Reinforced-Concrete Bridge Superstructures — Bridge Engineering & Design

Engineering story

The workhorse of the U.S. short-span inventory

Roughly one in four highway bridges in the National Bridge Inventory is a cast-in-place reinforced-concrete slab or T-beam bridge. They dominate the 20–80 ft span range because they are built on falsework with a single trade — reinforcing ironworkers and concrete finishers — with no shop drawings, no fabrication lead time, and no bearings on short spans (the deck is monolithic with the abutment). The material sits in compression under service loads and rarely fatigues. When they are detailed properly, they last a century.

This chapter walks through the full AASHTO design of the three reinforced-concrete superstructure families — solid slab, T-beam, and RC box girder. We build up the section analysis (Whitney stress block), add the effective flange width, layer in shear and torsion, and finish with two complete worked examples and a design challenge.

Chapter objectives

What you will be able to do

Learning objectives

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

1Choose between slab, T-beam, and RC box-girder superstructures based on span, width, and constructability.
2Apply AASHTO §4.6.2.6 to compute the effective flange width of a T-beam and use it in flexural design.
3Design longitudinal reinforcement for positive and negative moment using the Whitney stress block (§5.6.3).
4Check minimum reinforcement (§5.6.3.3), maximum reinforcement / ductility (§5.6.2.1), and crack control (§5.6.7).
5Design transverse shear reinforcement using the AASHTO simplified sectional method (§5.7.3.4.1).
6Add torsion design for curved or eccentric RC box girders (§5.7.3.6).
7Check fatigue on reinforcement (§5.5.3) and serviceability deflection (§2.5.2.6.2).
8Complete a full RC T-beam design — Strength I, Service I, Fatigue I — with every AASHTO check documented.

6.1 — RC superstructure families

Slab, T-beam, box

AASHTO LRFD §5.12.2, §5.12.3

The three RC superstructure families differ in how efficiently they place concrete in compression and steel in tension. A solid slab has the whole section engaged (inefficient for long spans because most of the concrete is unstressed near the neutral axis). A T-beam concentrates concrete near the compression face and steel near the tension face. A box girder adds a bottom slab that (a) provides a large compression flange for negative moment over interior supports and (b) closes the section to resist torsion.

Comparison of solid slab, T-beam, and RC box-girder bridge cross-sections with typical span ranges — Figure 6.1The three reinforced-concrete superstructure families. Slab bridges span 20–50 ft, T-beam bridges 40–80 ft, RC box-girder bridges 60–120 ft. Beyond 120 ft, prestressed concrete or steel takes over.

System	Span range	Depth ratio (§2.5.2.6.3)	Typical use
Solid slab	20 – 50 ft	1.2·(S+10)/30 (simple)	Culverts, low-traffic county roads
T-beam	40 – 80 ft	0.070 L (simple), 0.065 L (cont.)	State highway short spans
RC box girder	60 – 120 ft	0.060 L (simple), 0.055 L (cont.)	Interchange ramps (Caltrans standard)

6.2 — Effective flange width

How much of the slab helps the beam?

AASHTO LRFD §4.6.2.6

For a T-beam, shear lag causes the longitudinal stress in the slab flange to be highest directly over the web and to fall off toward the mid-panel. AASHTO §4.6.2.6.1 replaces the true non-uniform stress with an equivalent uniform stress acting over an effective flange width b_eff. For interior girders,b_eff is simply the girder spacing S.

T-beam cross-section with effective flange width, slab thickness, web width, and actual vs equivalent stress distribution — Figure 6.2AASHTO §4.6.2.6 effective flange width concept. The true stress varies across the flange (shear lag); the code replaces it with a uniform stress acting on b_eff.

(6.1)

Interior girder
$b_{eff,int} = S$
Exterior girder
$b_{eff,ext} = d_{o} + \min\left(\frac{S}{2},\ 6\cdot t_{s},\ \frac{L_{eff}}{8}\right)$

$b_{eff,int}$: effective flange width, interior girder [in]
$S$: center-to-center girder spacing [in]
$b_{eff,ext}$: effective flange width, exterior girder [in]
$d_{o}$: overhang length (fascia to fascia-girder CL) [in]

6.3 — Flexural design

Whitney stress block, φ, and ductility

AASHTO LRFD §5.6.2, §5.6.3

AASHTO adopts the Whitney rectangular stress block for ultimate flexural analysis. The actual parabolic compression stress distribution is replaced by a uniform stress of 0.85 f_c^' acting over a depth a = β₁·c.

Whitney rectangular stress block on a singly-reinforced concrete beam with strain diagram and force couple — Figure 6.3Whitney stress block. Force equilibrium (C = T) gives a = A_s f_y / (0.85 f'c b); moment equilibrium about the steel gives M_n = A_s f_y (d − a/2).

(6.2)

Formula
$\phi M_{n} = \phi \cdot A_{s} \cdot f_{y} \cdot \left(d - \frac{a}{2}\right) \ge M_{u}$

$M_{n}$: nominal flexural resistance [kip-in]
$a$: depth of Whitney block = β₁·c [in]
$A_{s}$: tension steel area [in²]
$d$: effective depth to tension steel [in]
$phi$: resistance factor = 0.90 for tension-controlled sections [—]

β₁ = 0.85 for f_c^' ≤ 4 ksi and drops 0.05 for each 1 ksi above 4 ksi, with a floor of 0.65 (AASHTO LRFD §5.6.2.2).

