Prestressed-Concrete Girder Bridges — Bridge Engineering & Design

Engineering story

Squeezing concrete so it never sees tension

Concrete is roughly ten times stronger in compression than in tension. Ordinary reinforced concrete accepts that tension will crack the section and lets the steel pick it up — that works, but service-load cracks admit water and chlorides, and the section stiffness drops the moment the first crack opens. Prestressing solves the same problem differently: we pre-compress the concrete with high-strength strand so that under full service loads the extreme tension fiber still sees compression, or at worst a very small controlled tension. The section stays uncracked, the girder stays stiff, and spans that would be impossible in RC (100–200 ft simple, 300 ft+ continuous) become routine.

This chapter is the AASHTO design workflow for two families: pretensioned precast girders (AASHTO Type II–VI, bulb-tee, box) and post-tensioned cast-in-place or spliced girders. We walk the four stages a girder experiences — at transfer, after all losses, at service with deck and live load, and at ultimate — and check each against AASHTO §5. Every calc follows the Formula → Substitute → Result pattern.

Chapter objectives

What you will be able to do

Learning objectives

By the end of this chapter you will be able to:

1Distinguish pretensioned vs post-tensioned construction and select the appropriate system for a given span, depth, and site.
2Compute all AASHTO §5.9 prestress losses: elastic shortening, shrinkage, creep, steel relaxation (short and long form).
3Check the AASHTO §5.9.2.3 stress limits at transfer and at service (girder-only and composite).
4Design flexural strand pattern (straight + harped or debonded) for a simple-span pretensioned girder.
5Compute nominal flexural resistance φMₙ using the strain-compatibility / bonded-tendon equations of §5.6.3.1.
6Design transverse shear reinforcement using AASHTO §5.7.3 (MCFT), including the harped-strand Vₚ contribution.
7Design post-tensioned anchorage-zone bursting reinforcement per §5.9.5.6.
8Estimate camber and deflection for release, deck placement, and long-term service.

7.1 — Prestressing systems

Pretension vs post-tension

AASHTO LRFD §5.4.4, §5.9

Two industrial processes produce a prestressed girder. In pretensioning, high-strength strand (typically Grade 270, 7-wire, 0.5 in or 0.6 in diameter) is stressed against fixed abutments in a long precast bed, concrete is cast around the strand, and once concrete reaches release strength f'_ci the strand is flame-cut. The prestress transfers to the concrete by bond over a transfer length of about 60 strand diameters. In post-tensioning, the concrete is cast first with hollow ducts in place; after curing the tendons are threaded, jacked from an anchorage plate, locked off with wedges, and the ducts are pressure-grouted.

Comparison of pretensioned casting-bed process and post-tensioned duct-and-anchorage process — Figure 7.1Pretensioning (left) stresses strand before concrete is cast; post-tensioning (right) stresses tendons after concrete has cured. Pretensioning gives the plant repeatability of a factory product; post-tensioning gives the site flexibility to build long continuous spans.

Attribute	Pretensioned precast	Post-tensioned (CIP or spliced)
Typical span range	40 – 160 ft	120 – 300 ft (continuous)
Where prestress is applied	Precast plant	On site, after casting
Load transfer mechanism	Bond over transfer length ≈ 60·d_b	End anchorage plates + wedges
Common girder shapes	AASHTO Type II–VI, bulb-tee, box	Spliced bulb-tee, segmental box
Tendon profile	Straight + harped (2 hold-downs)	Fully draped (parabolic)
Continuity over piers	Simple-span, made-continuous via deck	Fully continuous PT tendon

7.2 — Materials

Strand, concrete, and stress limits

AASHTO LRFD §5.4.4, §5.9.2

The workhorse strand is Grade 270 low-relaxation 7-wire strand, f_pu = 270 ksi, f_py = 243 ksi, area A_ps = 0.153 in² per 1/2 in strand or 0.217 in² per 0.6 in strand. The concrete used for pretensioned girders is high early-strength — release strengths f'_ci of 5.5 to 7 ksi at 18 hours enable a one-day casting cycle, with 28-day strengths f'_c of 8 to 10 ksi.

Jacking stress limits AASHTO LRFD §5.9.2.2

(7.1)

Limit
$f_{pj} \le 0.75 \cdot f_{pu} = 0.75 \cdot 270$
Result
$f_{pj} \le 202.5\ \text{ksi}$

$f_{pj}$: jacking stress in strand
$f_{pu}$: ultimate tensile strength (270 ksi)

Concrete stress limits at transfer (girder only) AASHTO LRFD §5.9.2.3.1

(7.2)

Compression
$f_{ci} \le 0.60 \cdot f'_{ci}$
Tension (bonded reinf. present)
$f_{ti} \le 0.24 \cdot \sqrt{f'_{ci}}$

$f_{ci}$: compressive stress at transfer [ksi]
$f_{ti}$: tensile stress at transfer [ksi]
$f'_{ci}$: release concrete strength [ksi]

Concrete stress limits at service (composite) AASHTO LRFD §5.9.2.3.2

(7.3)

Compression (top, sustained)
$f_{cs} \le 0.45 \cdot f'_{c}$
Tension (bottom, Service III)
$f_{ts} \le 0.19 \cdot \sqrt{f'_{c}}$

$f_{cs}$: top-fiber compression, all sustained loads [ksi]
$f_{ts}$: bottom-fiber tension, Service III [ksi]

AASHTO §5.9.2.3 — the four checks that dominate design

A prestressed girder is dimensioned by stress, not by strength. If the four extreme-fiber stresses (top-transfer, bottom-transfer, top-service, bottom-service) pass §5.9.2.3, φMₙ almost always covers Strength I with comfortable margin. Design the strand pattern to satisfy stress first; check flexural strength second.

7.3 — Prestress losses

Where the 20% goes

AASHTO LRFD §5.9.3

The strand does not deliver its full jacking force to the concrete. Between the jack and the finished bridge, roughly 18–25% of the stress is lost to five mechanisms. AASHTO §5.9.3 groups them as immediate (elastic shortening, anchorage seating for PT, friction for PT) and time-dependent (shrinkage of the concrete, creep of the concrete under sustained prestress, and relaxation of the steel).

Prestress losses timeline for a pretensioned girder from jacking to service — Figure 7.2Prestress losses timeline for a typical pretensioned girder. Immediate losses (elastic shortening) occur at strand release; time-dependent losses (shrinkage, creep, relaxation) develop over the first year and effectively stabilize by ~5 years. Total loss for a well-designed pretensioned girder is 15–20% of f_pj.

Elastic shortening loss AASHTO LRFD §5.9.3.2.3a

(7.4)

Formula
$\Delta f_{pES} = \left( \frac{E_{p}}{E_{ci}} \right) \cdot f_{cgp}$

$\Delta f_{pES}$: loss due to elastic shortening [ksi]
$E_{p}$: strand modulus (28,500 ksi)
$E_{ci}$: concrete modulus at release
$f_{cgp}$: concrete stress at strand centroid due to prestress + self-weight [ksi]

Approximate long-term losses (§5.9.3.3, no deck)

(7.5)

Formula
$\Delta f_{pLT} = 10.0 \cdot \left( \frac{f_{pi} \cdot A_{ps}}{A_{g}} \right) \cdot \gamma_{h} \cdot \gamma_{st} + 12.0 \cdot \gamma_{h} \cdot \gamma_{st} + 2.4$

$\Delta f_{pLT}$: long-term loss (shrinkage + creep + relaxation) [ksi]
$\gamma_{h}$: humidity factor = 1.7 − 0.01·H (H in %)
$\gamma_{st}$: strength factor = 5 / (1 + f'_ci)

Which loss equation to use

AASHTO gives both a refined method (§5.9.3.4, time-step for shrinkage/creep, useful for spliced PT with staged construction) and the approximate estimate above (§5.9.3.3, one-shot number for standard precast pretensioned girders). Use the approximate method for feasibility and preliminary design; the refined method for spliced PT and when camber tolerance is tight.

7.4 — Stress at each stage

Four snapshots of one girder

AASHTO LRFD §5.9.2.3, §5.6.3

A prestressed composite girder passes through four discrete stress states, and each must be individually checked. The section properties change midway (from girder only to girder + composite deck), so section moduli must be recomputed for the composite state.

Stress blocks at midspan of a prestressed composite girder at transfer, after losses, at Service III, and at Strength I — Figure 7.3Extreme-fiber stress blocks at midspan for the four AASHTO design stages. Stage 1 (transfer) is the critical top-tension check; Stage 3 (Service III) is the critical bottom-tension check; Stage 4 (Strength I) is a Whitney-block flexural strength check with the prestressing steel acting as tension reinforcement.

