1. Flexural Design For Prestress

Flexural Design for Prestress

Prepared by: Engr. Kenny B. Cantila Assistant Professor IV

0.85fc′

ϵcu

b

0.003

c ds dp

h

a = β1 c

de

Aps

ϵte Assumed strain

ϵte = ϵcu

de − a/2 T

As

𝐜 𝛜𝐜𝐮 = 𝐝𝐞 𝛜𝐜𝐮 + 𝛜𝐭𝐞

C

For As = 0, de = dp For Aps = 0, de = ds

de − c 1 = ϵcu −1 c c/de

where:

c = depth of compression stress block at nominal resistance ϵcu = assumed failure strain of concrete in compression

ϵte = tensile strain in the concrete at the level of the centroid of the resultant tensile force in the tension reinforcement, that is the level of de de = depth from the extreme compression fiber to the centroid of the tensile force resultant in the tension reinforcement (prestressed and/or non – prestressed). Note that, de = dp for As = 0 and de = ds for Aps = 0.

Moment Reduction Factor 0.90 ∅ = 0.70 + 0.20

0.70 0.65

1 5 − c/de 3

Spiral

∅ = 0.65 + 0.25

1 5 − c/de 3

Other

Compression controlled

Transition ϵs = 0.002 c = 0.600 de

Tension controlled ϵs = 0.005 c = 0.375 d𝑒

𝜖𝑠 > 0.005

Tension-controlled

𝜖𝑠 = 0.005

0.002 < 𝜖𝑠 < 0.005

Transition-controlled

𝜖𝑠 = 0.002 𝜖𝑠 < 0.002

𝜌 < 𝜌1

Compression-controlled

Tension-controlled

3 0.85𝑓 ′ 𝑐 𝛽1 𝜌1 = 8 𝑓𝑦 𝜌1 < 𝜌 < 𝜌2

Transition-controlled

3 0.85𝑓 ′ 𝑐 𝛽1 𝜌2 = 5 𝑓𝑦 𝜌2 < 𝜌

Compression-controlled

Background for Analysis of Sections at Ultimate Assumptions in the Analysis 1.

Plane sections remain plane under loading.

2.

Perfect bond exist between steel and concrete. This is not valid in unbonded tendons.

3.

The limiting strength of concrete 𝜖𝑐𝑢 is equal to 0.003.

4.

The tensile strength of concrete is neglected.

5.

The total force in the concrete compressive zone can be approximated by considering a uniform stress of magnitude 0.85f′c over a rectangular block of width b and depth 𝑎 = 𝛽1 𝑐, where c represents the depth of the neutral axis and 𝛽1 is a factor defined as follows: The factor 𝛃 shall be taken as follows:

a)

β = 0.85

0.05 ′ b) β = 0.85 − f c − 28 7 c) β = 0.65

for f′c ≤ 28 MPa but not less than 0.65 for f′c > 56 MPa

ϵcu

b

0.85fc′

f′c

0.003

𝑎/2

c ds dp

h

a = β1 c

de

Aps

fps fy

As Assumed strain

Actual stress diagram

ACI assumed

Stresses and Strains at Ultimate Behavior

T

C

6.

It is assumed that when reinforcing steel is used, its stress – strain curve is elastic perfectly plastic, and its stress at ultimate is its yield strength, 𝑓𝑦 . This is not the case of prestressing steel. Its ultimate, 𝑓𝑝𝑠 , is unknown and generally given by a code prediction equation. Using strain compatibility analysis, it is valid for the material properties currently used, namely, the stress – strain curve of the reinforcing steel, the effective prestress, stress – strain curve of the prestressing steel, and the ultimate compression strain of concrete.