6.3.1 Ductility — the tension-controlled requirement

AASHTO §5.6.2.1 requires the net tensile strain ε_t at the extreme tension steel to be at least 0.005 for the full φ = 0.90 to apply. Between 0.002 and 0.005 the section is a transition zone and φ is linearly interpolated. Below 0.002 the section is compression-controlled and φ = 0.75. Bridge girders are almost always designed tension-controlled by keeping c/d_t ≤ 0.375.

6.3.2 Minimum reinforcement — §5.6.3.3

(6.3)

Formula
$\phi M_{n} \ge \min\left(1.33\cdot M_{u},\ 1.2\cdot M_{cr}\right)$

M_cr is the cracking moment computed from the modulus of rupturef_r = 0.24·√f_c^' (ksi) per AASHTO LRFD §5.4.2.6.

6.3.3 Crack control — §5.6.7

(6.4)

Formula
$s_{max} = \frac{700\cdot \gamma_{e}}{\beta_{s}\cdot f_{ss}} - 2\cdot d_{c}$

$s_{max}$: maximum allowable bar spacing [in]
$gamma_{e}$: exposure factor: 1.0 (Class 1), 0.75 (Class 2) [—]
$f_{ss}$: service tensile stress in reinforcement [ksi]
$d_{c}$: cover to center of nearest bar [in]
$eta_{s}$: 1 + d_c / (0.7·(h − d_c)) [—]

6.4 — Shear design

AASHTO simplified sectional method

AASHTO LRFD §5.7.3

AASHTO §5.7.3 offers three shear procedures. For RC (non-prestressed) members, the simplified sectional method (§5.7.3.4.1) is by far the most common: it fixes β = 2.0 and θ = 45° and reduces to the familiar ACI-like form.

AASHTO simplified shear model with concrete contribution V_c and stirrup contribution V_s — Figure 6.4Simplified sectional model. Diagonal cracks form at 45°; concrete carries V_c across the compression zone and stirrups crossing the crack carry V_s.

(6.5)

Formula
$V_{n} = V_{c} + V_{s} = 0.0316\cdot \beta \cdot \sqrt{f_{c}'}\cdot b_{v}\cdot d_{v} + \frac{A_{v}\cdot f_{y}\cdot d_{v}}{s}$

$V_{n}$: nominal shear resistance [kip]
$V_{c}$: concrete contribution [kip]
$V_{s}$: steel contribution [kip]
$d_{v}$: effective shear depth = max(0.9·d, 0.72·h) [in]
$A_{v}$: area of one stirrup (2 legs) [in²]
$s$: stirrup spacing [in]

The 0.0316 factor converts √f_c^' from psi units to ksi. With β = 2.0 this reduces to V_c = 0.0632·√f_c^'·b_v·d_v.

6.4.1 Minimum shear reinforcement — §5.7.2.5

(6.6)

Formula
$A_{v,min} = \frac{0.0316\cdot \sqrt{f_{c}'}\cdot b_{v}\cdot s}{f_{y}}$

6.4.2 Maximum stirrup spacing — §5.7.2.6

If v_u < 0.125·f_c^': s_max = min(0.8·d_v, 24 in).
Otherwise: s_max = min(0.4·d_v, 12 in).

6.5 — Torsion in RC box girders

Closed-section torsion

AASHTO LRFD §5.7.3.6

RC box girder cross-section showing top and bottom slab reinforcement, web shear stirrups, and closed torsion hoops — Figure 6.5RC single-cell box girder. The closed section resists torsion by circulatory shear flow around the box; closed hoop stirrups plus longitudinal steel along all four faces are required.

(6.7)

Formula
$T_{n} = \frac{2\cdot A_{o}\cdot A_{t}\cdot f_{y}}{s\cdot \cot\theta}$

$T_{n}$: nominal torsional resistance [kip-in]
$A_{o}$: area enclosed by shear flow path ≈ 0.85·A_oh [in²]
$A_{t}$: area of one leg of closed torsion hoop [in²]
$heta$: diagonal-crack angle (default 45°) [deg]

Interaction check when both shear and torsion act (AASHTO LRFD §5.7.3.6.2): the equivalent shear stress from combined V_u and T_u must be below the concrete crushing limit 0.25·f_c^'.

6.5b — Worked example (torsion)

Torsional reinforcement design — RC single-cell box girder

AASHTO LRFD §5.7.2.1, §5.7.3.6

Given: single-cell RC box, overall b = 84 in, h = 60 in, wall thickness = 12 in, cover to hoop centerline = 2.5 in, f_c^' = 5 ksi, f_y = 60 ksi. Factored effects at the section: T_u = 3,600 kip-in, V_u = 240 kip,d_v = 54 in. Use φ = 0.90 (AASHTO LRFD §5.5.4.2), θ = 45°.

