressed concrete rectangular beam of size 500mm by 750mm has a simple span of 9.0m. The beam is subjected to prestressing force with an effective strength of 1700kN. Tendons are the parabolic set up and have eccentricity of 150mm at midspan. The beam carries a uniform distributed load of 40kN/m including beam weight. 1.) compute the total stresses at midspan section using: a.)Internal Resisting Couple Method, b.)Load Balancing Method .2.) Construct a stress diagram at the midspan section. SOLUTION: 1.)

A. 𝑀 =

𝑊𝐿2 8

=

40(90)2 8

= 405𝑘𝑁. 𝑚

𝑃 𝐴

𝑓=− =±

M = Ph ℎ=

𝑀 𝑃

=

net moment = Ce’ = 1700(0.088235) = 150kN.m

405(10ᶟ) 1700

= 238.235𝑚𝑚

=−

1700(10ᶟ) 500(750)

𝑓 = −4.53 ± 3.2

h = e + e’

Ft = -4.53 – 3.2 = -7.73MPa

e’ = h – e =238.235 – 150 =88.235mm

Fb = -4.53 + 3.2 = -1.33MPa

B. 𝑊𝑝 =

8𝑃𝑒 𝐿2

=

8(1700)(0.15) (9)2

=

25.185𝑘𝑁 𝑚

𝑊 ′ = 𝑊 − 𝑊𝑝 = 40 − 25.185 = 14.185𝑘𝑁/𝑚 𝑀′ =

𝑊′𝐿² 14.185(9)² = = 150𝑘𝑁. 𝑚 8 8

𝑃 6𝑀 1700(10ᶟ) 6(150)(106 ) 𝑓=− =± 2=− ± = 𝐴 𝑏ℎ 500(750) 500(750) 𝑓 = −4.53 ± 3.2 Ft = -4.53 – 3.2 = -7.73MPa Fb = -4.53 + 3.2 = -1.33MPa

2.)

6𝑀 𝑏ℎ2

±

6(150)(106 ) 500(750)

=

PS-LADORES-CARLOS-S.-JR. For the post-tensioned beam with a flanged section as shown, the profile of the CGS is parabolic with no eccentricity at the ends. The live moment due to service loads at midspan is 648 𝑘𝑁. 𝑚. The prestress after transfer is 1600𝑘𝑁. Assume 15% loss at service. Find the location of upper and lower kern points.

SOLUTION: 𝐴1 = (500)(200) 𝐴1 = 100,000 𝑚𝑚2 𝐴2 = (600)(150) 𝐴2 = 90,000 𝑚𝑚2 𝐴3 = (250)(200) 𝐴2 = 50,000 𝑚𝑚2 𝐴 𝑇 = 𝐴1 + 𝐴2 + 𝐴3 𝐴 𝑇 = 240,000 𝑚𝑚2

𝐒𝐨𝐥𝐯𝐞 𝐟𝐨𝐫 ȳ: 240,000ȳ = 100,000(900) + 90,000(500) + 50,000(100) ȳ = 583.3 mm

THEREFORE; 𝑦𝑏 = 583.3𝑚𝑚 𝑦𝑡 = 1000 − 583.3 𝑦𝑡 = 416.7𝑚𝑚 ECCENTRICITY: 𝑒 = ȳ − 150 𝑒 = 583.3 − 150 𝑒 = 433.3 𝑚𝑚 MOMENT OF INERTIA: 1 𝐼 = [ (500)(200)3 + 𝐴1 (900 − 583.3)2 ] + 12 1

[12 (150)(600)3 + 𝐴2 (583.3 − 500)2 ] + 1

[12 (250)(200)3 + 𝐴3 (583.3 − 100)2 ]

𝐼= 2.442𝑥1010 𝑚𝑚4 SQUARE THE RADIUS OF GYRATION 𝐼

𝑟2 = 𝐴 𝑟=

2.552 𝑥 1010 𝑚𝑚4 240,000𝑚𝑚2 5

𝑟 = 1.063 𝑥 10 𝑚𝑚2 𝑟 2 1.063𝑥 105 𝑚𝑚2 𝑘𝑡 = = 𝑦𝑏 583.3𝑚𝑚 𝑘𝑡 = 182.2 𝑚𝑚 𝑟 2 1.063𝑥 105 𝑚𝑚2 𝑘𝑏 = = 𝑦𝑡 416.7𝑚𝑚 𝑘𝑏 = 255.1 𝑚𝑚

PS-LAGRATA-ALBERT-F A beam 400mm x 800mm having a temporary stress at the top fiber at mid span of 2000Pa, support a distributed dead load including its weight of 7 kN/m throughout its 12 meter span and subjected to an initial prestressed force of 1200 kN. The beam is designed to support a live load of 25 kN/m.

a. Determine the eccentricity of steel tendon of the beam. b. Determine the stress at the top fiber of the beam at mid-span at final service load. c. Determine the stress at the bottom of the beam at the support at final service load.

SOLUTION:

a. Eccentricity of steel tendon of the beam

𝑓𝑡𝑜𝑝 = 2000 𝑃𝑎 STRESS DUE TO PRESTRESSED FORCE 𝑃

1200(103)

𝑓𝑃 = − 𝐴 = − (400)(800) = −3.75 𝑀𝑃𝑎 STRESS DUE TO ECCENTRICITY 𝑀 = 𝑃𝑒 = 1200𝑒 𝑓𝑝𝑒 =

6(1200𝑒)(106 ) (400)(8002 )

= 28.125𝑒

𝑓𝑡𝑜𝑝 = − 𝑓𝑃 + 𝑓𝑝𝑒 2 = -3.75 + 28.125e 28.125e = 5.75 e = 0.204m e = 204mm

b. Stress at the top fiber

STRESS DUE TO ECCENTRICITY 𝑀 = 𝑃𝑒 = 1200(0.204) = 244.8 𝑘𝑁. 𝑚 𝑓𝑝𝑒 =

6(244.8)(106) (400)(8002 )

= 5.738 𝑀𝑃𝑎

STRESS DUE TO LIVE LOAD

𝑀=

𝑤𝐿2 8

=

25(12)2 8

= 450 𝑘𝑁. 𝑚

6(450)(106 )

𝑓𝐿𝐿 = (400)((800)2 = 10.547 𝑀𝑃𝑎 STRESS DUE TO DEAD LOAD 𝑀=

𝑤𝐿2 8

=

7(12)2 8

= 126 𝑘𝑁. 𝑚

6(126)(106 )

𝑓𝐷𝐿 = (400)((800)2 = 2.953 𝑀𝑃𝑎 𝑓𝑡𝑜𝑝 = − 𝑓𝑃 + 𝑓𝑝𝑒 − 𝑓𝐿𝐿 − 𝑓𝐷𝐿 𝑓𝑡𝑜𝑝 = −3.75 + 5.738 − 10.547 − 2.953 𝒇𝒕𝒐𝒑 = −𝟏𝟏. 𝟓𝟏𝟐 𝑴𝑷𝒂

c. Stress at the bottom fiber

𝑓𝑏𝑜𝑡𝑡𝑜𝑚 = − 𝑓𝑃 − 𝑓𝑝𝑒 + 𝑓𝐿𝐿 + 𝑓𝐷𝐿 𝑓𝑏𝑜𝑡𝑡𝑜𝑚 = −3.75 − 5.738 + 10.547 + 2.953 𝒇𝒃𝒐𝒕𝒕𝒐𝒎 = 4.012 Mpa

PS-Legaspi-Olsen-A. PROBLEM] A symmetrical I section beam with total depth of 600 mm, moment of inertia of 5.0 x 109 mm4 and concrete area of 113,500 mm2 will be used on a 9.0 m simple span. Assume unit weight of concrete is equal to 24 kN/m3. The beam carries a concentrated live load of 30 kN at midspan. The beam will be pre-tensioned with multiple seven-wire strands below the centroid at constant eccentricity of 200 mm. The initial and final prestressing force is 1000 kN and 800 kN respectively. a. Calculate the stresses at midspan due to initial prestressing and uniformly distributed dead load. b. Calculate the stresses at midspan with full service load in place. c. Calculate the net deflection under service loads if E = 13734 MPa.