Typical stress – strain curves assumed for the reinforcement

Force Equilibrium C = T1 + T2 0.85f′c β1 cb = Aps fps + As fy

Depth of compression block: 0.85f′c β1 cb = Aps fps + As fy 0.85f′c ab = Aps fps + As fy Aps fps + As fy a= 0.85f′c b

a = β1 c β1 c= a

Moment capacity of the section: Mn = Aps fps Mn = Aps fps

a a dp − + As fy ds − 2 2 β1 c β1 c dp − + As fy ds − 2 2

a Mn = 0.85f′c ba de − 2 a Mn = (Aps fps + As fy ) de − 2

Aps = area of prestressing reinforcement in the tensile zone fps = stress in the prestressing steel at nominal flexural resistance of the section As = area of non – prestressed tension reinforcement

fy = specified yield strength of non–prestressed tensile reinforcement c = depth of neutral axis at ultimate

𝑎 = depth of rectangular stress block = β1 c 𝑑𝑝 = distance from the extreme compression fiber to centroid of tensile force in prestressed reinforcement 𝑑𝑠 = distance from the extreme compression fiber to centroid of tensile force in non–prestressed tensile reinforcement

Notations Aps = area of prestressed reinforcement in tensi0n zone, mm2 dp = distance from extreme compression fiber to centroid of prestressed reinforcement, mm f′ci = compressive strength of concrete at time of initial prestress, MPa fpc = average compressive stress in concrete due to effective prestress force only ( after allowance for all prestressed losses), MPa fp𝑠 = stress in prestressed reinforcement at nominal strength, MPa fpu = specified tensile strength of prestressing tendons, MPa fp𝑦 = specified yield strength of prestressing tendons, MPa

fse = Effective stress in prestressed reinforcement (after allowance for prestress losses), MPa lx = Length of prestressing tendon element from jacking end to any point n = number of mono strand anchorage devices in a group Ps = prestressing tendon force at jacking end Px = prestressing tendon force at any point x Psu = factored post – tensioned tendon force at the anchorage device

Notations (cont’d) 𝛼 γp

total angular change in prestressing tendon profile in radians from tendon jacking end to y to any point x = factor for type of prestressing tendon

0.55 for fpy /fpu not less than 0.80 (prestressing bars) 0.40 for fpy /fpu not less than 0.85 (stress – relieved strand)

μ = ρp = ω = ω′ = ωp =

0.28 for fpy /fpu not less than 0.90 (low relaxation strand) curvature friction coefficient ratio of prestressed reinforcement Reinforcement indices for flanged sections computed as ρfy /f′c for ω, ω′, and ωp except that b shall be the web width, ρ′fy /f′c and reinforcement shall be required to develop ρp fps /f′c compressive strength of web only.

Permissible Stresses in Concrete for Flexural Members

A. Stresses in concrete immediately after transfer (before time – dependent prestress losses) Extreme fiber stress in compression except as permitted by NSCP shall not exceed ………... …………………..….

0.60f′ci

Extreme fiber stress in compression at ends of simply supported member shall not exceed ………………………

0.70f′c

Where computed concrete tensile strength, 𝑓𝑡 , exceeds 0.5 𝑓′𝑐 at ends of simply supported members, or 0.25 𝑓′𝑐 at other locations, additional bonded reinforcement shall be provided in the tensile zone to resist the total tensile force in concrete computed with the assumption of uncracked section.

Permissible Stresses in Concrete for Flexural Members (cont’d)

B. Stresses in concrete at service loads based on uncracked section properties and the allowance for all prestress losses shall not exceed the following: Extreme fiber stress in compression due to prestress plus sustained loads …………………………...……………….

0.45f′c

Extreme fiber stress in compression due to prestress plus total loads ……………………...………………………….

0.60f′c

Permissible Stresses in Prestressing Tendons

Tensile stress in prestressing tendons shall not exceed the following: Due to prestressing tendon jack force ……........... 0.94 fpy ≤ 0.80 fpu Immediately after prestress transfer …..........…… 0.82 fpy ≤ 0.74 fpu

Post – tensioning tendons, at anchorage devices and coupler, immediately after force transfer ……