Step 1 — Section torsional properties

(Ex-6.5b.1)

Perimeter to hoop centerline
$p_{h} = 2\left[(b-2c)+(h-2c)\right]$
Substitute
$p_{h} = 2\left[(84-5)+(60-5)\right] = 268\ \text{in}$
Enclosed area of hoop centerline
$A_{oh} = (b-2c)(h-2c) = 79\cdot 55 = 4{,}345\ \text{in}^{2}$
Gross enclosed shear-flow area
$A_{o} \approx 0.85\,A_{oh} = 0.85(4{,}345) = 3{,}693\ \text{in}^{2}$

Step 2 — Is torsion large enough to design for? (AASHTO LRFD §5.7.2.1)

(Ex-6.5b.2)

Threshold torque
$T_{cr} = 0.126\,\lambda\sqrt{f'_{c}}\;\frac{A_{cp}^{2}}{p_{c}}$
A_cp = b·h, p_c = 2(b+h)
$A_{cp}=84\cdot 60=5{,}040\ \text{in}^{2},\ \ p_{c}=2(84+60)=288\ \text{in}$
Substitute (λ = 1.0)
$T_{cr} = 0.126(1.0)\sqrt{5}\,\frac{5{,}040^{2}}{288} = 24{,}870\ \text{kip-in}$
Design threshold
$\phi\,\tfrac{1}{4}\,T_{cr} = 0.90(0.25)(24{,}870) = 5{,}596\ \text{kip-in}$
Compare
$T_{u}=3{,}600 < 5{,}596 \Rightarrow \text{torsion could be neglected}$

For instructional purposes (and because eccentric live load on a curved box commonly pushesT_u above the threshold), we proceed with full torsion design.

Step 3 — Required transverse hoop steel A_t / s

(Ex-6.5b.3)

Formula (rearrange §5.7.3.6.2)
$\frac{A_{t}}{s} = \frac{T_{u}}{\phi\,2\,A_{o}\,f_{y}\,\cot\theta}$
Substitute
$\frac{A_{t}}{s} = \frac{3{,}600}{0.90\,(2)(3{,}693)(60)(1.0)}$
Result — one leg
$\frac{A_{t}}{s} = 0.00902\ \text{in}^{2}/\text{in per leg}$

Step 4 — Combine with shear stirrups

(Ex-6.5b.4)

Shear demand on stirrups
$\frac{A_{v}}{s} = \frac{V_{u}/\phi - V_{c}}{f_{y}\,d_{v}\,\cot\theta}$
V_c (β = 2.0, simplified)
$V_{c}=0.0316\,\beta\sqrt{f'_{c}}\,b_{v}\,d_{v}=0.0316(2.0)\sqrt{5}(24)(54)=183\ \text{kip}$
Substitute (b_v = two 12-in webs)
$\frac{A_{v}}{s} = \frac{240/0.90 - 183}{60(54)(1.0)} = 0.0258\ \text{in}^{2}/\text{in}$
Combined per web (one leg = A_v/2 + A_t)
$\frac{A_{v}+2A_{t}}{s} = 0.0258 + 2(0.00902) = 0.0438\ \text{in}^{2}/\text{in}$

Try #5 closed hoops (A_b = 0.31 in², two legs → 0.62 in²): s = 0.62 / 0.0438 = 14.2 in. Governing max spacing per AASHTO LRFD §5.7.2.6 is min(0.375·d_v, 12 in) = 12 in. Use #5 closed hoops @ 12 in.

Step 5 — Longitudinal torsion steel A_ℓ

(Ex-6.5b.5)

Formula (§5.7.3.6.3)
$A_{\ell} = \frac{A_{t}}{s}\,p_{h}\,\cot^{2}\theta\,\frac{f_{y}}{f_{y\ell}}$
Substitute
$A_{\ell} = (0.00902)(268)(1.0)^{2}(1.0) = 2.42\ \text{in}^{2}$
Distribute equally around perimeter
$A_{\ell}/\text{face} \approx 2.42/4 = 0.60\ \text{in}^{2}$

Provide 2 – #5 bars in each face (top slab, bottom slab, each web) at hoop corners: 4(0.31) = 1.24 in² per face — well above the 0.60 in² required. Space additional longitudinal bars ≤ 12 in around p_h.

Step 6 — Crushing (interaction) check

(Ex-6.5b.6)

Equivalent shear stress
$v_{u} = \sqrt{\left(\frac{V_{u}}{b_{v}\,d_{v}}\right)^{2} + \left(\frac{T_{u}\,p_{h}}{1.7\,A_{oh}^{2}}\right)^{2}}$
Substitute
$v_{u} = \sqrt{\left(\frac{240}{24\cdot 54}\right)^{2} + \left(\frac{3{,}600\cdot 268}{1.7(4{,}345)^{2}}\right)^{2}}$
Evaluate
$v_{u} = \sqrt{0.185^{2} + 0.030^{2}} = 0.187\ \text{ksi}$
Limit
$\phi\,0.25\,f'_{c} = 0.90(0.25)(5) = 1.125\ \text{ksi} \gg v_{u}\ \checkmark$

Step 7 — Detailing summary

Transverse: #5 closed hoops @ 12 in around the full box perimeter (135° hooks lapped in the corners).
Longitudinal: Continuous bars distributed around p_h with ≥ 1 bar in each corner and spacing ≤ 12 in; total ≥ 2.42 in².
Do not use open U-stirrups for torsion — the shear-flow loop must close through the top slab.

6.6 — Worked example 1

Interior RC T-beam — full flexural and shear design

AASHTO LRFD §4.6.2.2, §5.6.3, §5.7.3

Elevation and cross-section of a 60 ft simply-supported RC T-beam bridge with 5 girders at 8 ft, 8.5 in deck, 42 in web — Figure 6.6Design example. Simply-supported 60 ft span, 40 ft roadway, 5 cast-in-place RC T-beam girders at 8 ft o.c., 8.5 in deck slab, 14 in × 42 in web (50.5 in total depth), 3.5 ft overhangs, 32-in barriers. Materials: f'c = 4 ksi, f_y = 60 ksi.