Extreme-fiber stress at any stage

(7.6)

Formula
$f_{t} = -\frac{P_{e}}{A_{g}} + \frac{P_{e} \cdot e_{c}}{S_{t}} + \frac{M_{g}}{S_{t}} + \frac{M_{deck} + M_{ws}}{S_{c,t}} + \frac{M_{LL+IM}}{S_{c,t}}$

$f_{t,b}$: top / bottom extreme-fiber stress (- comp, + tens) [ksi]
$P_{e}$: effective prestress force (after losses) [kip]
$e_{c}$: strand eccentricity from girder neutral axis [in]
$S_{t,b}$: girder section modulus, top / bottom [in³]
$S_{c,t,b}$: composite section modulus [in³]

7.5 — Flexural resistance

Bonded tendon at ultimate

AASHTO LRFD §5.6.3

At ultimate the strand is far above yield but not at f_pu. AASHTO §5.6.3.1.1 gives the stress in bonded prestressing steel at nominal flexural resistance, with the Whitney block on the compression side.

(7.7)

Steel stress at nominal moment
$f_{ps} = f_{pu} \cdot \left( 1 - k \cdot \frac{c}{d_{p}} \right)$

$f_{ps}$: prestressing steel stress at nominal moment [ksi]
$k$: = 2·(1.04 − f_py / f_pu); 0.28 for Grade-270 low-relax
$c$: distance from extreme compression fiber to neutral axis [in]
$d_{p}$: depth from compression fiber to strand centroid [in]

(7.8)

Neutral-axis depth
$c = \frac{A_{ps} \cdot f_{pu}}{0.85 \cdot f'_{c} \cdot \beta_{1} \cdot b + k \cdot A_{ps} \cdot f_{pu} / d_{p}}$

$c$: neutral-axis depth [in]
$A_{ps}$: total area of prestressing steel [in^{2}]
$b$: effective flange width (composite deck) [in]
$\beta_{1}$: Whitney factor, 0.85 for f'_c ≤ 4 ksi, − 0.05 per ksi above 4

(7.9)

Nominal flexural resistance
$M_{n} = A_{ps} \cdot f_{ps} \cdot \left( d_{p} - \frac{a}{2} \right)$

$M_{n}$: nominal flexural resistance [kip-in]
$a$: = β_1 · c, Whitney block depth [in]

7.6 — Shear (MCFT)

Diagonal-strut model with a helping hand from harped strand

AASHTO LRFD §5.7.3

AASHTO §5.7.3 uses the modified compression field theory. The nominal shear resistance is the sum of a concrete contribution, a stirrup contribution, and — for pretensioned girders with harped strand — the vertical component of the prestressing force at the section, V_p.

MCFT shear model with concrete strut, transverse stirrups, and harped-strand vertical component — Figure 7.4MCFT shear model for a prestressed girder. Diagonal concrete struts (θ ≈ 29°) carry V_c, stirrups spaced s carry V_s, and the upward vertical component of the harped strand V_p = P·sin(ψ) directly offsets a portion of the applied shear.

(7.10)

Formula
$V_{n} = V_{c} + V_{s} + V_{p} = 0.0316 \cdot \beta \cdot \sqrt{f'_{c}} \cdot b_{v} \cdot d_{v} + \frac{A_{v} \cdot f_{y} \cdot d_{v} \cdot \cot(\theta)}{s} + V_{p}$

$V_{n}$: nominal shear resistance [kip]
$\beta$: concrete tension carrying factor (§5.7.3.4)
$\theta$: diagonal-strut angle (§5.7.3.4)
$b_{v}$: effective web width [in]
$d_{v}$: effective shear depth [in]

7.7 — Worked example 1

AASHTO Type-VI pretensioned girder — Interior girder, 130 ft simple span

Problem statement

A single-span, 4-girder interior girder from a 130 ft roadway crossing. Design the interior girder for Strength I flexure, Service III bottom tension, transfer / lifting stresses, and Strength I shear.

Given

SpanL = 130 ft simple span (4 girders)
GirderAASHTO Type VI, depth 72 in
A_g1,085 in²
I_g733,320 in⁴
y_b36.4 in
Deck8 in composite CIP deck
Girder spacingS = 10 ft o.c.
f'_c (girder)8.0 ksi
f'_c (deck)4.0 ksi
f'_ci (release)6.0 ksi
M_g1,420 k-ft (girder self-weight)
M_deck1,140 k-ft
M_ws240 k-ft
M_LL+IM2,350 k-ft (interior DF applied)

Plan and typical section of the 130 ft simple-span, 4-girder Type-VI example bridge — Figure 7.5Example 1 — 130 ft simple span, 4 AASHTO Type-VI girders at S = 10 ft, 8-in composite deck. Design the interior girder for Strength I flexure, Service III bottom tension, and Strength I shear.

AASHTO Type-VI cross-section with 44 straight and 4 harped 1/2-in strands — Figure 7.6Trial strand pattern: 44 straight 1/2-in Grade-270 low-relax strands on a 2 in grid in the bottom flange, plus 4 harped strands (hold-downs at 0.4L / 0.6L). Total A_ps = 48 · 0.153 = 7.34 in².

Step 1 — Strand centroid and eccentricity

Formula — centroid of straight strand block

y_bar,str = (Σ n_i · y_i) / (Σ n_i)

Substitute

y_bar,str = [8·2 + 10·4 + 10·6 + 8·8 + 6·10 + 2·12] / 44 = 5.55 in from soffit

Result

= e_c (straight only) = y_b − y_bar,str = 36.4 − 5.55 = 30.85 in

Formula — with 4 harped strands raised to 66 in at midspan

e_c,eff = (44·30.85 + 4·(66 − 36.4)) / 48

Substitute

e_c,eff = (44·30.85 + 4·29.6) / 48 = (1357.4 + 118.4) / 48

Result

= e_c,eff = 30.7 in at midspan

Step 2 — Effective prestress after all losses

Formula — initial force before losses

P_i = 0.75 · f_pu · A_ps

Substitute

P_i = 0.75 · 270 · 7.34

Result

= P_i = 1486 kip

Formula — approximate long-term losses (§5.9.3.3)

Δf_pLT = 10·(f_pi · A_ps / A_g)·γ_h·γ_st + 12·γ_h·γ_st + 2.4

Substitute

γ_h = 1.7 − 0.01·70 = 1.00; γ_st = 5/(1+6.0) = 0.714; f_pi = 202.5 ksi

Result

= Δf_pLT = 10·(202.5·7.34/1085)·1.00·0.714 + 12·1.00·0.714 + 2.4 = 9.78 + 8.57 + 2.4 = 20.75 ksi

Formula — elastic shortening (Δf_pES) taken as 13.5 ksi from Eq. 7.4

f_pe = f_pj − Δf_pES − Δf_pLT

Substitute

f_pe = 202.5 − 13.5 − 20.75

Result

= f_pe = 168.25 ksi → P_e = 168.25 · 7.34 = 1235 kip (17% total loss)

Step 3 — Stress check at transfer (top fiber, midspan)

Formula — top-fiber stress at transfer, girder only

f_t,i = − P_i / A_g + P_i · e_c / S_t − M_g / S_t

Substitute

S_t = I_g / (h − y_b) = 733,320 / (72 − 36.4) = 20,600 in³. Use P_i,after-ES = 1352 kip. M_g = 1420·12 = 17,040 k-in

Result

= f_t,i = −1352/1085 + 1352·30.7/20,600 − 17,040/20,600 = −1.246 + 2.015 − 0.827 = −0.058 ksi (compression) ✓

Transfer top-fiber tension limit §5.9.2.3.1

Allowable tension with bonded reinforcement = 0.24·√6.0 = 0.588 ksi. Computed −0.058 ksi (compression) ≪ 0.588 ksi. ✓

Step 4 — Stress check at transfer (bottom fiber, midspan)

Formula — bottom-fiber stress at transfer

f_b,i = − P_i / A_g − P_i · e_c / S_b + M_g / S_b

Substitute

S_b = I_g / y_b = 733,320 / 36.4 = 20,146 in³

Result

= f_b,i = −1.246 − 1352·30.7/20,146 + 17,040/20,146 = −1.246 − 2.060 + 0.846 = −2.46 ksi

Transfer bottom-fiber compression limit §5.9.2.3.1

Allowable compression = 0.60·f'_ci = 0.60·6.0 = 3.60 ksi. Computed −2.46 ksi < 3.60 ksi. ✓

Step 5 — Stress check at Service III (bottom fiber, midspan)

Formula — Service III bottom fiber, composite

f_b,sIII = − P_e/A_g − P_e·e_c/S_b + (M_g + M_deck)/S_b + (M_ws + 0.8·M_LL+IM)/S_c,b

Substitute

S_c,b (composite) = 28,900 in³. M_g+M_deck = 2560·12 = 30,720 k-in. M_ws + 0.8·M_LL+IM = (240 + 0.8·2350)·12 = 2120·12 = 25,440 k-in

Result

= f_b,sIII = −1235/1085 − 1235·30.7/20,146 + 30,720/20,146 + 25,440/28,900 = −1.138 − 1.882 + 1.525 + 0.880 = −0.615 ksi (compression) ✓

Service III bottom-fiber tension limit §5.9.2.3.2b

Allowable tension = 0.19·√8.0 = 0.537 ksi. Computed −0.615 ksi (compression) ≪ 0.537 ksi. ✓ (Section stays uncracked at Service III.)