0.74 fpu

Ultimate strength design

Determination of stress in prestressed reinforcement at nominal strength (𝐟𝐩𝐬 ) as an alterative to a more accurate determination based on strain compatibility, the following approximate values shall be permitted to be used if 𝐟𝐬𝐞 is not less than 𝟎. 𝟓𝐟𝐩𝐮. For members with bonded tendons

fps = fpu

γp fpu d 1− ρp + ω − ω′ β1 fc ′ dp

If compression steel is used: fpu d ρp + ω − ω′ fc ′ dp

≥ 0.17

and

d′ ≤ 0.15db

dp = Distance from extreme compression fiber to centroid of prestressed reinforcement, mm d = Distance from extreme compression fiber to centroid of non – prestressed tension reinforcement, mm β1 = 0.85 − 0.05 fc ′ − 28 ; 0.65 ≤ β1 ≤ 0.85 7 γp = factor for type of prestressing tendon = 0.55 for fpy /fpu not less than 0.80 = 0.40 for fpy /fpu not less than 0.85 = 0.28 for fpy /fpu not less than 0.90 Aps bd f ω = ρ y fc ′ f ω′ = ρ′ y fc ′

ρp =

For members with unbonded tendons a)

Members with unbonded tendons and with span – to – depth ratio of 35 or less: fps

b)

f′c = fse + 70 + 100ρp

Members with unbonded tendons and with span – to – depth ratio of greater than 35: fps

f′c = fse + 70 + 300ρp

But 𝐟𝐩𝐬 shall not be greater than 𝐟𝐩𝐲 , nor greater than (𝐟𝐬𝐞 + 𝟐𝟏𝟎)

Minimum Bonded Reinforcement 1.

A minimum area of bonded reinforcement shall be taken as: As = 0.004Act

where Act is the area of that part of cross section between the flexural tension face and center of gravity of gross cross section. 2.

Bonded reinforcement shall not be required in positive moment areas where ft , the extreme fiber stress in tension in the pre – compressed tensile zone at service load (after allowance for prestress losses) does not exceed 0.17 f′ci .

3.

In positive moment area where computed tensile stress in concrete at service exceeds 0.17 f′ci minimum area of bonded reinforcement shall be computed by: Nc As = 0.5fy

where fy should be less than 415 MPa; Nc is the tensile force in concrete due to unfactored dead load plus live load 4.

In negative moment areas at column supports, minimum area of bonded reinforcement As in the top of the slab in each direction shall be computed by:

As = 0.00075Acf where Acf is the larger gross cross – sectional area of the slab – beam strips in two orthogonal equivalent frames intersecting at a column in two – way slab. Bonded reinforcement required shall be distributed between lines that are 1.5h outside opposite faces of the column support. At least four bars or wires shall be provided in each direction. Spacing of bonded reinforcement shall not exceed 300 mm.

5.

In positive moment areas, minimum length of bonded reinforcement shall be one – third the clear span length, ln , and centered in positive moment area.

6.

In negative moment areas, bonded reinforcements shall extend one – sixth the clear span, ln , on each side of the support

Reinforcement Index, 𝛚𝐩 Under-reinforced ρp fps 0.85a ωp = , < 0.36β1 fc ′ d C=T 0.85fc ′ab = Aps fps Aps fps a= 0.85fc ′b a Mu = ∅T d − 2 a Mu = ∅Aps fps d − 2

Over-reinforced ρp fps 0.85a ωp = , > 0.36β1 fc ′ d

Aps fps a= 0.85fc ′b ρp bdfps ρp fps a= ; ωp = 0.85fc ′b fc ′ ωp d a= 0.85 0.36β1 d a= 0.85 36 a= β d 85 1 a Mu = ∅C d − 2 a Mu = ∅0.85fc ′ab d − 2

Example 1: Determine the nominal resistance of a partially prestressed concrete rectangular section with dimension given in the figure, for f′c = 5,000 psi, β1 = 0.80, fr = -530.3 psi, fpu = 270 ksi, fpe = 0.55 fpu = 148 ksi, fpy = 0.85 fpu , fy = 60 ksi, Aps = 0.918 in2 or six ½ - in diameter stress – relieved prestressing strands, As = 1.2 in2 or two #7 reinforcing bars, dp = 20.75 in, ds = 21.5 in, b = 12 in. The prestressing tendons are assumed bonded.