Step 1 — Loads per interior girder (Strength I)

Section geometry: interior girder tributary width = 8.0 ft. Self-weight of the deck (8.5 in × 0.150 kcf) + web (14 in × 42 in × 0.150 kcf) + barrier share.

Formula — DC per girder

w_DC = (t_s·S/12)·γ + (b_w·h_w/144)·γ + W_b/N

Substitute

= (8.5·8/12)·0.150 + (14·42/144)·0.150 + 0.470/5

Result

= w_DC = 0.850 + 0.613 + 0.094 = 1.56 klf

Formula — DW per girder (2 in FWS × girder width)

w_DW = t_ws·S·γ_ws

Substitute

= (2/12)·8·0.140

Result

= w_DW = 0.187 klf

Step 2 — Live-load distribution factor (§4.6.2.2.2b, Type (e))

Formula — moment DF, one lane loaded

g_M,1 = 0.06 + (S/14)^0.4·(S/L)^0.3·(K_g/12·L·t_s³)^0.1

Substitute

= 0.06 + (8/14)^0.4 · (8/60)^0.3 · 1.03

Result

= g_M,1 = 0.51

Formula — moment DF, two-or-more lanes (governs)

g_M,2 = 0.075 + (S/9.5)^0.6·(S/L)^0.2·(K_g/12·L·t_s³)^0.1

Substitute

= 0.075 + (8/9.5)^0.6 · (8/60)^0.2 · 1.03

Result

= g_M = 0.70 (governs)

Step 3 — Midspan moment envelope

Moment and shear envelope for 60 ft simply-supported T-beam under HL-93 — Figure 6.7Envelope diagrams. Positive moment peaks at midspan; shear peaks at the support.

Formula — HL-93 midspan (truck + lane)

M_LL = g_M·(1 + IM)·M_truck + g_M·M_lane

Substitute

= 0.70·1.33·806 + 0.70·288

Result

= M_LL+IM = 951 kip-ft

Formula — Strength I midspan

M_u = 1.25·M_DC + 1.50·M_DW + 1.75·M_LL+IM

Substitute

= 1.25·702 + 1.50·84 + 1.75·951

Result

= M_u = 878 + 126 + 1664 = 2668 kip-ft

Step 4 — Effective flange width (§4.6.2.6)

Formula — interior girder

b_eff = S = 8 ft

Substitute

= 96 in

Result

= b_eff = 96 in

Step 5 — Flexural design (§5.6.3)

T-beam rebar layout with 8 #10 bars in web and #5 stirrups — Figure 6.8Trial rebar layout: 8 #10 tension bars in two rows (A_s = 10.16 in²), 1.5 in clear cover, d = 46.2 in. #5 stirrups at 6 in wrap the web.

Formula — depth of Whitney block (assume rectangular section, a ≤ t_s)

a = (A_s·f_y) / (0.85·f_c^'·b_eff)

Substitute

= (10.16·60) / (0.85·4·96)

Result

= a = 1.87 in ≤ t_s = 8.5 in ⇒ T-beam behaves as a wide rectangle ✓

Formula — nominal moment

M_n = A_s·f_y·(d − a/2)

Substitute

= 10.16·60·(46.2 − 0.93)/12

Result

= M_n = 2299 kip-ft

Formula — factored resistance

φM_n = 0.90·M_n

Substitute

= 0.90·2299

Result

= φM_n = 2069 kip-ft — INSUFFICIENT (M_u = 2668)

Increase to 10 #10 bars (A_s = 12.70 in²): a = 2.34 in; φM_n = 0.90·12.70·60·(46.2 − 1.17)/12 = 2572 kip-ft — still short. Try 12 #10 in three rows (A_s = 15.24 in², revised d = 44.7 in): a = 2.80 in; φM_n = 2966 kip-ft ≥ 2668 ✓. Check c/d_t = 0.074 ≤ 0.375 (tension-controlled).

Step 6 — Minimum reinforcement (§5.6.3.3)

Formula — cracking moment

M_cr = f_r·I_g/y_t

Substitute

f_r = 0.24·√4 = 0.48 ksi ; using gross T-section I_g ≈ 190,000 in⁴, y_t = 34.6 in

Result

= M_cr = (0.48·190000)/(34.6·12) = 220 kip-ft

Formula — minimum required

φM_n ≥ min(1.33·M_u, 1.2·M_cr)

Substitute

min(1.33·2668 ; 1.2·220) = min(3549 ; 264) = 264 kip-ft

Result

= 2966 ≥ 264 ✓

Step 7 — Shear at the critical section (§5.7.3.2 — d_v from support face)

Formula — factored shear at x = d_v

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

Substitute

= 1.25·40.2 + 1.50·4.8 + 1.75·64.6

Result

= V_u = 50.3 + 7.2 + 113.1 = 170.6 kip

Formula — effective shear depth

d_v = max(0.9·d, 0.72·h)

Substitute

max(0.9·44.7 ; 0.72·50.5)

Result

= d_v = 40.2 in

Formula — concrete contribution (β = 2.0)