SOLUTION: a. fT and fB f=−

P Pec MDLc ± ± A I I Beam weight = 0.1135 x 24 = 2.724 kN/m wL2 2.724(9)2 MDL = = = 27.58 kN ∙ m 8 8

f=−

1000 x 103 1000 x 103 (200)(300) 27.58 x 106 (300) ± ± 113,500 5.0 x 109 5.0 x 109

f = −8.81 ± 12 ± 1.65 fT = −8.81 + 12 − 1.65 = +1.54 MPa fB = −8.81 − 12 + 1.65 = −19.16 MPa b. fT and fB f=−

P Pec MDLc MLLc ± ± ± A I I I MLL =

f=−

PL 30(9) = = 67.50 kN ∙ m 4 4

800 x 103 800 x 103 (200)(300) 27.58 x 106 (300) ± ± 113,500 5.0 x 109 5.0 x 109 ±

67.50 x 106 (300) 5.0 x 109

f = −7.05 ± 9.6 ± 1.65 ± 4.05 fT = −7.05 + 9.6 − 1.65 − 4.05 = −3.15 MPa

fB = −7.05 − 9.6 + 1.65 + 4.05 = −10.95 MPa c. Net deflection Due to prestressing force and eccentricity; yP =

ML2 8EI M = Pe = (800)(0.2) M = 160 kN∙m

160 x 106 (9000)2 yP = 8(13734)(5.0 x 109 ) yP =23.59 mm (Upward) Due to beam weight/dead load yDL = yDL

5wL4 384EI

5(2.724)(9000)4 = 384(13734)(5.0 x 109 )

yDL = 3.39 mm (Downward) Due to concentrated load; yLL =

PL3 48EI

yLL =

30 x 103 (9000)3 48(13734)(5.0 x 109 )

yLL = 6.63 mm (Downward) Then; yN = 23.59 − 3.39 − 6.63 yN = 13.57 mm (Upward) A simply supported beam 300 mm by 700 mm and span of 10 m is prestressed by a straight tendon with a force of 1500 KN at an eccentricity of 200 mm from the centroid. The beam supports a live load of 50 KN at the mid span and a dead load of 30 KN/m including the beam’s weight. A.) Calculate the final stress at the top fiber in MPa. B.) Calculate the final stress at the bottom fiber in MPa. Solution: Stress due to P: 𝑃 𝐴

𝑓𝑝 = − = −

1500 (1000) 300 (700)

𝑓𝑝 = - 7.14 MPa

Stress due to eccentricity “e” 𝑓𝑒 =

6𝑃𝑒 𝑏𝑑 2

=

6 (1500)(200)(1000) 300 (700)2

𝑓𝑒 = ± 12.24 MPa

Stress due to self weight: 𝑤𝐿2 8

𝑓𝑏 =

6𝑀 𝑏𝑑 2

𝑓𝑏 =

6 (375)(1000)2 300 (700)2

; M=

=

30 (10)2 8

= 375 KN.m

𝑓𝑏 = ± 15.306 MPa

Stress due to Live Load: 𝑓𝑙𝑙 =

6𝑀 𝑏𝑑 2

𝑓𝑙𝑙 =

; M=

𝑃𝐿 4

=

50 (10) 4

= 125 KN.m

6 (125)(1000)2 300 (700)2

𝑓𝑙𝑙 = ± 5.10 MPa

Final Stresses: 𝑓𝑡𝑜𝑝 = -7.14 + 12.24 – 15.306 – 5.10 𝒇𝒕𝒐𝒑 = -15.306 MPa 𝑓𝑏𝑜𝑡𝑡𝑜𝑚 = -7.14 – 12.24 + 15.306 + 5.10 𝒇𝒃𝒐𝒕𝒕𝒐𝒎 = 1.026 MPa

PS-Malabanan-Aldrin-John-V For simply supported beam, determine the maximum stresses at midspan section due to its own weight. With a uniform live load of 28.32 kN/m and an eccentric longitudinal compressive force of P= 1130.5 kN acting at an eccentricity e = 100mm. Use weight of concrete 23.50 kN/𝑚𝑚3 . a.) b.) c.) d.)

Using transform concrete into an elastic form. Using Internal Resisting Couple Method Using Load Balancing Method Stress Diagram at question a.

Given: L = 6.80 mm b = 280 mm d = 520 mm Ꝩc = 23.50 kN/m P = 1130.50 kN/m e = 100 mm(below N.A) WLL = 28.32 kN/m

Solution: w𝑙2

WB = Ꝩc b d

MB =

MLL =

WB = 23.50 (0.28) (0.52)

MB =

WB = 3.4216 kN/m

MB = 19.777 kN.m

8 3.4216(6.80)2 8

MLL =

w𝑙2 8 28.32(6.80)2 8

MLL = 163.6896 kN.m

a.) Using transform concrete into an elastic form. Formula: 𝑃

F = - 𝐴 +(-) F=-

6𝑃𝑒 2

𝑏𝑑

+(-)

1130.5(10)3 280(520)

6𝑀 2

𝑏𝑑

+(-)

6(1130.5)(10)3 (100) 280(520)

2

+(-)

6(19.777)(10)6 2

280(520)

+(-)

6(163.6896)(10)6 280(520)

2

F = -7.764 +(-) 8.959 +(-) 1.576 +(-) 12.972

Stress at Top

Stress at Bottom

Ftop = -7.764 + 8.959 - 1.576 - 12.972

Ftop = -7.764 - 8.959 + 1.576 + 12.972

Ftop = -13.344 or 13.344 MPa (Compression)

Ftop = -2.184 or 2.184 MPa (Compression)

b.) Using Internal Resisting Couple Method W = WB + WLL

Ph = M

W = 3.4216 + 28.32

h=

W = 31.7416 kN/m

h = 162. 287 mm

M= M=

w𝑙2

31.7416(6.80)2

𝑃

1130.5(10)3

𝑒′ = h - e

𝑒 ′ = 62.287 mm

M = 183.466 kN.m

F=-

𝑀

𝑒 ′ = 162.287 – 100

8

F = - 𝐴 +(-)

h=

183.466(10)6

h = 𝑒′ + e ;

8

𝑃

;

6𝑃𝑒 ′ 2

𝑏𝑑

1130.5(10)3 280(520)

+(-)

6(1130.5)(10)3 (62.287) 280(520)

2

F = -7.764 +(-) 5.58

Stress at Top

Stress at Bottom

Ftop = -7.764 – 5.58

Ftop = -7.764 + 5.58

Ftop = -13.344 or 13.344 MPa (Compression)

Ftop = -2.184 or 2.184 MPa (Compression)

c.) Using Load Balancing Method W = WB + WLL W = 3.4216 + 28.32 W = 31.7416 kN/m

𝑊 ′ = W – WP

WP =

𝑊 ′ = 31.7416 – 19.559

Wp =

8Pe 𝐿2 8(1130.5)(0.100) (6.80)2

𝑊 ′ = 12.1826 kN/m

𝑀′ = 𝑀′ =

WP = 19.559 kN/m

w𝑙2 8 12.1826(6.80)2 8

𝑀′ = 70.415 kN.m

Stress at Top

Stress at Bottom

Ftop = -7.764 + 8.959 - 1.576 - 12.972

Ftop = -7.764 - 8.959 + 1.576 + 12.972

Ftop = -13.344 or 13.344 MPa (Compression)

Ftop = -2.184 or 2.184 MPa (Compression)

d.) Stress Diagram at question a.