BONDED TENDONS

Problem 01: The beam is prestressed by using bonded (stress relieved tendons) with a specified tensile strength of prestressing tendons fpu =1,350 MPa. f′c = 35 MPa. The area of prestressed reinforcement in tension zone Aps = 1,000 mm2 . The specified yield strength of prestressing tendons fpy = 1160 MPa. 0.85f′c

Determine the following:

b. c.

Value of the stress in prestressed reinforcement at nominal strength fps Value of the depth of the compression block. Ultimate moment capacity of the prestressed beam.

430

a.

a

C

N. A.

d−

T = Aps fps 250 All dimensions in mm

a 2

Part a:

For bonded tendons:

fpy 1,160 = fpu 1,350 fpy = 0.86 ≥ 0.85 fpu

0.05 β1 = 0.85 − fc ′ − 28 7 0.05 β1 = 0.85 − 35 − 28 7 β1 = 0.80

Use: 𝛄 = 𝟎. 𝟒𝟎

γp fpu d fps = fpu 1 − ρp + ω − ω′ β1 fc ′ dp γp fpu fps = fpu 1 − ρ +0 β1 p fc ′ 0.40 1,350 fps = 1,350 1 − (0.0093) 0.80 35 𝐟𝐩𝐬 = 𝟏, 𝟏𝟎𝟕. 𝟗 𝐌𝐏𝐚

Aps ρp = bd 1,000 ρp = 430(250) 𝛒𝐩 = 𝟎. 𝟎𝟎𝟗𝟑

Depth of Compression Block Aps fps = 0.85f′c abw Aps fps a= 0.85f′c bw 1,000(1,107.9) a= 0.85(35)(250) 𝐚 = 𝟏𝟒𝟖. 𝟗𝟔 𝐦𝐦

Under – reinforced d ωp + ω − ω′ < 0.36β1 dp a a a Mn = Aps fps dp − + As fy d − + A′s fy − d′ 2 2 2

Over – reinforced d ωp + ω − ω′ > 0.36β1 dp ′

Mu = ∅f c bdp

2

0.36β1 − 0.08β1

2

ρp fps ωp = f′c 0.0093(1,107.9) ωp = 35 ωp = 0.294 d ωp + ω − ω′ = 0.294 + 0 dp d ωp + ω − ω′ = 0.294 dp

0.36β1 = 0.36(0.80) 0.36β1 = 0.288

d Since ωp + ω − ω′ > 0.36β1 dp (over − reinforced)

Nominal flexural strength:

Mn = f ′ c bw dp 2 0.36β1 − 0.08β1 2 Mn = 35 250 430 2 0.36 0.80 − 0.08 0.80 𝐌𝐧 = 𝟑𝟖𝟑. 𝟏 𝐤𝐍 ∙ 𝐦

2

/106

Problem 02: The prestressed I – beam shown in cross – section with bonded tendons is pretensioned using seven ordinary stress – relieved strands Grade 250 (fpu = 1,728 MPa) carrying an effective prestress fpe = 988 MPa, fpy = 1480 MPa, f’c = = 27.6 MPa. Total depth of the beam = 600 mm Distance from center to tendons to the top of the beam = 440 mm Aps = 650 mm2 Average flange thickness = 150 mm Width of the flange = 300 mm Thickness of the web = 110 mm Determine the following: a.

b. c. d.

Value of the prestress in prestressed reinforcement at nominal strength. Nominal flexural strength of the beam Design strength of the beam Is the beam over – reinforced or under – reinforced

300

Material Properties 150 ave. 300

440

150

600 e 300

h = 600

mm

dp = 440

mm

Aps = 650

mm2

t f = 150

mm

bf = 300

mm

t w = 110

mm

f′c = 27.6

MPa

fpe = 988

MPa

fpy = 1,480

MPa

fpu = 1,728

MPa

Solution: 0.5fpu = 0.5(1,728) 0.5fpu = 864 MPa < (fpe = 988 MPa)