V_c = 0.0316·2.0·√f_c^'·b_v·d_v

Substitute

= 0.0316·2.0·√4·14·40.2

Result

= V_c = 71.1 kip

Formula — required steel contribution

V_s = V_u/φ − V_c

Substitute

= 170.6/0.90 − 71.1

Result

= V_s,req = 118.5 kip

Formula — required stirrup spacing (θ = 45°)

s = A_v·f_y·d_v / V_s

Substitute

= 0.62·60·40.2 / 118.5 (#5 double-leg, A_v = 0.62 in²)

Result

= s_req = 12.6 in

Apply the §5.7.2.6 maximum: v_u = V_u/(φ·b_v·d_v) = 170.6/(0.9·14·40.2) = 0.336 ksi < 0.125·f'c = 0.500 ksi ⇒ s_max = min(0.8·40.2, 24) = 24 in. Provide #5 stirrups @ 6 in for the first quarter-span, @ 12 in in the middle half.

Step 8 — Crack control (§5.6.7, Class 2 exposure)

Formula — service moment steel stress

f_ss = M_s·(d − k·d/3) / (A_s·(d − k·d/3)·d)

Substitute

Service I M_s = 702 + 84 + 951 = 1737 kip-ft ; k·d = 8.3 in

Result

= f_ss = 30.9 ksi ≤ 0.6·f_y = 36 ksi ✓

Formula — allowable spacing

s_max = (700·γ_e)/(β_s·f_ss) − 2·d_c

Substitute

= (700·0.75)/(1.18·30.9) − 2·2.5

Result

= s_max = 9.4 in ≥ s_prov ≈ 2.4 in (12 bars in a 14 in web) ✓

Step 9 — Fatigue (§5.5.3)

Formula — fatigue live-load moment (Fatigue I, γ = 1.5)

M_f = γ·g_M·(1 + IM_fat)·M_{fatigue-truck}

Substitute

= 1.5·0.70·1.15·458

Result

= M_f = 553 kip-ft

Formula — stress range in bottom bar

Δf = M_f·(d − k·d/3)/(A_s·(d − k·d/3)·d)

Substitute

Using cracked section n = 8, k = 0.31

Result

= Δf = 9.8 ksi ≤ (ΔF)_TH = 26 − 22·(f_min/f_y) = 24 ksi ✓

Step 10 — Live-load deflection (§2.5.2.6.2)

Formula — allowable

Δ_allow = L/800

Substitute

= 60·12/800

Result

= Δ_allow = 0.90 in

Formula — actual (elastic cracked section, HL-93)

Δ = (5·w_LL·L⁴)/(384·E_c·I_cr)

Substitute

using I_cr = 88,000 in⁴, E_c = 3605 ksi

Result

= Δ = 0.42 in ≤ 0.90 ✓

AASHTO check summary — RC T-beam Example 1

Flexural strength (§5.6.3): φM_n = 2966 ≥ M_u = 2668 kip-ft ✓
Ductility (§5.6.2.1): c/d_t = 0.074 ≤ 0.375 (tension-controlled) ✓
Minimum reinforcement (§5.6.3.3): φM_n ≥ 1.2 M_cr = 264 kip-ft ✓
Crack control (§5.6.7): s_max = 9.4 ≥ 2.4 in provided ✓
Shear (§5.7.3): φ(V_c + V_s) = 0.9·(71.1 + 249) = 288 ≥ 170.6 kip ✓
Fatigue (§5.5.3): Δf = 9.8 ≤ 24 ksi ✓
Deflection (§2.5.2.6.2): Δ = 0.42 ≤ L/800 = 0.90 in ✓

Final section detailing (from computed A_s)

Interior RC T-beam — 14 in web × 42 in web depth (50.5 in total), S = 8 ft, L = 60 ft

Location	A_s required	Bars provided	Spacing / detail
Bottom (tension) longitudinal — midspan	A_s,req = 12.2 in²	8 – #11 (A_s = 12.48 in²)	2 rows of 4; 1½-in clear cover, 2 in clear between bars
Top flange (deck) longitudinal — over interior girder	0.15% of A_g,flange ≈ 1.22 in²/ft	#5 @ 6 in top, #5 @ 8 in bottom	Effective flange b_eff = 96 in (§4.6.2.6)
Skin steel — side face (§5.6.7)	0.012·(d − 30) = 0.20 in²/ft each face	#4 @ 12 in each face over lower h/2	Stops at mid-height of web
Shear stirrups — end 12 ft	A_v/s ≥ 0.055 in²/in	#4 U-stirrup, two-leg (A_v = 0.40 in²)	s = 6 in c/c
Shear stirrups — interior 36 ft	A_v/s ≥ 0.025 in²/in (min per §5.7.2.5)	#4 U-stirrup, two-leg	s = 12 in c/c (≤ 0.8·d_v = 32 in ✓)

Development: 8-#11 bottom bars extend past face of support with a standard 90° hook (l_dh ≈ 22 in per §5.10.8.2). Bar cutoffs staggered to satisfy the envelope of M_u plus the 1.3·M_cr minimum. Distribution steel in the deck follows Ch. 5 (§9.7.3.2).

6.7 — Worked example 2

Solid RC slab bridge — 30 ft simple span

AASHTO LRFD §4.6.2.3, §5.6.3, §5.12.2

Problem statement

Design a simply-supported solid concrete slab bridge. Determine the slab thickness, primary longitudinal reinforcement, distribution steel, and temperature / shrinkage steel per §4.6.2.3 and §5.6.