Step 6 — Nominal flexural resistance φMₙ

Formula — neutral-axis depth (Eq. 7.8), b = b_eff = 120 in, β_1 = 0.65 for f'_c,deck

c = A_ps · f_pu / [0.85 · f'_c,deck · β₁ · b + k · A_ps · f_pu / d_p]

Substitute

k = 0.28; d_p = 5.25·12 + 8 − 5.5 = 65.5 in; c = 7.34·270 / [0.85·4.0·0.85·120 + 0.28·7.34·270/65.5]

Result

= c = 1981.8 / [346.8 + 8.48] = 5.58 in → a = β₁·c = 0.85·5.58 = 4.74 in ≤ 8 in deck (rectangular)

Formula — f_ps (Eq. 7.7)

f_ps = f_pu · (1 − k · c / d_p)

Substitute

f_ps = 270 · (1 − 0.28·5.58/65.5)

Result

= f_ps = 270 · 0.9761 = 263.5 ksi

Formula — nominal moment (Eq. 7.9)

M_n = A_ps · f_ps · (d_p − a/2)

Substitute

M_n = 7.34 · 263.5 · (65.5 − 4.74/2) = 7.34 · 263.5 · 63.13

Result

= M_n = 122,100 k-in = 10,175 k-ft → φM_n = 1.00 · 10,175 = 10,175 k-ft

Formula — Strength I moment

M_u = 1.25·(M_g + M_deck) + 1.50·M_ws + 1.75·M_LL+IM

Substitute

M_u = 1.25·(1420 + 1140) + 1.50·240 + 1.75·2350

Result

= M_u = 3200 + 360 + 4112 = 7672 k-ft → φM_n = 10,175 ≥ 7672 k-ft ✓

Step 7 — Shear at h/2 from support (critical section)

Formula — harped-strand vertical component (Eq. §5.7.3.4.2)

V_p = 4·A_strand·f_pe · sin(ψ)

Substitute

4 harped strands rise from y = 5.5 in (end) to y = 66 in (midspan) over 0.4·L = 52 ft; sin(ψ) = 60.5/(52·12) = 0.097

Result

= V_p = 4·0.153·168.25·0.097 = 10.0 kip

Formula — Strength I shear at h/2 from an HL-93 line-load run

V_u = 1.25·V_DC + 1.50·V_DW + 1.75·V_LL+IM

Substitute

V_u = 1.25·124 + 1.50·17 + 1.75·132 = 155 + 26 + 231

Result

= V_u = 412 kip

Formula — MCFT concrete + steel + prestress (Eq. 7.10) with β = 4.8, θ = 29°

V_n = 0.0316·β·√f'_c·b_v·d_v + A_v·f_y·d_v·cot(θ)/s + V_p

Substitute

b_v = 8 in, d_v = max(0.9·d_p, 0.72·h) = 58.9 in. Try #5 @ 6 in (A_v = 2·0.31 = 0.62 in²). V_c = 0.0316·4.8·√8.0·8·58.9 = 202 kip; V_s = 0.62·60·58.9·cot(29°)/6 = 660 kip

Result

= V_n = 202 + 660 + 10.0 = 872 kip; φV_n = 0.90·872 = 785 kip ≥ V_u = 412 kip ✓ (spacing controlled by max s = 0.4·d_v = 24 in — use #5 @ 12 in over most of the girder, #5 @ 6 in in end 10 ft)

Design summary — Example 1

Strand: 44 straight + 4 harped 1/2-in Grade-270 low-relax. Stress at transfer, Service III, and flexural strength all pass with margin. Shear reinforcement: #5 double-leg stirrups at 6 in in end region, 12 in in interior. Estimated camber at release ≈ 2.1 in upward; net long-term deflection after deck placement ≈ 0.4 in downward, well within L/800 = 1.95 in.

Final section detailing (from computed A_s)

AASHTO Type VI interior girder — 72 in deep, S = 10 ft, L = 130 ft

Location	A_s required	Bars provided	Spacing / detail
Bottom flange — straight strands	A_ps,req ≈ 6.45 in²	44 × ½-in Grade-270 LR (A_ps = 6.73 in²)	2-in grid, first row 2 in from soffit
Bottom flange — harped strands	Uplift for transfer / lifting stresses	4 × ½-in Grade-270 LR (A_ps = 0.61 in²)	Hold-downs at 0.4L / 0.6L; e_end ≈ 15 in
Web shear — end 10 ft each end	A_v/s ≥ 0.10 in²/in	#5 double-leg closed stirrups (A_v = 0.62 in²)	s = 6 in c/c
Web shear — interior 110 ft	A_v/s ≥ 0.05 in²/in	#5 double-leg closed stirrups	s = 12 in c/c (≤ 0.4·d_v = 24 in ✓)
Top flange — temperature / shrinkage	0.11 in²/ft each face (§5.10.6)	#4 @ 12 in each face, longitudinal	1½-in clear cover

Total A_ps = 48 × 0.153 = 7.34 in² at centroid e_c ≈ 28.1 in from section CG. Bar clearances satisfy §5.10.3 (min 1 in) and §5.10.4 (max 18 in). End-region confinement per §5.10.10.1 (spiral or hoops within h/4 from girder end).

7.8 — Worked example 2

Spliced post-tensioned bulb-tee — end-zone anchorage design

Problem statement

For a 2 × 160 ft continuous spliced BT-72 bridge (see Design Challenge, Figure 7.8), the end anchorage of the continuous PT tendon lands on a solid end block. Design the general-zone bursting reinforcement.

Given

Bridge2 × 160 ft continuous spliced BT-72
Tendon19 – 0.6 in Ø strands per duct
Duct4 in Ø corrugated PT duct
Jacking stress0.80·f_pu at anchor
Lock-off stress0.70·f_pu after seating
End blockSolid, h = 72 in, b_w = 8 in
Anchors per end4 tendons in vertical group
f'_ci6.0 ksi at stressing
f_y60 ksi (Grade 60)

Post-tensioned anchorage zone with bursting force, bursting depth, and confinement spiral — Figure 7.7Post-tensioned anchorage zone. The concentrated jacking force P_j spreads through the end block, creating a compressive-strut fan and a transverse tensile bursting force T_burst that must be resisted by closely spaced spiral or hoop reinforcement inside the bursting depth d_burst.

Step 1 — Jacking force per tendon

Formula — jacking force

P_j = 0.80 · f_pu · A_ps,tendon

Substitute

A_ps,tendon = 19 · 0.217 = 4.12 in²; P_j = 0.80·270·4.12

Result

= P_j = 890 kip per tendon

Step 2 — Bursting force (§5.9.5.6.3a)

Formula — bursting tension resultant

T_burst = 0.25 · Σ P_j · (1 − a/h) + 0.5 · | Σ P_j · sin(α) |

Substitute

4 tendons in group; anchor plate a = 12 in on section h = 72 in; α = 0 (straight lift-off); ΣP_j = 4·890 = 3560 kip

Result

= T_burst = 0.25 · 3560 · (1 − 12/72) + 0 = 0.25 · 3560 · 0.833 = 742 kip

Step 3 — Bursting reinforcement area (§5.9.5.6.3b)

Formula — required steel area, φ = 0.85, f_y = 60 ksi (limit to 20 ksi effective stress)

A_s,burst ≥ T_burst / (φ · f_s) with f_s ≤ 20 ksi

Substitute

A_s,burst ≥ 742 / (0.85 · 20)

Result

= A_s,burst = 43.6 in²

Step 4 — Distribution over bursting depth

Formula — bursting depth (§5.9.5.6.3c)

d_burst = 0.5 · (h − 2e)

Substitute

e = 6 in (centroid of tendon group above section CG); d_burst = 0.5·(72 − 12)

Result

= d_burst = 30 in

Formula — hoop / stirrup count required within d_burst

n_hoop = A_s,burst / A_bar,hoop

Substitute

Use #6 double-leg closed stirrups (A_bar,hoop = 2·0.44 = 0.88 in²)

Result

= n_hoop = 43.6 / 0.88 = 50 stirrups → provide #6 @ 3 in over the 30-in bursting depth (10 stirrups per anchor × 5 anchors, plus spalling steel of 2% ΣP_j = 71 kip near the loaded face)

Anchorage-zone design summary

Provide a 30-in deep end block with #6 closed stirrups at 3 in c/c, plus #4 spiral confinement (2.5 in pitch) inside each anchor's local zone per the anchorage-device supplier's approved general-zone reinforcement. Verify that the anchor-plate bearing stress ≤ 0.7·f'_ci · √(A/A_b) per §5.9.5.6.5.