Therefore, use fps = fpu

γp fpu d 1− ρp + ω − ω′ β1 fc ′ dp

Aps ρp = bd 650 ρp = 300(440) 𝛒𝐩 = 𝟎. 𝟎𝟎𝟒𝟗𝟐𝟒 fpy 1480 = = 0.856 fpu 1728 Sincefpy /fpu > 0.85, use γ = 0.40

0.40 1,728 fps = 1,728 1 − 0.004924 +0 0.85 27.6 𝐟𝐩𝐬 = 𝟏, 𝟒𝟕𝟕. 𝟑𝟏 𝐌𝐏𝐚

Depth of Compression Block Check whether the stress block depth is greater or less than the average flange thickness of 150 mm.

Aps fps = 0.85f′c abw Aps fps a= 0.85f′c bw 650(1,477.31) a= 0.85(27.6)(300) a = 136.44 mm < 150 mm

Test the following (for Bonded Tendon 𝑎 ≤ ℎ𝑓 ) Under – reinforced d ωp + ω − ω′ < 0.36β1 dp Mn = Aps fps

a a a dp − + As fy d − + A′s fy − d′ 2 2 2

Over – reinforced d ωp + ω − ω′ > 0.36β1 dp ′

Mu = ∅f c bdp

2

0.36β1 − 0.08β1

2

ρp fps ωp = f′c 0.004924(1,477.31) ωp = 27.6 ωp = 0.263 d ωp + ω − ω′ = 0.263 + 0 dp d ωp + ω − ω′ = 0.263 dp 0.36β1 = 0.36(0.85) 0.36β1 = 0.306 d Sinceωp + ω − ω′ < 0.36β1 dp (under − reinforced)

Note: When stress block depth is less than the average flange thickness and is under – reinforced, use the formula:

Nominal flexural strength: Mn = Aps fps

a d− 2

136.44 Mn = (650)(1,477.31) 440 − 2 Mn = 357 × 106 N ∙ mm 𝐌𝐧 = 𝟑𝟓𝟕 𝐤𝐍 ∙ 𝐦 Flexural design strength: Mu = ∅Mn Mu = 0.90(357) 𝐌𝐮 = 𝟑𝟐𝟏. 𝟑 𝐤𝐍 ∙ 𝐦

Problem 03: The prestressed I – beam shown is pretensioned by using ordinary stress relieved strands having a specified tensile strength fpu = 1,750 MPa, fpy = 1,480 MPa.

Total depth of the beam = 600 mm Distance from center to tendons t the top of the beam = 440 mm Aps = 650 mm2 Average flange thickness = 127 mm Width of the flange = 300 mm Thickness of the web = 100 mm Determine the following: a.

b. c. d.

Value of the prestress in prestressed reinforcement at nominal strength. Nominal flexural strength of the beam Design strength of the beam Is the beam over – reinforced or under – reinforced?

300

Material Properties hf(ave) = 127 100

600

mm

d = 440

mm

Aps = 650

440

Aps = 650

h = 600

mm2

All dimensions in mm

mm2

t f = 127

mm

bf = 300

mm

t w = 100

mm

f′c = 27.6

MPa

fpe = 988

MPa

fpy = 1,480

MPa

fpu = 1,750

MPa

Part a: fpy 1480 = = 0.845 fpu 1750 fpy Since 0.80 < < 0.85, use γ = 0.55 fpu Aps ρp = bd 650 ρp = 300(440) 𝛒𝐩 = 𝟎. 𝟎𝟎𝟒𝟗 γp fpu d fps = fpu 1 − ρp + ω − ω′ β1 fc ′ dp 0.55 1,750 fps = 1,750 1 − 0.004924 +0 0.85 27.6 𝐟𝐩𝐬 = 𝟏, 𝟑𝟗𝟔. 𝟒𝟕 𝐌𝐏𝐚

Nominal flexural strength Check whether the stress block depth is greater or less than the average flange thickness of 150 mm.