Given

SpanL = 30 ft simply-supported
RoadwayW = 30 ft, two 12 ft lanes
f'_c4.0 ksi
f_y60 ksi (Grade 60)
Wearing surface2 in bituminous, DW = 0.025 ksf
Barriers32-in F-shape, 0.470 klf each side
Cover2.0 in top, 1.0 in bottom (§5.10.1)

Required

Compute h, primary A_s, distribution A_s,dist, and T & S steel.

Step 1 — Minimum thickness (§2.5.2.6.3, Table)

Formula — simple-span solid slab minimum

h_min = 1.2·(S + 10)/30

Substitute

= 1.2·(30 + 10)/30

Result

= h_min = 1.60 ft = 19.2 in ⇒ try h = 20 in

Step 2 — Equivalent strip width (§4.6.2.3)

Formula — one-lane, single-lane loaded

E₁ = 10 + 5·√(L₁·W₁)

Substitute

L₁ = min(L, 60) = 30 ft ; W₁ = min(W, 30) = 30 ft

Result

= E₁ = 10 + 5·√900 = 160 in = 13.3 ft

Formula — multi-lane

E_m = 84 + 1.44·√(L₁·W₁) ≤ 12·W/N_L

Substitute

= 84 + 1.44·√900 = 127 in = 10.6 ft ; cap 12·30/2 = 180 in

Result

= E_m = 10.6 ft (governs)

Step 3 — Design moments per strip

Formula — self-weight moment per foot

M_DC = (w_s·L²)/8

Substitute

w_s = (20/12)·0.150 = 0.250 klf/ft ; M_DC = 0.250·30²/8

Result

= M_DC = 28.1 kip-ft/ft

Formula — HL-93 truck moment on E strip

M_LL+IM = (1 + IM)·M_truck/E_m

Substitute

M_truck at midspan = 224 kip-ft ; IM = 0.33

Result

= M_LL+IM = 1.33·224/10.6 = 28.1 kip-ft/ft

Formula — Strength I moment

M_u = 1.25·M_DC + 1.50·M_DW + 1.75·M_LL+IM

Substitute

= 1.25·28.1 + 1.50·3.5 + 1.75·28.1

Result

= M_u = 84.6 kip-ft/ft

Step 4 — Primary reinforcement

Formula — required A_s (assume ρ, iterate)

A_s = M_u/(φ·f_y·(d − a/2))

Substitute

d = 20 − 1 − 0.625 = 18.4 in ; assume a = 2.0 in

Result

= A_s = 84.6·12/(0.9·60·17.4) = 1.08 in²/ft

Provide #9 @ 8 in (A_s = 1.50 in²/ft). Recompute a = 1.50·60/(0.85·4·12) = 2.21 in; φM_n = 0.9·1.50·60·(18.4 − 1.10)/12 = 117 kip-ft/ft ≥ 84.6 ✓.

Step 5 — Distribution reinforcement (§5.12.2.1)

Formula — percentage of main steel

% = 100/√L ≤ 50 %

Substitute

= 100/√30

Result

= 18.3 % of main steel

Formula — required transverse A_s

A_s,dist = 0.183·A_s,main

Substitute

= 0.183·1.50

Result

= A_s,dist = 0.275 in²/ft ⇒ #5 @ 12 in (0.31 in²/ft) ✓

Step 6 — Shrinkage & temperature (§5.10.6)

Formula — each face

A_s,st = 1.30·b·h/(2·(b + h)·f_y)

Substitute

b = 360 in per 30 ft, h = 20 in

Result

= A_s,st = 0.13 in²/ft ⇒ #4 @ 18 in top and bottom ✓

AASHTO check summary — RC slab Example 2

Minimum thickness (§2.5.2.6.3): h = 20 ≥ 19.2 in ✓
Flexural strength (§5.6.3): φM_n = 117 ≥ M_u = 84.6 kip-ft/ft ✓
Distribution steel (§5.12.2.1): #5 @ 12 in ✓
Shrinkage & temperature (§5.10.6): #4 @ 18 in each face ✓
Deflection (§2.5.2.6.2): Δ_LL = 0.28 in ≤ L/800 = 0.45 in ✓

Final section detailing (from computed A_s)

Solid RC slab bridge — 20 in thick × 30 ft roadway, L = 30 ft

Location	A_s required	Bars provided	Spacing / detail
Primary bottom (longitudinal) — midspan	A_s,req ≈ 1.60 in²/ft	#9 @ 7 in bottom (A_s = 1.71 in²/ft)	1-in bottom cover; hooked at abutments
Primary top (longitudinal) — over abutment	0.5·A_s,bot for continuity / temperature	#6 @ 12 in top	Extends 0.3·L past support
Distribution steel (transverse, bottom) — §5.12.2.1	A_s,dist = 100/√L · A_s,pri ≤ 50% = 0.79 in²/ft	#5 @ 12 in bottom (A_s = 0.31 in²/ft → increase to #6 @ 8 in = 0.66 in²/ft)	Placed above the primary bars
Shrinkage & temperature (transverse, top) — §5.10.6	A_s,T&S ≥ 0.11 in²/ft each face	#4 @ 18 in each face	Both directions on top mat

Effective strip widths: E₁ = 13.3 ft (single-lane), E_m = 10.6 ft (multi-lane, governs). All bars Grade 60; 2 in top cover, 1 in bottom cover. Bar clearances and cutoffs per §5.10.8; extend at least ⅓ of positive steel into the support (§5.10.8.1.2b).