Final end-block detailing (from computed A_s,burst)

BT-72 solid end block — h = 72 in, b_w = 8 in, 4-tendon anchor group

Location	A_s required	Bars provided	Spacing / detail
General-zone bursting steel (within d_burst = 30 in from loaded face)	A_s,burst ≥ 43.6 in²	#6 closed hoops, double-leg (A_bar = 0.88 in²) × 50 hoops	s = 3 in c/c over 30 in behind anchor
Local-zone confinement (each anchor)	Per anchor supplier's approved GZR	#4 spiral, 8-in Ø × 20 in long	2.5 in pitch
Spalling steel — loaded face	T_spall ≈ 0.02·ΣP_j = 71 kip → A_s ≥ 1.2 in²	2 layers #5 orthogonal mat (A_s = 1.24 in²)	Within 0.5·h = 36 in of loaded face
Skin / longitudinal side face	0.012·(h − 30) per §5.6.7 = 0.50 in²/ft each face	#5 @ 12 in each face, longitudinal	Full depth of end block

Bearing check: σ_b = P_j / A_b ≤ 0.7·f'_ci·√(A/A_b) per §5.9.5.6.5. First hoop within 1½ in of the loaded face; extend closed hoops at least 1.0·h behind the last anchor before transitioning to normal web reinforcement.

7.9 — Spliced post-tensioned girders

How you get to 300 ft with precast components

Three-stage erection sequence of a spliced post-tensioned bulb-tee bridge — Figure 7.8Spliced post-tensioned bulb-tee erection sequence. Segment sizes are chosen so each haul is within highway load and length limits (typically 130–160 ft precast segments). Closure pours field-splice the ducts, and one continuous PT tendon is stressed after the closure concrete reaches strength.

Spliced construction combines the plant quality of pretensioning with the span reach of post-tensioning. Design implications: (i) the section carries different amounts of prestress at different times, so a time-step loss analysis (§5.9.3.4) is required; (ii) closure pours introduce a locally different concrete age and modulus — check stress transfer across the joint; (iii) the continuous PT tendon changes the continuity moments over the pier, and secondary moments M_sec (from prestress restraint) must be added to the DL + LL envelope.

Figure 7.9Harped strand geometry reference — used to size the harping angle ψ for shear V_p and to check top-fiber tension at the hold-down (where eccentricity is maximum but girder self-weight moment is not yet fully developed).

7.10 — Camber and deflection

What the girder does between plant and paving

AASHTO LRFD §5.6.3.5, §2.5.2.6.2

Camber (upward deflection under prestress) is calculated at release, before deck placement, and long term after all sustained loads. The PCI multiplier method scales the elastic release camber by empirical factors to represent the effect of creep over time.

(7.11)

Formula
$\Delta_{p,rel} = \frac{P_{i} \cdot e_{c} \cdot L^{2}}{8 \cdot E_{ci} \cdot I_{g}} - \frac{5 \cdot w_{g} \cdot L^{4}}{384 \cdot E_{ci} \cdot I_{g}}$

$\Delta_{p,rel}$: upward camber at release [in]
$P_{i}$: initial prestress force (after ES loss) [kip]

Stage	PCI multiplier (no composite topping)	PCI multiplier (with composite topping)
Prestress camber at release	1.80	1.80
Girder DL deflection at release	1.85	1.85
Prestress camber, long-term (final)	2.45	2.20
Deck / topping DL, long-term	—	2.40

7.10b — Design Example 6

Prestressed interior concrete girder (simple span, L = 80 ft)

AASHTO LRFD Strength I flexure + Service I stresses at release and after all losses

This worked example — reproduced from Simplified LRFD Bridge Design (Design Example 6) — walks the complete AASHTO LRFD workflow for a pretensioned interior I-girder on an 80 ft simple span in central New York. Every check is presented as Equation → Substitute → Result so the formula is stated first before any numbers are entered.

Figure 7.11Fig. 2.63 — Elevation and cross section. Five prestressed I-girders (51 in deep × 24 in bottom flange) spaced at S = 7.5 ft with 3.75 ft overhang each side; 8-in composite deck (0.5 in integral wearing surface); 80 ft simple span.

Problem statement

Design the interior prestressed concrete girder of a two-lane simple-span highway bridge (AASHTO HL-93) for Strength I flexure and Service I concrete stresses at release and after all losses.

Given

Span L80 ft (simple span)
Girder spacing S7.5 ft (5 girders)
Overhang3.75 ft
Slab thickness t_s8 in (7.5 in structural + 0.5 in integral wearing)
Girder depth51 in (I-beam, 18-in top flange, 24-in bottom flange)
Basic beam A_g762 in²
Basic beam I_g212,450 in⁴
S_nc,top / S_nc,bottom7,692 in³ / 9,087 in³
y_b (basic beam)23.38 in
Strand pattern A_ps44 × ½-in 7-wire Grade 270 = 6.732 in²
Strand centroid g5 in from bottom → e_m = 18.38 in
Girder f'_c / f'_ci6.5 ksi / 6.0 ksi
Slab f'_cs4.5 ksi
f_pu, E_p270 ksi, 28,500 ksi
FWS3-in bituminous @ 0.140 kip/ft³
Barrier / parapet0.506 kip/ft (each of 2 barriers)
Ambient RH70 % (central NY)

Required

Determine the composite section properties, the factored midspan moment M_u (Strength I), the girder capacity φM_n (= M_r), and the concrete stresses at midspan at release and after all long-term losses.

Step 1

Composite section properties (§4.6.2.6, §5.4.2.4)

Because the girder concrete ( $f'_{cg}=6.5$ ksi) is stiffer than the deck slab ( $f'_{cs}=4.5$ ksi), the slab width is transformed by the modular ratio $n=E_{cg}/E_{cs}$ before adding it to the girder.

Minimum depth check (Table 2.5.2.6.3-1)

Minimum depth for precast prestressed I-beam, simple span

d_{min} \;=\; 0.045\,L

Substitute

d_{min} \;=\; 0.045\,(80\;\text{ft})(12\;\text{in/ft})

Result

d_{min} \;=\; 43.2\;\text{in} \;<\; d = 7.5 + 51 = 58.5\;\text{in}\;\;[\text{OK}]

Effective flange width (interior girder) — equals beam spacing

b_e \;=\; S

Substitute

b_e \;=\; (7.5\;\text{ft})(12\;\text{in/ft})

Result

b_e \;=\; 90\;\text{in}

Concrete modulus (girder), AASHTO Eq. 5.4.2.4-1

E_{cg} \;=\; 33{,}000\,w_c^{1.5}\,\sqrt{f'_{cg}}

Substitute

E_{cg} \;=\; 33{,}000\,(0.15)^{1.5}\,\sqrt{6.5}

Result

E_{cg} \;=\; 4{,}890\;\text{ksi}

Concrete modulus (slab)

E_{cs} \;=\; 33{,}000\,w_c^{1.5}\,\sqrt{f'_{cs}}

Substitute

E_{cs} \;=\; 33{,}000\,(0.15)^{1.5}\,\sqrt{4.5}

Result

E_{cs} \;=\; 4{,}070\;\text{ksi}

Modular ratio (girder / slab)

n \;=\; \dfrac{E_{cg}}{E_{cs}}

Substitute

n \;=\; \dfrac{4{,}890}{4{,}070}

Result

n \;=\; 1.2

Transformed slab area (structural thickness = 7.5 in)

A_{s,\text{tr}} \;=\; t_s \cdot \dfrac{b_e}{n}

Substitute

A_{s,\text{tr}} \;=\; (7.5)\left(\dfrac{90}{1.2}\right)

Result

A_{s,\text{tr}} \;=\; 562.5\;\text{in}^2

Composite area

A_c \;=\; A_g + A_{s,\text{tr}}

Substitute

A_c \;=\; 762 + 562.5

Result

A_c \;=\; 1{,}324.5\;\text{in}^2

Figure 7.12Fig. 2.64 — Composite section for the interior girder: 90-in effective flange × 7.5-in structural slab acting with the 51-in prestressed I-girder. 44 × ½-in strands are grouped in the bottom flange with pattern centroid g = 5 in from the soffit → e_m = 23.38 − 5 = 18.38 in.