Apf fps = 0.85f ′ c bw − t w hf 0.85f ′ c bw − t w hf Apf = fps 0.85(27.6) 300 − 100 (127) Apf = 1396.47 Apf = 426.70 mm2 Apw = Aps − Apf Apw = 650 − 426.70 Apw = 223.3 mm2

Apw fps = 0.85f′c abw Apw fps a= 0.85f′c bw 223.3(1,396.47) a= 0.85(27.6)(100) a = 132.92 mm > 127 mm

Test the following (for Bonded Tendon 𝑎 > ℎ𝑓 ) Under – reinforced d ωpw + ωw − ωw ′ < 0.36β1 dp

Mn = Apw fps

Apw = Aps fps + As fy − 0.85f ′ c b − bw hf a hf ′ dp − + As fy d − dp + 0.85f c b − bw hf dp − 2 2

Over – reinforced d ωpw + ωw − ωw ′ > 0.36β1 dp

Mn = f ′ c bw dp 2 0.36β1 − 0.08β1 2 + 0.85𝑓 ′ 𝑐 𝑏 − 𝑏𝑤 ℎ𝑓 (𝑑𝑝 − 0.5ℎ𝑓 )

ρp fps ωp = f′c 0.0049 (1,396.47) ωp = 27.6 ωp = 0.248 d ωp + ω − ω′ = 0.248 + 0 dp d ωp + ω − ω′ = 0.248 dp 0.36β1 = 0.36(0.85) 0.36β1 = 0.306

d ω − ω′ < 0.36β1 dp (under − reinforced)

Since ωp +

Note: When stress block depth is greater than the average flange thickness and is over – reinforced, use the formula: a hf ′ Mn = Apw fps dp − + As fy d − dp + 0.85f c b − bw hf dp − 2 2 132.92 Mn = 223.3 1,396.47 440 − +0 2 127 + 0.85(27.6) 300 − 100 (127) 440 − 2 𝐌𝐧 = 𝟑𝟒𝟎. 𝟖𝟑 𝐤𝐍 ∙ 𝐦 Flexural design strength: Mu = ∅Mn Mu = 0.90(340.83) 𝐌𝐮 = 𝟑𝟎𝟔. 𝟕𝟓 𝐤𝐍 ∙ 𝐦

Problem 04: A prestressed beam is reinforced with a bonded tendon having an area Ap𝑠 = 1,580 mm2 , f′c = 34.5 MPa, fpu =1,862 MPa, and the effective stress after losses fpe = 1102 MPa. fpy = 1,763 MPa. Span of beam is 6 m. Determine the following: Value of the stress in prestressed reinforcement at nominal strength fps Ultimate moment capacity Live load 1,220

N. A.

0.85f′c a

80 600

a. b. c.

C a d− 2

Aps T = Aps fps 250 All dimensions in mm

Part a:

fpy 1713 = fpu 1862 fpy = 0.92 ≥ 0.90 fpu Use: 𝛄 = 𝟎. 𝟐𝟖 Check fse 0.5fpy = 0.5(1,862) 0.5fpy = 931 MPa < fse Aps ρp = bd 1,580 ρp = 1220(600) 𝛒𝐩 = 𝟎. 𝟎𝟎𝟐𝟏𝟔

For bonded tendons: 0.05 β1 = 0.85 − fc ′ − 28 7 0.05 β1 = 0.85 − 34.5 − 28 7 β1 = 0.8036

γp fpu d fps = fpu 1 − ρp + ω − ω′ β1 fc ′ dp γp fpu fps = fpu 1 − ρ +0 β1 p fc ′ 0.28 1,862 fps = 1,862 1 − (0.00216) 0.8036 34.5 𝐟𝐩𝐬 = 𝟏, 𝟕𝟖𝟔. 𝟒 𝐌𝐏𝐚

Part b: Checking the depth of compression block:

C=T 0.85f′c ab = Aps fps (1,580)(1,786) a= 0.85(34.5)(1,220) a = 78.9 mm < 90 mm (ok!) ρp fps ωp = f′c 0.00216(1,786.4) ωp = 34.5 ωp = 0.118

d ωp + ω − ω′ = 0.118 + 0 dp d ωp + ω − ω′ = 0.118 dp

0.36β1 = 0.36(0.8036) 0.36β1 = 0.289

d Since ωp + ω − ω′ < 0.36β1 dp (under − reinforced)