PS-Manlusoc-Alfredo-R. A 5m prestressed concrete simple beam having a cross section of 200mm x 400mm is subjected to prestressing force with an effective strength of 1500Kn. Tendons are in parabolic set up and having eccentricity of 100mm at midspan. The beam carries a total uniform distributed load of 40kn/m including beams weight. a. Det. the total stresses at midspan using load balancing method. b. Det. The total stresses at midspan using Internal Resisting Method.

a. Det. the total stresses at midspan using load balancing method. Sol’n : Wp =

8 𝑃𝑒 𝐿2

=

8 (1500)(0.10) 52

= 48kn/m

W’ = W – Wp = 40 – 48 = 8kn/m (upward) M’ =

𝑊 ′ 𝐿2 8

=

𝑃

8 (5)2 8

= 25kn⋅m

6𝑀

Ftop = - 𝐴 + 𝑏𝑑2 = -

1500𝑥103 200𝑥400

+

6(25)(106 ) 200𝑥4002

= -18.75 + 4.6875 = -14.0625 Mpa Ftop = -

𝑃 𝐴

+

6𝑀 𝑏𝑑 2

=-

1500𝑥103 200𝑥400

-

6(25)(106) 200𝑥4002

= -18.75 - 4.6875 = -23.4375 Mpa b. Det. The total stresses at midspan using Internal Resisting Method. Sol’n: M’ =

𝑊 ′ 𝐿2 8

=

40 (5)2 8

= 125 kn⋅m

M = Ph h=

𝑀 𝑃

=

125𝑥106 1500𝑥103

= 83.33 mm

h = e + e’ = 100 + e’ e’ = 16.67 𝑃

ftop = - 𝐴 + 𝑃

ftop = - 𝐴 -

6𝑃𝑒′ 𝑏𝑑 2 6𝑃𝑒′ 𝑏𝑑 2

====-

1500𝑥103 200𝑥400 1500𝑥103 200𝑥400

+ -

6(1500)(16.67(103 ) 200𝑥4002

6(1500)(16.67(103 ) 200𝑥4002

= -14.06 Mpa

= -23.438 Mpa

PS-Mendoza-Giselle-Ann-L. A rectangular prestressed beam 150mm wide and 300mm deep is used over an effective span of 10m. The cable with zero eccentricity at the supports, and linearly varying to 50mm at the center, carries an effective prestressing force of 500kN. Find the magnitude of the concentrated load P located at the center of the span for the following conditions at the center of span section: (a) If the load counteract the bending effect of the prestressing force(neglecting selfweight of beam) and (b) If the pressure line passes through the upper kern of the section under the action of the external load, self - weight and prestress. SOLUTION: A = (150 x 300) = 45x103 mm2 𝑍=

150 𝑥 300² 6

= 225 𝑥10⁴ mmᵌ

Self – weight of beam, g = 0.15 x 0.3 x 24 = 1.08 kN/m Where; P = 500 kN

e = 50mm

If the inclination of the cable to the horizontal is Ɵ, and P = concentrated load at the center of the span, for load balancing, (a) P = 2P sin Ɵ = 2Ptan Ɵ =

2 𝑥 500 𝑥 50 5 𝑥 1000

= 10𝑘𝑁

(b) Moment due to self – weight = 0.125 x 1.08x102 = 13.5 kN.m Stressed due to self – weight =

13.5 𝑥106 225 𝑥104 𝑃

Stresses due to prestressing = +

= +6𝑁/𝑚𝑚²

𝑃𝑒

𝐴 𝑍 500 𝑥103

= 45 𝑥103 +

500 𝑥103 (50) 225 𝑥104

Stress at the bottom fiber = 22.22 N/mm² If P = concentrated load at the center of the span, moment at the center of the span M=

𝑃 𝑥 10 4

= 2.5𝑃

Bending stress =

2.5𝑃 𝑥 106 225 𝑥 104

Then, 2.5𝑃 𝑥 106 225 𝑥 104

+ 6 = 22.22

P = 14.60kN

PS-MoisesDavidJonathanJ. Problem: A prestressed concrete simple beam having a cross-section of 250x600 mm is subjected to a prestressing force in steel immediately after prestressing which eventually reduced to 1600kN due to losses acting on the parabolic tendons. The beam carries a total uniformly distributed load of 30kN/m including beams weight. a] Determine rhe location of the lower kern point? b] Location of upper kern point?

PS-Munar-Corina-May-H. A pre-tensioned concrete beam, 100 mm wide and 300 mm deep, is prestressed by straight wires carrying an initial force of 150 KN at an eccentricity of 50 mm. The modulus of elasticity of steel and concrete are 210 and 35 KN/mm2 respectively. Estimate the percentage loss of stress in steel due to elastic deformation of concrete if the area of steel wires is 188 mm 2. Given: b = 100 mm

d = 300 mm P = 150 KN e = 50 mm Es= 210 KN/mm2 Ec = 35 KN/mm2 As = 188 mm2 Solution: Ab = b x h = 100 mm (300 mm) Ab = 30000 mm2 I=

𝑏ℎ3 12

=

100(300)3 12

I = 225 x 106 Es

210

Ec

35

𝛼𝑐 = ( ) = 𝛼𝑐 = 6 Initial stress in steel, 𝛿𝑠 =

𝑃 As

𝑃

Stress in concrete, 𝑓𝑐 = 𝐴 + 𝑏

𝑓𝑐 =

150 𝑥 103

=

188

𝑃𝑒𝑐 I

150 𝑥 103 3𝑥

= 800 N/mm2

104

+

150 𝑥 103 (50)(50) 225 𝑥 106

𝑓𝑐 = 6.66 N/mm2 Loss of stress due to elastic deformation of concrete = 𝛼𝑐 𝑓𝑐 = 6 (6.66) Loss of stress due to elastic deformation of concrete = 40 N/mm 2 Percentage loss of stress in steel =

𝑙𝑜𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑒𝑠𝑠 𝑡𝑜 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝛿𝑠

=

40 (100) 800

Percentage loss of stress in steel = 5 %

(100 %)

PS-NAVARROJOANNA-MARIE-D. A post-tension beam has a midspan cross section with a duct of 50𝑚𝑚 𝑥 75𝑚𝑚 to house the wires as shown. It is prestressed with 516𝑚𝑚2 of steel to an initial stress of 1034 N/𝑚𝑚2 . Immediately after transfer, the stress is reduced to 5% owing to anchorage loss and elastic shortening of concrete. a) Compute the moment of inertia of the net concrete section. b) Compute the stress in the upper top fiber in the concrete at transfer after loss of stress. c) Compute the stress in the lower top fiber in the concrete at transfer after loss of stress.