Locate the composite centroid by taking moments about the basic girder centroid ( $y_t = 27.62\;\text{in}$ to the top of the girder), then find the composite section moduli.

Shift of centroid from basic beam c.g.

\bar y \;=\; \dfrac{A_{s,\text{tr}}\,(y_t + t_s/2)}{A_c}

Substitute

\bar y \;=\; \dfrac{562.5\,(27.62 + 3.75)}{1{,}324.5}

Result

\bar y \;=\; 13.32\;\text{in}

Composite y'_b (bottom fiber to composite c.g.)

y'_b \;=\; y_b + \bar y

Substitute

y'_b \;=\; 23.38 + 13.32

Result

y'_b \;=\; 36.7\;\text{in}

Composite y'_t (slab top to composite c.g.)

y'_t \;=\; (h_g + t_s) - y'_b

Substitute

y'_t \;=\; (51 + 7.5) - 36.7

Result

y'_t \;=\; 21.8\;\text{in}

Composite moment of inertia (parallel-axis on both parts)

I_c \;=\; I_g + A_g\,\bar y^{\,2} + \dfrac{(b_e/n)\,t_s^{\,3}}{12} + A_{s,\text{tr}}\left(y'_t - \dfrac{t_s}{2}\right)^{2}

Substitute

I_c \;=\; 212{,}450 + 762\,(13.32)^2 + \dfrac{75\,(7.5)^3}{12} + 562.5\,(21.8-3.75)^2

Result

I_c \;=\; 533{,}546.5\;\text{in}^4

Composite section modulus — bottom fiber

S_{bc} \;=\; \dfrac{I_c}{y'_b}

Substitute

S_{bc} \;=\; \dfrac{533{,}546.5}{36.7}

Result

S_{bc} \;=\; 14{,}538\;\text{in}^3

Composite section modulus — slab top

S_{tc} \;=\; \dfrac{I_c}{y'_t}

Substitute

S_{tc} \;=\; \dfrac{533{,}546.5}{21.8}

Result

S_{tc} \;=\; 24{,}474.6\;\text{in}^3

Figure 7.13Fig. 2.65 — Area-transformed section. The 90-in slab shrinks to 90/1.2 = 75 in so that girder and deck share one elastic (n = 1) reference. Composite centroid drops to y'_b = 36.7 in from the soffit.

Step 2

Live-load, dead-load and factored midspan moments

AASHTO LRFD §3.6, §4.6.2.2, §1.3, Table 3.4.1-1

Dynamic load allowance $IM = 33\%$ applies to the truck / tandem axles (not to the lane load). Interior-beam distribution factors use the stiffness parameter term $(K_g/12Lt_s^3)^{0.1}=1.09$ for cross-section type (k).

DFM — one lane loaded (interior) — §4.6.2.2.2b

DFM_{si} \;=\; 0.06 + \left(\dfrac{S}{14}\right)^{0.4}\left(\dfrac{S}{L}\right)^{0.3}\left(\dfrac{K_g}{12Lt_s^3}\right)^{0.1}

Substitute

DFM_{si} \;=\; 0.06 + \left(\dfrac{7.5}{14}\right)^{0.4}\left(\dfrac{7.5}{80}\right)^{0.3}(1.09)

Result

DFM_{si} \;=\; 0.477

DFM — two or more lanes loaded (interior) — governs

DFM_{mi} \;=\; 0.075 + \left(\dfrac{S}{9.5}\right)^{0.6}\left(\dfrac{S}{L}\right)^{0.2}\left(\dfrac{K_g}{12Lt_s^3}\right)^{0.1}

Substitute

DFM_{mi} \;=\; 0.075 + \left(\dfrac{7.5}{9.5}\right)^{0.6}\left(\dfrac{7.5}{80}\right)^{0.2}(1.09)

Result

DFM_{mi} \;=\; 0.664 \;\;[\text{controls}]

Figure 7.14Fig. 2.66 — Live-load positions for maximum midspan moment: (a) HS-20 truck (8, 32, 32 kip axles at 14 ft), (b) design tandem (2 × 25 kip at 4 ft), (c) 0.64 kip/ft uniform lane load.

Truck moment — take moments about support B with the middle 32-kip axle placed at midspan.

Reaction at A (ΣM_B = 0)

R_A \;=\; \dfrac{8\,(26) + 32\,(40) + 32\,(54)}{80}

Substitute

R_A \;=\; \dfrac{208 + 1{,}280 + 1{,}728}{80}

Result

R_A \;=\; 40.2\;\text{kip}

Truck midspan moment

M_{tr} \;=\; R_A\,(L/2) - 32\,(14)

Substitute

M_{tr} \;=\; 40.2\,(40) - 32\,(14)

Result

M_{tr} \;=\; 1{,}160\;\text{ft-kip}

Tandem midspan moment

M_{tandem} \;=\; R_A\,(L/2)\;\;\text{with}\;\; R_A=\dfrac{25(36)+25(40)}{80}=23.75\;\text{kip}

Substitute

M_{tandem} \;=\; 23.75\,(40)

Result

M_{tandem} \;=\; 950\;\text{ft-kip} \;<\; M_{tr}\;\;[\text{truck controls}]

Lane midspan moment (w = 0.64 kip/ft)

M_{ln} \;=\; \dfrac{w L^{2}}{8}

Substitute

M_{ln} \;=\; \dfrac{0.64\,(80)^2}{8}

Result

M_{ln} \;=\; 512\;\text{ft-kip}

Distributed LL + IM per girder — truck governs

M_{LL+IM} \;=\; DFM_{mi}\left[M_{tr}\,(1+IM) + M_{ln}\right]

Substitute

M_{LL+IM} \;=\; 0.664\,[1{,}160\,(1.33) + 512]

Result

M_{LL+IM} \;=\; 1{,}364.38\;\text{ft-kip / girder}

Dead loads — beam self-weight and slab dead load act on the basic (non-composite) section; barrier and future wearing surface act on the composite section and are distributed equally over the 5 girders.

Beam self-weight (A_g × w_c)

w_g \;=\; A_g \cdot w_c

Substitute

w_g \;=\; \dfrac{762}{144}\,(0.15)

Result

w_g \;=\; 0.79\;\text{kip/ft} \;\Rightarrow\; M_g \;=\; \dfrac{0.79\,(80)^2}{8} \;=\; 632\;\text{ft-kip}

Slab dead load (t_s × S × w_c, full 8-in slab)

w_D \;=\; t_s \cdot S \cdot w_c

Substitute

w_D \;=\; \tfrac{8}{12}\,(7.5)\,(0.15)

Result

w_D \;=\; 0.75\;\text{kip/ft} \;\Rightarrow\; M_D \;=\; 600\;\text{ft-kip}

Total non-composite DC

M_{DC} \;=\; M_g + M_D

Substitute

M_{DC} \;=\; 632 + 600

Result

M_{DC} \;=\; 1{,}232\;\text{ft-kip}

Future wearing surface DW (3 in bituminous)

w_{DW} \;=\; \dfrac{t_{FWS}}{12}\,w_{FWS}\,S

Substitute

w_{DW} \;=\; \tfrac{3}{12}\,(0.140)\,(7.5)

Result

w_{DW} \;=\; 0.263\;\text{kip/ft} \;\Rightarrow\; M_{DW} \;=\; 210.4\;\text{ft-kip}

Strength I factored midspan moment (η = 1.0)

M_u \;=\; 1.25\,M_{DC} + 1.50\,M_{DW} + 1.75\,M_{LL+IM}

Substitute

M_u \;=\; 1.25\,(1{,}232) + 1.50\,(210.4) + 1.75\,(1{,}364.38)

Result

M_u \;=\; 4{,}243.3\;\text{ft-kip}

Step 3

Girder moment capacity φM_n (= M_r)

AASHTO LRFD §5.7.3.1, §5.7.3.2

Assume rectangular behavior — the neutral axis is expected to sit inside the 7.5-in slab (verify after). The distance from the extreme compression fiber (top of slab) to the strand centroid is:

Effective depth to prestressing steel

d_p \;=\; (h_g - g) + t_s

Substitute

d_p \;=\; (51 - 5) + 8

Result

d_p \;=\; 54\;\text{in}

Whitney stress block factor β₁ (§5.7.2.2)

\beta_1 \;=\; 0.85 - 0.05\,(f'_c - 4)

Substitute

\beta_1 \;=\; 0.85 - 0.05\,(6.5 - 4)

Result

\beta_1 \;=\; 0.725

Neutral axis depth c — Eq. 5.7.3.1.1-4 (rectangular, no A_s, A'_s)

c \;=\; \dfrac{A_{ps}\,f_{pu}}{0.85\,f'_c\,\beta_1\,b_e + k\,A_{ps}\,\dfrac{f_{pu}}{d_p}}

Substitute

c \;=\; \dfrac{6.732\,(270)}{0.85\,(6.5)\,(0.725)\,(90) + 0.28\,(6.732)\,(270/54)}

Result

c \;=\; 4.91\;\text{in} \;<\; t_s = 8\;\text{in} \;\;[\text{rectangular assumption OK}]

Average prestressing steel stress — Eq. 5.7.3.1.1-1

f_{ps} \;=\; f_{pu}\left(1 - k\,\dfrac{c}{d_p}\right)

Substitute

f_{ps} \;=\; 270\left(1 - 0.28\,\dfrac{4.91}{54}\right)

Result

f_{ps} \;=\; 263.1\;\text{ksi}

Depth of equivalent rectangular stress block

a \;=\; \beta_1\,c

Substitute

a \;=\; 0.725\,(4.91)

Result

a \;=\; 3.56\;\text{in}

Nominal flexural resistance — Eq. 5.7.3.2.2 (neglect mild steel)

M_n \;=\; A_{ps}\,f_{ps}\left(d_p - \dfrac{a}{2}\right)

Substitute

M_n \;=\; \dfrac{6.732\,(263.1)\,(54 - 3.56/2)}{12}

Result

M_n \;=\; 7{,}725.2\;\text{ft-kip}

Strength I flexural check

Factored resistance (φ = 1.0 for prestressed tension-controlled, §5.5.4.2)

M_r \;=\; \phi\,M_n

Substitute

M_r \;=\; 1.0\,(7{,}725.2)

Result

M_r \;=\; 7{,}725.2\;\text{ft-kip} \;>\; M_u = 4{,}243.3\;\text{ft-kip}\;\;[\text{OK}]

Step 4

Concrete stresses at midspan at release of prestress

AASHTO LRFD §5.9.4.1, §5.9.5.2.3

At release only the beam self-weight $M_g$ acts against the initial prestress force. Allowable stresses at time of transfer use $f'_{ci}=6.0$ ksi.

Figure 7.15Fig. 2.67 — Basic I-girder section at midspan. Strand pattern centroid g = 5 in from the soffit gives an eccentricity e_m = y_b − g = 23.38 − 5 = 18.38 in.

Allowable compression at transfer (§5.9.4.1.1)

f_{ci,\text{allow}} \;=\; 0.6\,f'_{ci}

Substitute

f_{ci,\text{allow}} \;=\; 0.6\,(6.0)

Result

f_{ci,\text{allow}} \;=\; 3.6\;\text{ksi (compression)}

Allowable tension at transfer (§5.9.4.1.2)

f_{ti,\text{allow}} \;=\; 0.24\,\sqrt{f'_{ci}}

Substitute

f_{ti,\text{allow}} \;=\; 0.24\,\sqrt{6.0}

Result

f_{ti,\text{allow}} \;=\; 0.588\;\text{ksi} \;\approx\; 0.563\;\text{ksi (per source, uses 5.5)}

Concrete modulus at transfer, E_ci (Eq. 5.4.2.4-1)

E_{ci} \;=\; 33{,}000\,w_c^{1.5}\,\sqrt{f'_{ci}}

Substitute

E_{ci} \;=\; 33{,}000\,(0.15)^{1.5}\,\sqrt{6.0}

Result

E_{ci} \;=\; 4{,}696\;\text{ksi}

Jacking and pre-transfer stresses (Table 5.9.3-1)

f_{pj} \;=\; 0.75\,f_{pu} \;;\quad f_{pbt} \;=\; 0.9\,f_{pj}

Substitute

f_{pj} \;=\; 0.75\,(270) = 203\;\text{ksi} \;;\quad f_{pbt} \;=\; 0.9\,(203)

Result

f_{pbt} \;=\; 182.3\;\text{ksi}

Elastic-shortening loss Δf_pES — Eq. C5.9.5.2.3a-1 (avoids iteration)

\Delta f_{pES} \;=\; \dfrac{A_{ps}\,f_{pbt}\left(I_g + e_m^{2}A_g\right) - e_m\,M_g\,A_g}{A_{ps}\left(I_g + e_m^{2}A_g\right) + \dfrac{A_g\,I_g\,E_{ci}}{E_p}}

Substitute

\Delta f_{pES} \;=\; \dfrac{6.732\,(182.3)\,[212{,}450 + (18.38)^2(762)] - 18.38\,(632)(762)}{6.732\,[212{,}450 + (18.38)^2(762)] + \dfrac{762\,(212{,}450)\,(4{,}696)}{28{,}500}}

Result

\Delta f_{pES} \;=\; 15.77\;\text{ksi}

Prestress force after transfer

P \;=\; (f_{pbt} - \Delta f_{pES})\,A_{ps}

Substitute

P \;=\; (182.3 - 15.77)\,(6.732)

Result

P \;=\; 1{,}121.1\;\text{kip}

Superpose the three stress components — axial $-P/A_g$ , bending from eccentricity $Pe_m/S$ , and bending from self-weight $M_g/S$ . Compression is negative.

Top-fiber stress at midspan (release)

f_t \;=\; -\dfrac{P}{A_g} + \dfrac{P\,e_m}{S_{nc,top}} - \dfrac{M_g}{S_{nc,top}}

Substitute

f_t \;=\; -\dfrac{1{,}121.1}{762} + \dfrac{1{,}121.1\,(18.38)}{7{,}692} - \dfrac{632\,(12)}{7{,}692}

Result

f_t \;=\; -1.47 + 2.68 - 0.98 \;=\; +0.23\;\text{ksi (tension)} \;<\; 0.563\;\text{ksi}\;\;[\text{OK}]

Bottom-fiber stress at midspan (release)

f_b \;=\; -\dfrac{P}{A_g} - \dfrac{P\,e_m}{S_{nc,bottom}} + \dfrac{M_g}{S_{nc,bottom}}

Substitute

f_b \;=\; -\dfrac{1{,}121.1}{762} - \dfrac{1{,}121.1\,(18.38)}{9{,}087} + \dfrac{632\,(12)}{9{,}087}

Result

f_b \;=\; -1.47 - 2.27 + 0.83 \;=\; -2.91\;\text{ksi (compression)} \;<\; 3.6\;\text{ksi}\;\;[\text{OK}]

Figure 7.16Fig. 2.68 — Concrete stresses at midspan at release of prestress. Top fiber: −1.47 + 2.68 − 0.98 = +0.23 ksi tension. Bottom fiber: −1.47 − 2.27 + 0.83 = −2.91 ksi compression. Both within the transfer allowables.

Step 5

Final concrete stresses at midspan after all losses (except friction)

AASHTO LRFD §5.9.5.3, §5.9.4.2

Long-term losses use the approximate expression of Eq. 5.9.5.3-1 for low-relaxation strand: humidity factor $\gamma_h=1.7-0.01H$ and strength factor $\gamma_{st}=5/(1+f'_{ci})$ .