Ultimate moment capacity: a Mu = ∅T d − 2 a Mu = 0.90Aps fps d − 2 Mu = 0.90(1,580)(1,786) 600 −

78.9 2

Mu = 1,424 × 106 N ∙ mm

Part c: wu L2 Mu = 8 8Mu wu = 2 L 1,424 8 wu = 62 𝐰𝐮 = 𝟑𝟏𝟔. 𝟒𝟒 𝐤𝐍/𝐦

wDL = [1.22 0.08 + 0.25 0.6 ](2.4)(9.8) 𝐰𝐃𝐋 = 𝟓. 𝟖𝟑 𝐤𝐍/𝐦 wu = 1.2wDL + 1.6wLL 316.44 = 1.2(5.83) + 1.6wLL 𝐰𝐋𝐋 = 𝟏𝟗𝟑. 𝟒 𝐤𝐍/𝐦

UNBONDED TENDONS

Problem 05: A 250 mm × 500 mm prestressed beam has a simple span of 6 m. The beam uses stress relieved tendons with fpu = 1,350 MPa and an effective stress in prestressed reinforcement after allowable for all prestress losses fse = 760 MPa. The prestressed beam is not to be grouted (unbonded) after the application of the prestress. If the area of prestress reinforcement in tension zone is Aps = 1,000 mm2 applied 70 mm above the bottom of the beam and f’c = 35 MPa, determine the following: 0.85f′c

b.

c.

Reinforcement indices for the prestressed beam Safe ultimate moment capacity of the beam Safe live load that the prestressed beam could carry if concrete weighs 23.54 kN/m3.

a

C

430

a.

d− Aps = 1,000 mm2

T = Aps fps 250 All dimensions in mm

a 2

ρp fps ωp = f′c 0.0093(868) ωp = 35 ωp = 0.231

Aps ρp = bd 1,000 ρp = 250(430) 𝛒𝐩 = 𝟎. 𝟎𝟎𝟗𝟑

0.05 fc ′ − 28 7 0.05 β1 = 0.85 − 35 − 28 7 β1 = 0.80

L 6,000 = d 500 − 70 L = 13.95 < 35 d

β1 = 0.85 −

Use:

fps

f ′c = fse + 70 + 100ρp

fps = 760 + 70 + 𝐟𝐩𝐬 = 𝟖𝟔𝟖 𝐌𝐏𝐚

35 100(0.0093)

Limiting value of ωp

0.36β1 = 0.36(0.80) 0.36β1 = 0.288 ωp = 0.231 < 0.36β1 (under − reinforced)

Part b: Safe ultimate moment capacity of the beam

Depth of compression block: C=T 0.85f′c ab = Aps fps (1,000)(868) a= 0.85(35)(250) 𝐚 = 𝟏𝟏𝟔. 𝟕𝟏 𝐦𝐦

Ultimate moment capacity: a Mu = ∅T d − 2 a Mu = 0.90Aps fps d − 2

Mu = 0.90 1,000 868 𝐌𝐮 = 𝟐𝟗𝟎. 𝟑 𝐤𝐍 ∙ 𝐦

116.71 430 − /106 2

Part c: Safe live load wu L2 Mu = 8 wu 6 2 290.3 = 8 𝐰𝐮 = 𝟔𝟒. . 𝟓𝟏 𝐤𝐍/𝐦 wDL = (0.20)(0.5)(1)(23.54) 𝐰𝐃𝐋 = 𝟐. 𝟗𝟒 𝐤𝐍/𝐦 wu = 1.2wDL + 1.6wLL 64.51 = 1.2(2.94) + 1.6wLL 𝐰𝐋𝐋 = 𝟑𝟖. 𝟏𝟏𝐤𝐍/𝐦

Problem 06: The I – beam with flange width of 460 mm and a thickness of 175 mm has a total depth of 900 mm. Thickness of web is 140 mm. The prestressed beam is provided by unbonded tendons with an effective prestress after all losses fse = 1,100 MPa, fpy = 1,500 MPa, f’c = 36 MPa. Determine the following: a.

b.

c.