SOLUTION: a) MOMENT OF INERTIA OF THE NET CONCRETE SECTION 𝐴1 = (200)(300) 𝐴1 = 60,000 𝑚𝑚2 𝐴2 = (50)(75) 𝐴2 = 3,750 𝑚𝑚2 𝐴 = 𝐴1 − 𝐴2 𝐴 = 60,000 − 3,750 𝐴 = 56,250 𝑚𝑚2 𝐴ȳ = 𝐴1 𝑦1 − 𝐴2 𝑦2 56,250 ȳ = 60,000(150) − 3,750(75) ȳ = 155𝑚𝑚 𝐼=

(200)(300)2

(50)(75)3

+ [(200)(300)(5)2 ] − [

12

6

𝐼 = 425.74 𝑥 10 𝑚𝑚

12

4

b) STRESS IN THE UPPER TOP FIBER IN THE CONCRETE AT TRANSFER AFTER LOSS OF STRESS 𝑃

𝑓𝑡 = − 𝐴 +

𝑀𝐶 𝐼

𝑃

𝑃𝑒 𝐶

𝐴

𝐼

𝑓𝑡 = − +

𝑃 = 𝐴𝑠𝑝 𝑓𝑝𝑠 (0.95) 𝑃 = (516)(1034)(0.95) 𝑃 = 506,867 𝑁 𝑓𝑡 = −

506,867 56,250

+

506,867(80)(145) 425.74 𝑥 106

𝑓𝑡 = 4.80 𝑁/𝑚𝑚 6 (𝑇𝐸𝑁𝑆𝐼𝑂𝑁) c) STRESS IN THE LOWER TOP FIBER IN THE CONCRETE AT TRANSFER AFTER LOSS OF STRESS

𝑓𝑏 = − 𝑓𝑏 = −

𝑃 𝑃𝑒 𝐶 − 𝐴 𝐼 506,867 56,250

−

506,867(80)(155) 425.74 𝑥 106

𝑓𝑏 = −23.77 𝑁/𝑚𝑚 6 (𝐶𝑂𝑀𝑃𝑅𝐸𝑆𝑆𝐼𝑂𝑁)

]

PS-Parica-Dhenize-DC. The prestressed concrete T-Beam shown is 8.5 meters and to be prestressed with tendons located 100mm from the bottom of the beam at midspan. Determine the following: a. Location of the Neutral Axis from the top of the beam b. Eccentricity of the prestressing tendons c. Ratio of effective prestressing force to the total uniform load if the resulting moment at the midspan is zero. d. The maximum uniform live load it can safety carry if P=2MN after losses, and allowable tensile stress after losses is 2.92MPa and allowable compressive stress is 15.53MPa.

Solutions: a. A1 = (300)(600) = 180 000 mm2 A2 = (600)(200) = 120 000 mm2 AT = 180 000 + 120 000 = 300 000 mm2

AT𝑦̅ = A1y1 + A2y2 300 000 𝑦̅ = 180 000 (500) + 120 000 (100) 𝑦̅ = 340mm

b. e = 200 + 600 – 𝑦̅ – 100 e = 200 + 600 – 340 – 100 e = 360 mm

c. P = Presstressing Force W = Total Uniform Load If w = Uniform Load (kN/m) W = wL +M =

𝑤𝐿2 8

=

𝑊𝐿 8

=

𝑊 (8.5) 8

= 1.0625W

-M = Pe = P(0.360) = 0.360P +M = -M 1.0625W = 0.360P

𝑃 𝑊

= 2.951

d. Ft = 2.92 MPa Fb = -15.53 MPa

I = I1 + A1d12 + I2 + A2d22 ; d1 = 800 – 340 – 300 d1 = 160 mm d2 = 340 – 100 d2 = 240 mm

I= I=

𝑏1 ℎ12

𝑏2 ℎ22

+ A1d12 +

12

300(600)2

12

+ A2d22

+ 180 000 (160)2 +

12

600(200)2 12

+ 120 000 (240)2

I = 1.732x1010 mm4 𝑃

𝑃𝑒𝑐

Ft = − 𝐴 − 𝐼𝐹𝑡 𝐶

𝐼

+

𝑀𝐶

; C = 700 – 340 = 360 mm

𝐼

𝐼𝑃

= − 𝐴𝑐 − 𝑃𝑒 + 𝑀

M=

𝐼𝐹𝑡

𝐼𝑃

+ 𝐴𝑐 + 𝑃𝑒

𝐶

M=[

(1.732𝑥1010)(2.92) 360

+

(1.732𝑥1010 )(2𝑥106) 300 000 (360)

1

+ (2𝑥106 )(360] [10002]

M = 1181.225 kN.m M=

𝑤𝐿2 8

1181.225 =

𝑤(8.5)2 8

w = 130.79 w = DL + LL ; DL = WcAt = 23.5(300 000)/10002) = 7.05kN/m

w = DL + LL 130.79 = 7.05 + LL LL = 123.74 kN/m 𝑃

Fb = − 𝐴 + 𝐼𝐹𝑏 𝐶

𝑃𝑒𝑐 𝐼

−

𝑀𝐶

; C = 𝑦̅ = 340 mm

𝐼

𝐼𝑃

= − 𝐴𝑐 + 𝑃𝑒 − 𝑀

M=−

𝐼𝐹𝑡

M = [−

𝐶

𝐼𝑃

− 𝐴𝑐 + 𝑃𝑒

(1.732𝑥1010)(−15.53) 340

M = 1171.509 kN.m

−

(1.732𝑥1010)(2𝑥106 ) 300 000 (340)

1

+ (2𝑥106 )(360] [10002 ]

M=

𝑤𝐿2 8

1171.509 =

𝑤(8.5)2 8

w = 129.72 kN/m w = DL + LL ; DL = WcAt = 23.5(300 000)/10002) = 7.05kN/m 129.72 = 7.05 + LL LL = 122.67 kN/m

Use LL = 122.67 kN/m

PS-ReyesCharizze-Anne-P. A beam with b = 300 mm and depth = 600 mm is to be pre stressed. Considering a 15% pre stress loss, compute the value of the final pre stressing force P and eccentricity e. 1. If the compressive stress of 21 MPa. 2. If the compressive stress at the bottom fiber is 12 MPa and the tensile stress at the top fiber is 2 MPa. 3. If the compressive stress at the top fiber is 16 MPa and zero at the bottom fibers. Solution: 𝑃

1. f = - 𝐴 𝑃

-21 = - 300(600) P = 3780 kN e=0

2. 𝑃

6𝑃𝑒

𝑓𝑡 = − 𝐴 + 𝑏𝑑2 𝑃

6𝑃𝑒 300(6002 )

+ 2 = - 300(600) + 2=−

𝑃 300(600)

[1 −

𝑃

6𝑒 600

] -- Eq. 1

6𝑃𝑒

𝑓𝑏 = − 𝑏𝑑 [1 − 𝑏𝑑2 ] 𝑃

6𝑃𝑒

- 12 = - 300(600) − 300(6002 ) - 12 = −

𝑃 300(600)

[1 −

6𝑒 600

Divide Eq. 1 by Eq. 2 −

2 600 − 6𝑒 = 12 600 + 6𝑒

−

1 600 − 6𝑒 = 6 600 + 6𝑒

-600 – 6e = 6(600 – 6e) 30e = 3600 + 600

] -- Eq. 2

e = 140 mm 2=− 2=−

𝑃 300(600) 𝑃 300(600)

[1 − [1 −

6𝑒 600

]

6(140) 600

]

P = 900000 kN or 900 kN

3. 𝑓𝑏 = −

𝑃 6𝑃𝑒 + 𝐴 𝑏𝑑 2

0= −

𝑃 6𝑃𝑒 + 300(600) 300(6002 )

0=− 1=

𝑃 300(600)

[1 −

6𝑒 600

]

6𝑒 600

e = 100 mm

𝑓𝑡 = −

𝑃 6𝑃𝑒 − 𝐴 𝑏𝑑 2

- 16 = − - 16 = −

𝑃 300(600) 𝑃 300(600)

[1 + [1 +

6𝑒 600

]

6(100) 600

P = 1440000 N or 1440 kN

]

PS-Sandoval-Kenneth-Joseph-A. A 300 mm x 400 mm concrete beam has a span of 6 m. a post-tension force of 600 kN was applied 60 mm above the bottom of the beam. Assume concrete won’t crack in tension. f’c = 20.7 Mpa, unit weight of concrete is 23.5 kN/m3.

a. Compute the deflection due to pre-stressing force of 250 kN. b. Compute the net deflection of the beam immediately after transfer. c. Compute the safe uniform load that may be imposed on the beam so that there will be a net deflection upward of 5 mm.