Humidity factor (H = 70 %)

\gamma_h \;=\; 1.7 - 0.01\,H

Substitute

\gamma_h \;=\; 1.7 - 0.01\,(70)

Result

\gamma_h \;=\; 1.0

Strength factor at transfer

\gamma_{st} \;=\; \dfrac{5}{1 + f'_{ci}}

Substitute

\gamma_{st} \;=\; \dfrac{5}{1 + 6.0}

Result

\gamma_{st} \;=\; 0.833

Long-term loss Δf_pLT — Eq. 5.9.5.3-1 (Δf_pR = 2.4 ksi for low-relaxation)

\Delta f_{pLT} \;=\; 10.0\,\dfrac{f_{pi}\,A_{ps}}{A_g}\,\gamma_h\,\gamma_{st} \;+\; 12.0\,\gamma_h\,\gamma_{st} \;+\; \Delta f_{pR}

Substitute

\Delta f_{pLT} \;=\; 10.0\,\dfrac{203\,(6.732)}{762}\,(1.0)(0.833) + 12.0\,(1.0)(0.833) + 2.4

Result

\Delta f_{pLT} \;=\; 25.4\;\text{ksi}

Total prestress loss and effective stress

\Delta f_{pT} \;=\; \Delta f_{pES} + \Delta f_{pLT} \;;\quad f_{se} \;=\; f_{pj} - \Delta f_{pT}

Substitute

\Delta f_{pT} \;=\; 15.77 + 25.4 = 41.17\;\text{ksi} \;;\quad f_{se} \;=\; 203 - 41.17

Result

f_{se} \;=\; 161.3\;\text{ksi} \;\Rightarrow\; P_e = f_{se}\,A_{ps} = 1{,}085.9\;\text{kip}

Now split the fibre stresses in two stages: (a) prestress + non-composite DC on the basic beam, and (b) superimposed dead load + LL on the composite section (its stiffer denominator is the composite section modulus).

Stage (a) — bottom of basic beam (P_e + M_g + M_D)

f_{b,a} \;=\; -\dfrac{P_e}{A_g} - \dfrac{P_e\,e_m}{S_{nc,bottom}} + \dfrac{M_g+M_D}{S_{nc,bottom}}

Substitute

f_{b,a} \;=\; -\dfrac{1{,}085{,}870}{762} - \dfrac{1{,}085{,}870\,(18.38)}{9{,}087} + \dfrac{(1{,}232)(12{,}000)}{9{,}087}

Result

f_{b,a} \;=\; -1{,}994.4\;\text{psi (compression)}

Stage (a) — top of basic beam

f_{t,a} \;=\; -\dfrac{P_e}{A_g} + \dfrac{P_e\,e_m}{S_{nc,top}} - \dfrac{M_g+M_D}{S_{nc,top}}

Substitute

f_{t,a} \;=\; -\dfrac{1{,}085{,}870}{762} + \dfrac{1{,}085{,}870\,(18.38)}{7{,}692} - \dfrac{(1{,}232)(12{,}000)}{7{,}692}

Result

f_{t,a} \;=\; -752.3\;\text{psi (compression)}

Superimposed DC + DW per girder (2 barriers over 5 girders + 3-in FWS)

w_s \;=\; \dfrac{2\,BW}{n_g} + \dfrac{t_{FWS}}{12}\,w_{FWS}\,S

Substitute

w_s \;=\; \dfrac{2\,(0.506)}{5} + \tfrac{3}{12}\,(0.14)\,(7.5)

Result

w_s \;=\; 0.4649\;\text{kip/ft} \;\Rightarrow\; M_s = \dfrac{0.4649\,(80)^2}{8} = 371.9\;\text{ft-kip}

Stage (b) — bottom of composite (M_s + M_LL+IM on S_bc)

\Delta f_{b,b} \;=\; \dfrac{M_s + M_{LL+IM}}{S_{bc}}

Substitute

\Delta f_{b,b} \;=\; \dfrac{(371.9 + 1{,}364.38)(12{,}000)}{14{,}538}

Result

\Delta f_{b,b} \;=\; +1{,}433\;\text{psi (tension at bottom)}

Stage (b) — top of slab

\Delta f_{t,\text{slab}} \;=\; -\dfrac{M_s + M_{LL+IM}}{S_{tc}}

Substitute

\Delta f_{t,\text{slab}} \;=\; -\dfrac{(371.9 + 1{,}364.38)(12{,}000)}{24{,}474.6}

Result

\Delta f_{t,\text{slab}} \;=\; -851.2\;\text{psi (compression)}

Figure 7.17Fig. 2.69 — Final concrete stresses at midspan after all losses. Basic-beam stresses (−752.3 top / −1,994.4 bottom psi) superpose on composite-stage stresses (−851.2 slab top / +1,433 bottom psi) to give final stresses −1,311.1 psi at girder top, −561.4 psi at girder bottom (all compression). Units: lbf/in²; “−” = compression.

Service I limit-state stress checks (after all losses)

Girder-top compression allowable

f_{cs,\text{allow}} \;=\; 0.45\,f'_c

Substitute

f_{cs,\text{allow}} \;=\; 0.45\,(6.5)

Result

f_{cs,\text{allow}} \;=\; 2.93\;\text{ksi} \;>\; |{-1.311}|\;\text{ksi}\;\;[\text{OK}]

Girder-bottom tension allowable (§Tbl. 5.9.4.2.2-1)

f_{ts,\text{allow}} \;=\; 0.19\,\sqrt{f'_c}

Substitute

f_{ts,\text{allow}} \;=\; 0.19\,\sqrt{6.5}

Result

f_{ts,\text{allow}} \;=\; 0.484\;\text{ksi} \;>\; |{-0.561}|\;\text{ksi}\;\;[\text{OK — actually compression here}]

Design Example 6 — Summary

Results at a glance

Check	Demand	Capacity / Allowable	Result
Strength I flexure φM_n	M_u = 4,243 ft-kip	M_r = 7,725 ft-kip	OK (1.82×)
Transfer — top tension	+0.23 ksi	0.563 ksi	OK
Transfer — bottom compression	−2.91 ksi	3.60 ksi	OK
Service I — girder top comp.	−1.311 ksi	2.93 ksi	OK
Service I — girder bottom	−0.561 ksi (comp.)	0.484 ksi tension limit not reached	OK
Total prestress loss	Δf_pT = 41.2 ksi (20.3 % of f_pj)	—	f_se = 161.3 ksi

7.11 — Concept checkpoint

Three quick self-checks

Why does the top-fiber stress at transfer usually govern strand pattern selection at the girder ends rather than at midspan?
A designer proposes to eliminate the harped strand and use debonded straight strand instead. What two design concerns does this shift onto other reinforcement?
The Service III bottom-fiber tension check just barely passes with the trial strand pattern. Name three levers you can pull before adding a strand.

7.12 — Design-check summary

Every AASHTO check for a PS girder

Check	AASHTO	Governs
Jacking stress	§5.9.2.2	Strand selection
Transfer top tension	§5.9.2.3.1	Debond / harp length
Transfer bottom compression	§5.9.2.3.1	Release strength f'_ci
Service III bottom tension	§5.9.2.3.2b	Strand quantity
Service I top compression	§5.9.2.3.2a	Deck / girder f'_c
Flexural strength φMₙ	§5.6.3	Strand + section
Minimum reinforcement	§5.6.3.3	φMₙ ≥ 1.33·M_u or M_cr
Fatigue on strand	§5.5.3	Stress range at Fatigue I
Shear (MCFT)	§5.7.3	Stirrup spacing
Anchorage zone (PT)	§5.9.5.6	End-block reinforcement
Deflection / camber	§2.5.2.6.2	L/800 live-load limit

7.13 — Mini design challenge

2 × 160 ft continuous spliced BT-72 bridge

A rural arterial requires a two-span continuous prestressed girder bridge. Roadway width 44 ft out-to-out, 5 bulb-tees BT-72 at 9 ft spacing, 9 in composite deck,f'_c = 8.0 ksi girder / 4.0 ksi deck. Design each interior BT-72 as a spliced post-tensioned girder with a continuous draped tendon (low at midspan, high over pier).

Design-challenge bridge: 2 × 160 ft continuous, 5 BT-72 girders at S = 9 ft, draped PT tendon — Figure 7.10Design-challenge geometry. Live load HL-93; wearing surface 30 psf; barrier 300 plf per side. Provide plan, elevation, cross-section, tendon profile, strand-pattern schedule, and all §5.9 / §5.7 / §5.9.5.6 checks.

Deliverables

Full HL-93 moment and shear envelopes (Strength I, Service I, Service III, Fatigue I).
Strand pattern (straight + harped or full PT tendon layout) with computed e_c(x).
All §5.9.2.3 stress checks at transfer, deck placement, and Service III at midspan and over pier.
φMₙ check at midspan (positive) and over pier (negative — deck reinforcement contributes to tension side).
MCFT shear check at h/2 from support, at hold-down, and at pier face.
Anchorage-zone bursting design at each end (§5.9.5.6).
Camber schedule at release, 60 days, and long term; check L/800 live-load deflection.

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Section 2

Fully Worked Examples

Complete AASHTO LRFD solutions with knowns, assumptions, step calculations, verification, and design commentary. Difficulty rises from basic to consulting-grade.

Worked Example 1

Concept: what does prestress actually do to a beam?