Stress in prestressed reinforcement at nominal strength if it has a span of 28 m. Number of 16 mm ∅ bonded steel reinforcement needed. Place the steel bars 50 mm above the bottom of the beam. Ultimate moment capacity of the beam if it is supplied with an unbonded tendons placed at 115 mm above the bottom of the beam in addition to that of the steel reinforcement needed. Aps = 1,750 mm2.

460 175

550

140

d = 850

175

50

All dimensions in mm

Part a:

Aps ρp = bd 1,750 ρp = 460(850) 𝛒𝐩 = 𝟎. 𝟎𝟎𝟒𝟒𝟖 fps

f ′c = fse + 70 + 100ρp

36 fps = 1,100 + 70 + 100(0.00448) 𝐟𝐩𝐬 = 𝟏𝟐𝟓𝟎. 𝟒 𝐌𝐏𝐚 < 𝐟𝐩𝐲 (𝐨𝐤!)

fse + 415 = 1,515

𝐟𝐩𝐬 < 𝐟𝐬𝐞 + 𝟒𝟏𝟓 (𝐨𝐤!) Use: 𝐟𝐩𝐬 = 𝟏𝟐𝟓𝟎. 𝟒 𝐌𝐏𝐚

Part b: A = 460 − 140 175 + 450(140) A = 119,000 mm2 As = 0.004A As = 0.004(119,000) As = 476 mm2

460 175 450 550

Using 16 mm ∅ As N= Ao 476 N=π 2 16 4 N = 2.36 𝐍 = 𝟑 𝐛𝐚𝐫𝐬

N. A. 140 450

175

50 All dimensions in mm

460 0.85f′c C1

C2

a 785 − 2

a 850 − 2

a

175

550

785 140 Aps

175

0.85f′c

60 65 50

As

T1 = Aps fps

T2 = As fy

All dimensions in mm

T1 = Aps fps T1 = 1,750(1,250.4) 𝐓𝟏 = 𝟐, 𝟏𝟖𝟖, 𝟐𝟎𝟎 𝐍

T2 = As fy

π 16 2 (415) 4 𝐓𝟐 = 𝟐𝟓𝟎, 𝟑𝟐𝟐 𝐍 T2 = 3 ×

T = T1 + T2 T = 2,188,200 + 250,322 𝐓 = 𝟐, 𝟒𝟑𝟖, 𝟓𝟐𝟐 𝐍

460 175

550

C = 0.85f′c abf

785 − 𝑎/2

a

785 140 Aps

175

60 65 50

As

850 − 𝑎/2

0.85f′c

T1 = Aps fps T2 = As fy

All dimensions in mm

T1 = Aps fps T1 = 1,750(1,250.4) 𝐓𝟏 = 𝟐, 𝟏𝟖𝟖, 𝟐𝟎𝟎 𝐍

T2 = As fy

π 16 2 (415) 4 𝐓𝟐 = 𝟐𝟓𝟎, 𝟑𝟐𝟐 𝐍 T2 = 3 ×

T = T1 + T2 T = 2,188,200 + 250,322 𝐓 = 𝟐, 𝟒𝟑𝟖, 𝟓𝟐𝟐 𝐍

Depth of compression block:

C=T 0.85f′c abf = T

2,438,522 a= 0.85(36)(460) a = 173.2 mm < 175 mm (ok!)

Ultimate moment capacity: a a Mu = ∅T1 d1 − + ∅T2 d2 − 2 2 173.2 0.90 2,188,200 785 − + 2 Mu = 173.2 0.90 250,322 850 − 2 Mu = 1,547 × 106 N ∙ mm 𝐌𝐮 = 𝟏, 𝟓𝟒𝟕 𝐤𝐍 ∙ 𝐦