Solution: a. e = 200-80 e = 120 mm M = Pe =600(.12) M = 72 kN.m 𝐼=

𝑏ℎ3 300(4003 ) = = 1600 𝑥 106 𝑚𝑚4 12 12

𝐸 = 4700√𝑓′𝑐 = 4700 √20.7 = 21384 𝑀𝑃𝑎 𝛿=

𝑀𝑙2 72 𝑥 106 (60002 ) = 8𝐸𝐼 8(1600𝑥106 )(21384)

𝜹 = 𝟗. 𝟒𝟕 𝒎𝒎

b. 𝑤 = (. 3)(. 4)(23.5) = 2.82

𝛿=

𝑘𝑁 𝑚

5𝑤𝑙2 5(2.82)(6)(63 )(1𝑥106 )(1𝑥106 ) = = 1.39 𝑚𝑚 384𝐸𝐼 384(21384)((1600𝑥106 )

𝛿𝑛𝑒𝑡 = 𝛿𝑖 − 𝛿 = 9.47 − 1,39 𝜹𝒏𝒆𝒕 = 𝟖. 𝟎𝟖 𝒎𝒎

c. 𝛿𝑛𝑒𝑡 = 𝛿𝑖 − 𝛿 = 8.08 − 5 = 3.08 𝑚𝑚

𝛿=

5𝑤𝑙2 384𝐸𝐼

3.08 =

5𝑤(60004 ) 384(21384)((1600𝑥106 )

𝒘 = 𝟔.24 kN/m

A pre-stressed beam having a cross section of 500mm x 750mm is subjected to a pre-stressing force of 1600kn acting on the parabolic tendons shown. The beam carries a total uniform load of 40kn/m including its own weight. Solve for the maximum stress at the top and bottom using conventional method

P 150 mm

P 225 mm

SOLUTION: Using Conventional Method 𝑊𝐿2 8 40(8)2 𝑀𝑑 = 8 𝑀𝑑 = 320 106 𝑀𝑑 =

6𝑀 𝑏𝑑 2 6(320)106 𝑓𝑑 = ± 500(750)2 𝑓𝑑 = ±6.83𝑀𝑃𝑎 𝑓𝑑 =

Due to Prestress Stress at the Top 𝑃 6𝑃𝑒 + 2 𝑏𝑑 𝑏𝑑 1.6𝑥106 6(1.6𝑥106 )(150) 𝑓𝑡 = − + 500(750) 500(750)2 𝑓𝑡 = −4.27 + 5.12 𝑓𝑡 = +0.85 𝑀𝑃𝑎 𝑓𝑡 = −

Stress at the Bottom 𝑃 6𝑃𝑒 𝑓𝑏 = − − 2 𝑏𝑑 𝑏𝑑 1.6𝑥106 6(1.6𝑥106 )(150) 𝑓𝑏 = − − 500(750) 500(750)2 𝑓𝑏 = −4.27 − 5.12 𝑓𝑏 = −9.39 𝑀𝑃𝑎 Max Stress at the top: 𝑓𝑡 = −6.83 + 0.85 𝑓𝑡 = −5.98 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛) Max Stress at the bottom: 𝑓𝑏 = +6.83 − 9.39 𝑓𝑏 = −2.56 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

PS-Sta-Ana-Limuel-F. A 6 meters rectangular concrete beam of cross-section 300 mm x 800 mm is subjected to pre stressing force with an effective strength of 1600 kN which has eccentricity of 150 mm below the C.G. The beam carries a uniform distributed load of 15 kN/m including its own weight. a. Determine the total stresses at midspan. b. Determine the total stresses at midspan using Internal Resisting Couple Method. c. Construct stress diagram.

SOLUTION: a. Stresses at midspan. 𝑓=−

𝑀=

𝑃 6𝑃𝑒 6𝑀 ± ± 𝐴 𝑏𝑑 2 𝑏𝑑 2

𝑤𝐿2 15(6)2 = = 67.5 𝑘𝑁. 𝑚 8 8

𝑓𝑡𝑜𝑝 = −

𝑃 6𝑃𝑒 6𝑀 + − 𝐴 𝑏𝑑 2 𝑏𝑑 2

𝑓𝑡𝑜𝑝 = −

1600(1000) 6(1600)(150)(1000) 6(67.5)(1000)2 + − 300(800) 300(800)2 300(800)2

𝑓𝑡𝑜𝑝 = −1.276 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛) 𝑓𝑏𝑜𝑡 = −

𝑃 6𝑃𝑒 6𝑀 − + 𝐴 𝑏𝑑 2 𝑏𝑑 2

𝑓𝑏𝑜𝑡 = −

1600(1000) 6(1600)(150)(1000) 6(67.5)(1000)2 − + 300(800) 300(800)2 300(800)2

𝑓𝑏𝑜𝑡 = −12.057 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

b. Total stresses at midspan using Internal Resisting Couple Method. 𝑀 = 𝑃ℎ 67.5 = 1600ℎ ℎ = 42.1875 𝑚𝑚 ℎ = 𝑒 + 𝑒′ 42.1875 = 150 + 𝑒 ′ 𝑒 ′ = 107.8125 𝑚𝑚 (𝑏𝑒𝑙𝑜𝑤 𝐶. 𝐺, ) 𝑓=−

𝑃 6𝑃𝑒′ ± 𝐴 𝑏𝑑 2

𝑓𝑡𝑜𝑝 = −

𝑃 6𝑃𝑒′ + 𝐴 𝑏𝑑 2

𝑓𝑡𝑜𝑝 = −

1600(1000) 6(1600)(107.8125)(1000) + 300(800) 300(800)2

𝑓𝑡𝑜𝑝 = −1.276 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛) 𝑓𝑏𝑜𝑡 = −

𝑃 6𝑃𝑒′ − 𝐴 𝑏𝑑 2

𝑓𝑏𝑜𝑡 = −

1600(1000) 6(1600)(107.8125)(1000) − 300(800) 300(800)2

𝑓𝑏𝑜𝑡 = −12.057 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

c. Stress diagram. -6.667

+5.391

-1.276

+

-6.667

=

-5.391

-12.391

PS-Verdadero-Jayvee-O. Problem No.1. The beam of a 2 storey warehouse has dimension shown below. The beams are simply supported on a span of 10 m and it is prestressed by a parabolic tendons( 15-5mmØ wires ) with an eccentricity of 100 mm at midspan. The total loss of prestress due to elastic shortening of concrete is Δfs = 50 kN and the beam carried a liveload of 20 kN/m. Use 24 kN/m³ weight if concrete and modular ratio of 12,500 for shortening of concrete. 300 mm

700 mm

  

Determine the Initial Prestressing Force Determine the Stress at the midspan immediately after the initial prestressed is applied. Determine the stress at midspan at final prestressding condition.

Solution: From the formula of elastic shortening of Concrete: 𝑛ὸ 𝐴𝑔

Δfs=

where: Δfs is the loss of prestressed (kN) N is the modular ratio ὸ initial stress of cable Ag is the gross area of section

100 kN =

12500ὸ 300 𝑥 700

ὸ = 840 MPa ;

ὸ=

𝑃 𝐴

P= ὸAst P= 840x

𝜋(5)² 4

x 15

P = 247 kN answer. 