6.7b — Design Example 5

Reinforced-concrete deck slab on five steel girders (approximate strip method)

AASHTO LRFD §4.6.2.1, §9.7, §5.7.3

This worked example extracts the full solution from Simplified LRFD Bridge Design(Taylor & Francis, 2008), Example 5. A 48 ft wide, 44 ft 6 in clear-roadway cast-in-place concrete deck is continuous across five W12×65 steel girders spaced at 10 ft with 4 ft overhangs. Every calculation follows the Formula → Substitute → Resultpattern using AASHTO Art. 4.6.2.1 (approximate strip method) and HL-93 live load.

Problem statement

Design the transverse reinforcement for the cast-in-place deck slab of a five-girder simple-span composite bridge using the AASHTO approximate strip method.

Given

Bridge width, overall48 ft
Clear roadway width44 ft 6 in
Slab thickness (structural + IWS)t = 8.5 + 0.5 = 9.0 in
Girder spacing, S10 ft (four spans)
Deck overhang4 ft (curb/parapet 21 in wide, 32 in tall, A = 3.37 ft²)
Future wearing surface, wFWS0.03 kip/ft² (3 in)
Concrete unit weight, wc0.150 kip/ft³
f'c (28-day)4.5 ksi
Fy (reinforcement)60 ksi
Live loadAASHTO HL-93 (axles only per §3.6.1.3.3), IM = 33%

Figure 6.10 (source Fig. 2.50)Concrete deck slab design example. Overall bridge cross-section (top) and cantilever curb-and-parapet geometry (bottom). The parapet centroid is x̄ = 0.66 ft from the exterior girder centerline.

Step 1 — Minimum slab thickness (§9.7.1.1)

The AASHTO minimum concrete deck depth is 7.0 in. Adopt structural depth 8.5 in plus 0.5 in integral wearing surface = t = 9.0 in.

Step 2 — Dead loads on a 1 ft transverse strip

Slab self-weight

w_{slab} = w_c \cdot t

Substitute

= (0.150\text{ kip/ft}^3)(9.0\text{ in})\left(\dfrac{1\text{ ft}}{12\text{ in}}\right)

Result

w_{slab} = 0.113\ \text{kip/ft}^2

Future wearing surface

w_{FWS} = 0.03\ \text{kip/ft}^2\ \text{(given)}

Curb + parapet per side (line load along span)

w_{C\&P} = w_c \cdot A_{C\&P}

Substitute

= (0.150\text{ kip/ft}^3)(3.37\text{ ft}^2)

Result

w_{C\&P} = 0.506\ \text{kip/ft of bridge span}

Step 3 — Dead-load force effects (approximate elastic strip, §4.6.2.1)

Model a 1 ft-wide transverse strip as a continuous beam over five rigid girder supports with 4 ft overhangs. Report the reaction R_A at the exterior support and moments at A (support), B (0.4S = 4 ft into the first interior span, maximum + moment) and C (first interior support, maximum − moment).

Figure 6.11 (source Fig. 2.51)Idealized transverse strip. Point A = exterior support, B = 0.4S from A (peak positive DL moment), C = first interior support (peak negative DL moment).

Fig. 2.52Slab self-weight applied over spans between girders only (cantilever excluded). Results: MA = 0, MB = +0.872, MC = −1.211 kip-ft/ft, RA = 0.444 kip/ft.

Fig. 2.53Slab dead load acting only on the two cantilevers. Results: MA = −0.904, MB = −0.439, MC = +0.258 kip-ft/ft, RA = 0.568 kip/ft.

Fig. 2.54Curb + parapet (0.506 kip/ft, applied at x̄ = 0.66 ft). Results: MA = −1.689, MB = −0.821, MC = +0.483 kip-ft/ft, RA = 0.723 kip/ft.

Fig. 2.55FWS (0.03 kip/ft²) between curbs only. Results: MA = −0.076, MB = +0.195, MC = −0.300 kip-ft/ft, RA = 0.195 kip/ft.

Step 4 — Live-load force effects (§3.6.1.3, §4.6.2.1.3)

Under the approximate strip method only the design-truck axles act on the deck (§3.6.1.3.3). Each wheel P = 16 kips, spaced 6 ft transversely. Wheels stay ≥ 1.0 ft from the face of curb (overhang design) and ≥ 2.0 ft from the design-lane edge (interior design). Number of design lanes across the 44.5 ft roadway:

Number of design lanes (§3.6.1.1.1)

N_L = \operatorname{int}\!\left(\dfrac{w}{12\text{ ft}}\right)

Substitute

= \operatorname{int}\!\left(\dfrac{44.5}{12}\right) = \operatorname{int}(3.71)

Result

N_L = 3\ \text{lanes}\ \ (m = 1.2,\ 1.0,\ 0.85\ \text{for 1, 2, 3 lanes})

4a — Maximum negative LL moment on the overhang

Fig. 2.56Wheel placed 1 ft from face of curb → X = 15 in = 1.25 ft from centerline of girder A.