Basic

Problem

Compute the initial bottom-fiber concrete stress at midspan from prestress alone (no external load).

Step-by-Step

P_j = A_ps·f_pj = 4.34·202.5 = 879 kip

f_b = −P/A − P·e/S_b = −879/560 − 879·12/10,540 = −1.57 − 1.00 = −2.57 ksi (compression)

Result

−2.57 ksi (comp)

Design Verification

Negative = compression by AASHTO convention. This 2.57 ksi is the "credit" the beam has before dead + live load try to tension the bottom fiber.

Discussion

The whole prestressing game: buy compression cheap at midspan bottom, spend it on tension caused by service loads.

Worked Example 2

Simple hand calc: transformed section vs. gross section

Basic

Problem

Find S_b,tr and % change from gross.

Step-by-Step

ΔA = (n−1)·A_ps = 5.45·4.34 = 23.7 in²

ȳ_tr = (A_g·y_g + ΔA·y_ps)/(A_g + ΔA) shift ≈ 0.8 in down.

Design Verification

Only ~3% change — for hand design, gross section is close enough. Refined LRFD uses transformed at all stages.

Discussion

Once bonded, strands act with the beam. §5.9.1.1 lets you use either gross or transformed provided you're consistent.

Worked Example 3

AASHTO approximate prestress losses (§5.9.3.3)

Intermediate

Problem

Compute ΔfpLT (ksi).

Step-by-Step

ΔfpLT = 10·(f_pi·A_ps/A_g)·γ_h·γ_st + 12·γ_h·γ_st + ΔfpR

= 10·1.57·1.00·0.77 + 12·1.00·0.77 + 2.4 = 12.1 + 9.24 + 2.4 = 23.7 ksi

Result

ΔfpLT ≈ 23.7 ksi

Design Verification

Long-term stress f_pe = 0.75·270 − 23.7 − ES ≈ 202.5 − 23.7 − 18 = 160.8 ksi ≈ 0.60 f_pu — typical bridge girder.

Discussion

Approximate method OK for standard beams with typical strand patterns. Refined per §5.9.3.4 for post-tensioned or unusual sections.

Worked Example 4

Complete flexural design: interior BT-72 girder, 130-ft span

Intermediate

Problem

Find required strand count and check limit states.

Step-by-Step

f_b = −P/A − P·e/S_b,nc + (M_DC1+DC2+DW)/S_bc-ish + 0.8·M_LL/S_bc = ...

P_e = 44·0.217·160 = 1,529 kip. f_b = −1.57·(44/20) − 1.62 + 3.30 = −0.31 ksi < 0.19√f′_c = 0.537 ksi allowable (Service III).

Design Verification

Bottom compression at transfer f_bi ≈ −3.6 ksi ≤ 0.65·f′_ci = −3.9 ✓; top tension +180 psi ≤ 3√f′_ci = 210 psi ✓ (§5.9.2.3.1a).

Discussion

Iterated on strand count to hit Service III (tension) with the smallest N. Strength never controls modern high-strength I-girders.

Worked Example 5

Multiple limit states: shear (§5.7.3) with prestress

Advanced

Problem

Compute V_n and confirm V_u ≤ φV_n.

Step-by-Step

V_c = 0.0316·β·√f′_c·b_v·d_v = 0.0316·3.20·√8·6·68 = 116 kip

A_v = 2·0.20 = 0.40 in²; s = 12; V_s = A_v·f_y·d_v·cot θ / s = 0.40·60·68·1.96/12 = 267 kip

Design Verification

Vertical-strand component V_p carries ~30% of V_u — dropping it (straight strands) would need heavier stirrups (#5 @ 8 in).

Discussion

Harping vs debonding is a fabricator preference; both work but harping cleans up end-zone stress and reduces stirrup steel near supports.

Worked Example 6

Design optimization: strand pattern trade study

Advanced

Problem

Pick the better of two strand strategies.

Step-by-Step

A: 44·130 = 5,720 ft. B: 40·130 + 8·(130−30) = 5,200 + 800 = 6,000 ft.

A: harping reduces prestress at ends → OK. B: debonded strands controlled to 25% of total, ≤ 25% per row per §5.9.4.3.3 ✓

Design Verification

Both pass all limits. Modern precast plants prefer debonding — faster casting, less bed reconfiguration.

Discussion

Design optimization at the material level is small money; the real wins come from choosing the right girder shape (BT-72 vs Type VI vs FIB-96) for the span.

Worked Example 7

Construction-stage: transfer, storage, and shipping stresses

Advanced

Problem

Verify no stage exceeds AASHTO transfer and hauling limits.

Step-by-Step

f_t = +P/A − P·e/S_t + M_sw/S_t = 1.51 − 1.65 + 0.62 = +0.48 ksi ≤ 0.24√5.5 = 0.563 ksi ✓

Additional creep + shrinkage losses ≈ 10 ksi. f_b ≈ −2.4 ksi (still compression) ✓

Design Verification

All hauling stresses clear with margin — bracing not required for L_lift < 140 ft on BT-72.

Discussion

Construction-stage governs the top-fiber tension check at release for every prestressed beam — never skip it.

Worked Example 8

Consulting: partial-length post-tensioning to add capacity

Consulting

Problem

Size the external PT force to close the capacity gap.

Step-by-Step

ΔM = 3,800 − 3,150 = 650 k·ft

f_ps ≈ f_se + 10 ksi (external unbonded) = 170 ksi at ULS.

Design Verification

Check secondary moments (statically determinate girders → secondary = 0) and post-strengthening shear at supports.

Discussion

External PT retrofit is a standard consulting move to defer replacement 15–20 yr. Anchorage detailing and corrosion protection dominate cost.

Worked Example 9

Case-study reproduction: FIU-style construction-stage risk analysis

Consulting

Problem

Compute demand vs capacity of the diagonal-to-deck cold joint.

Step-by-Step

V_ni = c·A_cv + μ·(A_vf·f_y + P_c) = 0.075·288 + 0.6·(3.52·60 + 0) = 21.6 + 126.7 = 148.3 kip

V_u,joint = D·sin(45°) = 155 kip > 148.3 kip. Interface shear NG.

Result

FAILS interface shear

Design Verification

Reproduces NTSB finding: construction-stage interface shear at Node 11/12 was ~ 50% underdesigned.

Discussion

Every construction stage is a load case. Peer review must catch temporary conditions as rigorously as final ones.

Worked Example 10

Comprehensive: full prestressed girder design and drawing package

Consulting

Problem

Produce a bid-quality precast beam design.

Step-by-Step

g_int,strength = 0.075 + (S/9.5)^0.6 · (S/L)^0.2 · (K_g/12·L·t_s³)^0.1 → 0.63

M_LL+IM,strength = 3,400 k·ft/girder; M_DC1+DC2+DW = 2,880 k·ft; M_u = 9,600 k·ft.

Design Verification

Cross-checks: total strand steel 350 lb/beam; typical for 140-ft BT-72. Camber 6.5 in matches PCI camber curves ✓.

Discussion

This is the deliverable a structural EIT is expected to produce end-to-end. The real complexity is not the math — it's coordinating fabricator, hauler, and erector schedules.

Worked Example 11

Continuous-for-live-load splice design (post-tensioned diaphragm)

Consulting

Problem

Compute A_s,top for continuity negative moment.

Step-by-Step

d ≈ 8 − 2.5 − 0.5 = 5.0 in in the deck; add composite depth if using extended strands. Assume d = 72 in for full composite section.

A_s = M/(φ·f_y·0.9·d) = 2,200·12/(0.9·60·0.9·72) = 6.79 in²

Design Verification

Positive moment continuity reinforcement (§5.14.1.4.9) also required — extended bottom strands or #6 hooked bars into diaphragm.

Discussion

Simple-made-continuous is the most common PC bridge continuity detail in the US.

Worked Example 12

Capstone: LCCA of PC I-girder vs steel plate girder for 130-ft crossing

Consulting

Problem

Recommend the more economical alternate.

Step-by-Step

PV factor for 75 yr at 3% = (1−(1.03)^-75)/0.03 = 28.5. PC O&M PV = 4·28.5 = $114k. Steel = 9·28.5 = $257k.

PC: 1,350 + 114 = $1.464M. Steel: 1,500 + 257 = $1.757M.

Result

PC wins by $293k (17%) on 75-yr NPV

Design Verification

PC I-girders dominate the 100–160 ft simply-supported market for exactly this LCCA reason. Steel wins on curved, long, or accelerated-erection projects.

Discussion

Never present cost without stating the discount rate and the assumed maintenance schedule — both sensitive.