Determine the stress at midspan immediately after the initial prestressed is applied. Due to P f=

247𝑥 10 ³ 300 𝑥 700

f = 1.18 MPa Due to e f=

6(247)(100) 300 𝑥 700²

f= 1.01 MPa Ftop Ftop = -1.18 + 1.01 Ftop = 0.17 MPa Fbot Ftop = -1.18 - 1.01 Ftop = 2.19 MPa 

Determine the stress at midspan at final condition Due to P f=

197𝑥 10 ³ 300 𝑥 700

f = 0.94MPa Due to e f=

6(197𝑥10³)(100) 300 𝑥 700²

f= 0.804 MPa Due to Moment DL = 24 kN/m³ ( .3)(.7) DL = 5.04 kN/m LL = 20 kN/m Wtotal = 25.04 kN/m M=

25.04(10)² 8

M= 313 kN.m. f=

6(313) 300 𝑥 700²

f = 12.78 MPa Ftop Ftop = -.94+ .804 – 12.78 Ftop = -12.97 MPa Ftop Ftop = -.94- .804 + 12.78 Ftop = 11.04 MPa

PS-Vicencio-Angelica-H. A prestressed concrete simple beam having a cross – section of 255 x 620mm is subjected to prestressing force in steel immediately after prestressing which eventually reduce to 1700 kN due to losses acting on the parabolic tendons. The beam carries a total uniformly distributed load of 30 KN/m (not including beam`s weight). The eccentricity is 110 mm and length of beam is 8.0 m. Determine the following: a. Bending moment at midspan. b. Total stresses at midspan. c. Total stresses at end section of the beam.

Given:

L = 8. 0 m

Cross-section: 225 x 620 mm

Required: a. M @ midspan b. ftop & fbottom @ midspan c. f @ end section

P = 1700 kN DL = 30 kN/m e = 110mm Solution: (82 )[23.54(0.255)(0.62)]

a. MDL =

8

MDL= 29.77 kN.m MLL =

30(8)2 8

MLL = 240 kN.m M = MDL + MLL M = 29.77 + 240 M = 269.77 kN.m (answer)

P

b. f = − A ± f=

−1700(103) 255(620)

±

Pec

∓

Mc

;I=

I I 1700(110)(310)(103) I

𝐟𝐭𝐨𝐩 = −𝟏𝟓. 𝟖𝟐 𝐌𝐏𝐚 (answer)

255(6203 )

∓

= 5064470000 mm2

12 269.77(106)(310) I

𝐟𝐛𝐨𝐭𝐭𝐨𝐦 = −𝟓. 𝟔𝟗 𝐌𝐏𝐚 (answer)

P

c. f = − A f=−

1700

255(620)

𝐟 = −𝟏𝟎. 𝟕𝟓 𝐌𝐏𝐚

*no eccentricity and no moment at end section because of parabolic tendons

A beam with width b = 250 mm and depth d = 450 mm is pre stressed by an initial force of 600 kN. Total loss of pre stress at service loads is 15%. 1. Calculate the resulting final compressive stress if the pre stressing force is applied at the centroid of the beam section. 2. Calculate the final compressive stress if the pre stressing force is applied at an eccentricity of 100 mm below the centroid of the beam section. 3. Calculate the eccentricity at which the pre stressing force can be applied so that the resulting tensile stress at the top fiber of the beam is zero. Solution: 1. Final compressive stress if the pre stressing force is applied at the centroid of the beam section.

𝜕= =

𝑃 𝐴

600(0.85)(1000) 250(450)

𝑀𝐴 = −4.53 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

2. Final compression stress if the pre stressing force is applied at an eccentricity of 100 mm below the centroid of the beam section.

𝑓𝑡 =

𝑃 𝑃𝑒 𝐶 + 𝐴 𝐼

𝑓𝑡 =

𝑃 6𝑃𝑒 + 𝑏𝑑 𝑏𝑑 2

𝑓𝑡 =

600000(0.85) 250(450) 6(600000)(0.85)(100) − 250(450)2

𝑓𝑡 = −10.58 𝑀𝑃𝑎 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

3. Eccentricity at which the pre stressing force can be applied so that the resulting tensile stress at the top fiber of beam is zero.

𝑓𝑡 = 0=

𝑃 6𝑃𝑒 + 𝑏𝑑 𝑏𝑑 2 −600000 6(600000)𝑒 + 250(450) 250(450)2

600000 6(600000)𝑒 = 250(450) 250(450)2 𝑒 = 75𝑚𝑚

PROBLEM] A symmetrical I section beam with total depth of 600 mm, moment of inertia of 5.0 x 109 mm4 and concrete area of 113,500 mm2 will be used on a 9.0 m simple span. Assume unit weight of concrete is equal to 24 kN/m3. The beam carries a concentrated live load of 30 kN at midspan. The beam will be pre-tensioned with multiple seven-wire strands below the centroid at constant eccentricity of 200 mm. The initial and final prestressing force is 1000 kN and 800 kN respectively. a. Calculate the stresses at midspan due to initial prestressing and uniformly distributed dead load. b. Calculate the stresses at midspan with full service load in place. c. Calculate the net deflection under service loads if E = 13734 MPa. SOLUTION: a. fT and P Pec MDLc f=− ± ± A I I Beam weight = 0.1135 x 24 = 2.724 kN/m wL2 2.724(9)2 MDL = = = 27.58 kN ∙ m 8 8 1000 x 103 1000 x 103 (200)(300) 27.58 x 106 (300) f=− ± ± 113,500 5.0 x 109 5.0 x 109 f = −8.81 ± 12 ± 1.65 fT = −8.81 + 12 − 1.65 = +1.54 MPa fB = −8.81 − 12 + 1.65 = −19.16 MPa b. fT and fB P Pec MDLc MLLc f=− ± ± ± A I I I PL 30(9) MLL = = = 67.50 kN ∙ m 4 4 800 x 103 800 x 103 (200)(300) 27.58 x 106 (300) 67.50 x 106 (300) f=− ± ± ± 113,500 5.0 x 109 5.0 x 109 5.0 x 109 f = −7.05 ± 9.6 ± 1.65 ± 4.05 fT = −7.05 + 9.6 − 1.65 − 4.05 = −3.15 MPa fB = −7.05 − 9.6 + 1.65 + 4.05 = −10.95 MPa c. Net deflection Due to prestressing force and 5(2.724)(9000)4 y = DL eccentricity; 384(13734)(5.0 x 109 ) 2 ML yDL = 3.39 mm (Downward) yP = 8EI Due to concentrated load; M = Pe = (800)(0.2) PL3 y = M = 160 kN∙m LL 48EI 160 x 106 (9000)2 30 x 103 (9000)3 yP = y = LL 8(13734)(5.0 x 109 ) 48(13734)(5.0 x 109 ) yP =23.59 mm (Upward) yLL = 6.63 mm (Downward) Due to beam weight/dead load Then; 5wL4 yN = 23.59 − 3.39 − 6.63 yDL = 384EI yN = 13.57 mm (Upward)

fB

A rectangular prestressed beam 150mm wide and 300mm deep is used over an effective span of 10m. The cable with zero eccentricity at the supports, and linearly varying to 50mm at the center, carries an effective prestressing force of 500kN. Find the magnitude of the concentrated load P located at the center of the span for the following conditions at the center of span section: (a) If the load counteract the bending effect of the prestressing force(neglecting selfweight of beam) and (b) If the pressure line passes through the upper kern of the section under the action of the external load, self - weight and prestress. SOLUTION: A = (150 x 300) = 45x103 mm2 𝑍=

150 𝑥 300² 6

= 225 𝑥10⁴ mmᵌ

Self – weight of beam, g = 0.15 x 0.3 x 24 = 1.08 kN/m Where; P = 500 kN

e = 50mm

If the inclination of the cable to the horizontal is Ɵ, and P = concentrated load at the center of the span, for load balancing, (a) P = 2P sin Ɵ = 2Ptan Ɵ =

2 𝑥 500 𝑥 50 5 𝑥 1000

= 10𝑘𝑁

(b) Moment due to self – weight = 0.125 x 1.08x102 = 13.5 kN.m Stressed due to self – weight =

13.5 𝑥106 225 𝑥104 𝑃

Stresses due to prestressing = 𝐴 + =

= +6𝑁/𝑚𝑚²

𝑃𝑒

𝑍 500 𝑥103 45 𝑥103

+

500 𝑥103 (50) 225 𝑥104

Stress at the bottom fiber = 22.22 N/mm² If P = concentrated load at the center of the span, moment at the center of the span M=

𝑃 𝑥 10 4

= 2.5𝑃

Bending stress =

2.5𝑃 𝑥 106 225 𝑥 104

Then, 2.5𝑃 𝑥 106 225 𝑥 104

+ 6 = 22.22

P = 14.60kN

A Single-T prestressed concrete beam shown below is simply supported having a span of 10 m. It carries a superimposed live load of 15.08 kN/m in addition to the weight of beam. It is prestressed with 700 mm2 of steel to an initial stress of 1034 N/mm2 located 400 mm from the topmost fiber of the beam section. Immediately after transfer, the stress is reduced by 12 %. Determine the stresses at L/4 from the support due to losses in prestress and final service loads. Use concrete weight equals to 24 kN/m 3. Properties of Single-T section AT = 205,000 mm2 I NA = 4.78 mm x 10 9 mm4 y = 303.66 mm Solution:

C.G.

P’= 0.88 x 700 ( 1034 ) = 636.944 kN

y

e = 303.66 – 100 = 203.66 mm Weight = 24 (

205000 10002

) = 4.92 kN/m

Total Weight = 4.92 + 15.08 = 20 kN/m

C

𝑥 =

𝐿 10 = = 2.5 𝑚 4 4

𝑅𝐴 = 𝑅𝐵 = 100 𝑘𝑁 𝑀𝑐 = 100 (2.5) − 20 (

2.52 ) 2

𝑀𝑐 = 187.5 𝑘𝑁. 𝑚 𝑃

f top = − 𝐴 + f top = −

𝑃𝑒𝑐 𝐼

−

𝑀𝑐 𝐼

636.944 (103 ) 205000

+

636.944 (103 )(203.66)(196.34) 4.78 𝑥 109

−

187.5 (106 )(196.34) 4.78 𝑥 109

f top = - 5.48 MPa (compression)

f bottom = −

636.944 (103 ) 205000

−

636.944 (103 )(203.66)(303.66) 4.78 𝑥 109

+

187.5 (106 )(303.66) 4.78 𝑥 109

f bottom = 0.564 MPa (tension)

A beam with width b = 300 mm and depth d = 600 mm is to be prestressed. Considering a 15% prestress loss, compute the value of initial prestressing force P and eccentricity e.

A) If the compressive stress is 21 MPa. B) If the compressive stress at the bottom fiber is 12 MPa and a tensile stress at the fiber is 2 MPa. C) If he compressive stress at the top fiber is 16 MPa and zero at the bottom fiber.

Solution: A) b = 300 mm

15% loss

d = 600 mm

Pἱ =? e =? P = 0.855Pἱ --- equation 1 e=0

𝑃

F=𝐴

P

P

𝑃

-21 = − 300(600) P = 3780 KN Substitute P to equation 1 3780 = 0.85 Pἱ Pἱ = 4447.06 KN, e = 0 (ANSWER) B) Top: −𝑃

6𝑃𝑒

2 = 300(600) + 300(600)2

e = 140

Bottom: -12 =

, P = 900 KN

−𝑃 300(600)

--

900 = 0.85 Pἱ Pἱ = 1059 KN (ANSWER)

6𝑃𝑒 300(600)2

C) Top: −𝑃

6𝑃𝑒

-16 = 300(600) - 300(600)2

, P = 1440 KN

e = 100

Bottom: −𝑃

6𝑃𝑒

0 = 300(600) + 300(600)2

1440 = 0.85 Pἱ = 1694 KN (ANSWER)

A concrete beam with cross-sectional area of 32 x 103 mm2 and the radius of gyration is 72 mm is prestressed by a parabolic cable carrying an effective stress of 1000 N/mm 2. The span of the beam is 8 m. the cable composed of 6 wires of 7 mm diameter has an eccentricity of 50 mm at the center and zero at the supports. Neglecting all losses, find the central deflection of the beam as follows: a) Self-weight + prestress b) Self-weight + prestress + live load of 2 kN/m Given: A = 32 x 103 mm2 E = 38 kN/mm2 DC = 24 kN/mm3 I = 72 mm L = 8 m or 8000 mm e = 50 mm Solution:

I = Ai

2

= 32 x 103 (72) = 166 x 106 mm4 Prestressing force

P = (6)(38.5)(1000) = 231000 N or 231 kN 32 x 103

Self- weight, g = (24) 106 g = 0.77 kN/m g = 0.00077 kN/mm 5𝑔𝐿4

Downward deflection due to self- weight = 384 𝐸𝐼

5(0.00077)(8000)4

= 384(38)(166 𝑋 106) = 6.5 mm Upward deflection dur to prestressing force = =

5𝑃𝑒𝐿4

48 𝐸𝐼 5(231)(50)(8000)4 48(38)(166 𝑋 106 )

= 12.2 mm Downward deflection due to live load =

Downward deflection due to self− weight Self− weight

6.5

(2) = 0.77 = 16.9 mm a) Self-weight + prestress = 12.2 – 6.5 = 5.7 mm (↑) b) Self-weight + prestress + live load of 2 kN/m = 6.5 – 12.2 + 16.9 = 11.2 mm (↓)

(2)

A pre-tensioned concrete beam, 100 mm wide and 300 mm deep, is prestressed by straight wires carrying an initial force of 150 KN at an eccentricity of 50 mm. The modulus of elasticity of steel and concrete are 210 and 35 KN/mm 2 respectively. Estimate the percentage loss of stress in steel due to elastic deformation of concrete if the area of steel wires is 188 mm2. Given: b = 100 mm d = 300 mm P = 150 KN e = 50 mm Es= 210 KN/mm2 Ec = 35 KN/mm2 As = 188 mm2 Solution: Ab = b x h = 100 mm (300 mm) Ab = 30000 mm2 𝑏ℎ3

100(300)3

I = 12 = 12 I = 225 x 106 Es 210 𝛼𝑐 = ( ) = Ec 35 𝛼𝑐 = 6 𝑃

Initial stress in steel, 𝛿𝑠 = As = 𝑃

Stress in concrete, 𝑓𝑐 = 𝐴 + 𝑏

150 𝑥 103

𝑃𝑒𝑐

150 𝑥 103

188

= 800 N/mm2

I 150 𝑥 103 (50)(50)

𝑓𝑐 = 3 𝑥 104 + 225 𝑥 106 2 𝑓𝑐 = 6.66 N/mm Loss of stress due to elastic deformation of concrete = 𝛼𝑐 𝑓𝑐 = 6 (6.66) Loss of stress due to elastic deformation of concrete = 40 N/mm2 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑒𝑠𝑠 𝑡𝑜 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 Percentage loss of stress in steel = (100 %) 𝛿 40 (100)

= 800 Percentage loss of stress in steel = 5 %

𝑠

A simply supported beam 300 mm by 700 mm and span of 10 m is prestressed by a straight tendon with a force of 1500 KN at an eccentricity of 200 mm from the centroid. The beam supports a live load of 50 KN at the mid span and a dead load of 30 KN/m including the beam’s weight. A.) Calculate the final stress at the top fiber in MPa. B.) Calculate the final stress at the bottom fiber in MPa. Solution: Stress due to P: 𝑃

𝑓𝑝 = − 𝐴 = −

1500 (1000) 300 (700)

𝑓𝑝 = - 7.14 MPa

Stress due to eccentricity “e” 6𝑃𝑒

𝑓𝑒 = 𝑏𝑑2 =

6 (1500)(200)(1000) 300 (700)2

𝑓𝑒 = ± 12.24 MPa

Stress due to self weight: 6𝑀

𝑓𝑏 = 𝑏𝑑2 ; M = 𝑓𝑏 =

𝑤𝐿2 8

=

30 (10)2 8

= 375 KN.m

6 (375)(1000)2 300 (700)2

𝑓𝑏 = ± 15.306 MPa

Stress due to Live Load: 6𝑀

𝑓𝑙𝑙 = 𝑏𝑑2 ; M = 𝑓𝑙𝑙 =

𝑃𝐿 4

=

50 (10) 4

= 125 KN.m

6 (125)(1000)2 300 (700)2

𝑓𝑙𝑙 = ± 5.10 MPa

Final Stresses: 𝑓𝑡𝑜𝑝 = -7.14 + 12.24 – 15.306 – 5.10 𝒇𝒕𝒐𝒑 = -15.306 MPa 𝑓𝑏𝑜𝑡𝑡𝑜𝑚 = -7.14 – 12.24 + 15.306 + 5.10 𝒇𝒃𝒐𝒕𝒕𝒐𝒎 = 1.026 MPa

A concrete beam with cross-sectional of 32 𝑥 103 𝑚𝑚2 & the radius of gyration is 72 𝑚𝑚 is 𝑁 prestressed by a parabolic cable carrying an effective stresses1000 𝑚𝑚2 .The span of the beam is 8 𝑚. The cable, composed of 6 wires of 7𝑚𝑚 diameter, has an eccentricity of 50 𝑚𝑚 at the center & zero at the supports. Neglecting all losses, find the central deflection of the beam as follows: a. Self-weight + prestressed 𝑘𝑁 b. Self-weight + prestressed + live load of 2 𝑚 . Solution: Cross-sectional area of beam, 𝐴 = 32𝑥103 𝑚𝑚2 𝑘𝑁

Modulus of elasticity, 𝐸 = 38 𝑚𝑚2 Dead weight of concrete,𝐷𝐶 = 24𝑘𝑁/𝑚𝑚2 Radius of gyration, 𝑖 = 72𝑚𝑚 Span, 8𝑚 = 8000𝑚𝑚 Eccentricity, 𝑒 = 50𝑚𝑚 𝐼 = 𝐴𝑖 2 = (32𝑥103 𝑥722 ) = 166𝑥106 𝑚𝑚4 𝑃 = (6𝑥38.5𝑥1000) = 23100 𝑁 = 231 𝑘𝑁 32𝑥103 𝑘𝑁 0.00077𝑘𝑁 𝑔=( 𝑥24) = 0.77 = 6 10 𝑚 𝑚𝑚 5gL4 5𝑥0.00077𝑥80004 Downward deflection due to self weight = ( )=( ) = 6.5 𝑚𝑚 384𝐸𝐼 384𝑥38𝑥166𝑥106 5𝑃𝑒𝐿4

5𝑥231𝑥50𝑥80004

Upward deflection due to prestressing force = ( 48𝐸𝐼 ) = ( 48𝑥38𝑥166𝑥106 ) = 12.2𝑚𝑚 6.5

Downward deflection due to live load = (0.77 x2) = 16.9mm a. Deflection due to (self-weight + prestress) = (12.2-6.5) = 5.7mm (upward) b. Deflection due to (self-weight + prestress + live load) = 6.5-12.2+16.9) = 11.2 (downward)

Problem No.1. The beam of a 2 storey warehouse has dimension shown below. The beams are simply supported on a span of 10 m and it is prestressed by a parabolic tendons( 155mmØ wires ) with an eccentricity of 100 mm at midspan. The total loss of prestress due to elastic shortening of concrete is Δfs = 50 kN and the beam carried a liveload of 20 kN/m. Use 24 kN/m³ weight if concrete and modular ratio of 12,500 for shortening of concrete. 300 mm

700 mm

  

Determine the Initial Prestressing Force Determine the Stress at the midspan immediately after the initial prestressed is applied. Determine the stress at midspan at final prestressding condition.

Solution: From the formula of elastic shortening of Concrete: Δfs=

𝑛ὸ 𝐴𝑔

where: Δfs is the loss of prestressed (kN) N is the modular ratio ὸ initial stress of cable Ag is the gross area of section

12500ὸ

100 kN = 300 𝑥 700 ὸ = 840 MPa ;

𝑃

ὸ= 𝐴

P= ὸAst P= 840x

𝜋(5)² 4

x 15

P = 247 kN answer.



Determine the stress at midspan immediately after the initial prestressed is applied.

Due to P 247𝑥 10 ³

f= 300 𝑥 700 f = 1.18 MPa Due to e f=

6(247)(100) 300 𝑥 700²

f= 1.01 MPa Ftop Ftop = -1.18 + 1.01 Ftop = 0.17 MPa Fbot Ftop = -1.18 - 1.01 Ftop = 2.19 MPa



Determine the stress at midspan at final condition Due to P 197𝑥 10 ³

f= 300 𝑥 700 f = 0.94MPa Due to e f=

6(197𝑥10³)(100) 300 𝑥 700²

f= 0.804 MPa Due to Moment DL = 24 kN/m³ ( .3)(.7) DL = 5.04 kN/m LL = 20 kN/m Wtotal = 25.04 kN/m 25.04(10)²

M= 8 M= 313 kN.m. 6(313)

f= 300 𝑥 700² f = 12.78 MPa Ftop Ftop = -.94+ .804 – 12.78 Ftop = -12.97 MPa Ftop Ftop = -.94- .804 + 12.78 Ftop = 11.04 MPa

A pre-tensioned beam of size 200 mm x 300 mm is eccentrically pre-stressed with 520 mm2 wires anchored to bulkheads with a fi = 1035 MPa. The cgs is 100 mm above the bottom of the beam. Assuming n = 6, compute the stresses in concrete and steel immediately after transfer due to pre-stress only.

A concrete beam is prestressed by 6 parabolic cables, each carrying a 40kN force. The beam is having a rectangular section 150mm wide by 330mm deep supports a uniformly distributed load of 5.5kN/m, which includes the self-weight of the beam. The effective span of the beam is 7 meters and the eccentricity is at the center of the span which is 60mm. Calculate the total stress at midspan and determine the upper kern points location.

Given: b d L P e

= = = = =

150mm 330mm 7m 6(40) = 240kN 60mm

Solution: (Using Internal Resisting Couple Method)

SOLVING FOR THE TOTAL STRESS @ MIDSPAN 𝑤𝐿2

𝑀=

8

=

5.5(7)2 8

= 33.688 𝑘𝑁 − 𝑚

𝑀 = 𝑃ℎ ℎ=

𝑀 𝑃

=

33.688(10)3 240

= 140.367𝑚𝑚

ℎ = 𝑒 + 𝑒′ 𝑒 ′ = 140.367 − 60 = 80.367𝑚𝑚

𝑃

𝑓 = −𝐴± 𝑓=−

6𝑃𝑒′ 𝑏𝑑 2

240(10)3 150(330)

±

6(240)(10)3 (80.367) 150(330)2

𝑓 = −4.848 ± 7.085

𝑓𝑇𝑂𝑃 = −4.848 − 7.085 𝑓𝑇𝑂𝑃 = −11.933𝑀𝑃𝑎

𝑓𝐵𝑂𝑇𝑇𝑂𝑀 = −4.848 + 7.085 𝑓𝐵𝑂𝑇𝑇𝑂𝑀 = 2.237𝑀𝑃𝑎

SOLVING FOR LOC. OF UPPER KERN POINT 𝑘𝑡 =

𝑟2 𝑐𝑡

since the section is a rectangle:

ℎ

𝑘𝑡 = 6 =

330 6

𝑘𝑡 = 55𝑚𝑚