Equivalent overhang strip width E (Tbl. 4.6.2.1.3-1)

E = 45.0 + 10.0\,X\ \text{[in]}

Substitute

= 45.0 + 10.0(1.25\text{ ft}) = 57.5\text{ in}

Result

E = 4.79\ \text{ft}

Negative LL moment per foot (m = 1.2, one lane)

M_A = -\dfrac{m\cdot P\cdot X}{E}

Substitute

= -\dfrac{(1.2)(16.0\text{ kips})(1.25\text{ ft})}{4.79\text{ ft}}

Result

M_A = -5.01\ \text{ft-kip/ft}

Fig. 2.57Free body used to compute MA: P = 16 kips at X = 1.25 ft, cut at A.

4b — Maximum positive LL moment in the first interior span

Fig. 2.58Single-lane loading: leading wheel at point B (0.4S = 4 ft), trailing wheel 6 ft away.

Equivalent strip width for interior positive M (Tbl. 4.6.2.1.3-1)

E_{+M} = 26.0 + 6.6\,S\ \text{[in]}

Substitute

= 26.0 + 6.6(10\text{ ft}) = 92\text{ in}

Result

E_{+M} = 7.67\ \text{ft}

Structural analysis of the strip under one-lane truck axles:

Formula

R_A = 8.160\ \text{kips};\quad M_B = 32.64\ \text{ft-kip}

Adjust for strip width and m = 1.2 (one lane)

R'_A = \dfrac{m\,R_A}{E_{+M}},\qquad M'_B = \dfrac{m\,M_B}{E_{+M}}

Substitute

R'_A = \dfrac{(1.2)(8.16)}{7.67},\qquad M'_B = \dfrac{(1.2)(32.64)}{7.67}

Result

R'_A = 1.28\ \text{kip/ft},\qquad M'_B = +5.11\ \text{ft-kip/ft}

Fig. 2.59Two-lane check (m = 1.0). Second axle pair placed at X = 28/34 ft to maximize the effect at B.

Two-lane check (m = 1.0)

M'_B = \dfrac{m\,M_B}{E_{+M}}

Substitute

= \dfrac{(1.0)(34.01)}{7.67}

Result

M'_B = +4.43\ \text{ft-kip/ft}\ <\ 5.11\ \Rightarrow\ \text{single lane governs}

4c — Maximum negative LL moment at the first interior support

Fig. 2.60Truck straddling support C, one lane loaded, produces the maximum negative moment MC.

Equivalent strip width for interior negative M (Tbl. 4.6.2.1.3-1)

E_{-M} = 48.0 + 3.0\,S\ \text{[in]}

Substitute

= 48.0 + 3.0(10\text{ ft}) = 78\text{ in}

Result

E_{-M} = 6.5\ \text{ft}

Formula

M'_C = \dfrac{m\,M_C}{E_{-M}}

Substitute

= \dfrac{(1.2)(-27.48)}{6.5}

Result

M'_C = -5.07\ \text{ft-kip/ft}

4d — Maximum live-load reaction at support A

Fig. 2.61Governing wheel placement (1 ft from curb) for the maximum LL reaction RA. Same E = 4.79 ft as the overhang strip.

Formula

R'_A = \dfrac{m\,R_A}{E}

Substitute

= \dfrac{(1.2)(25.36\text{ kips})}{4.79\text{ ft}}

Result

R'_A = 6.35\ \text{kip/ft}

Load case (per ft width)	R_A	M_A	M_B	M_C
Slab DL (excl. cantilever)	0.444	0.000	+0.872	−1.211
Slab DL (cantilever only)	0.568	−0.904	−0.439	+0.258
Curb + parapet	0.723	−1.689	−0.821	+0.483
FWS (DW)	0.195	−0.076	+0.195	−0.300
LL max −M overhang	—	−5.010	—	—
LL max +M / R first span	1.280	—	+5.110	—
LL max −M first support C	—	—	—	−5.070
LL max reaction at A	6.350	—	—	—

Step 5 — Strength I limit state (§1.3.2, §3.4.1)

η_D = η_R = η_I = 1.0, so η_i = 1.0. DC includes slab, cantilever slab and C&P (γ_p = 1.25 max, 0.90 min); DW = FWS only (γ_p = 1.50/0.65); LL uses γ = 1.75 with IM = 33 %.

Strength I combination (§1.3.2.1)

Q = \sum \eta_i\,\gamma_i\,Q_i = 1.0\!\left[\gamma_{DC}\,DC + \gamma_{DW}\,DW + 1.75\,(1{+}IM)\,LL\right]

Reaction at A

R_A = 1.25(R_{slab}+R_{cant}+R_{C\&P}) + 1.5\,R_{FWS} + 1.75(1.33)\,R_{LL}

Substitute

= 1.25(0.444+0.568+0.723) + 1.5(0.195) + 1.75(1.33)(6.35)

Result

R_A = 17.24\ \text{kip/ft}

Moment at A (overhang, γDC max, γDW max, LL max)

M_A = 1.25(M_{cant}+M_{C\&P}) + 1.5\,M_{FWS} + 1.75(1.33)\,M_{LL,ov}

Substitute

= 1.25(-0.904-1.689) + 1.5(-0.076) + 1.75(1.33)(-5.010)

Result

M_A = -15.02\ \text{ft-kip/ft}

Moment at B (γp = 0.90 on cantilever + C&P since they reduce +M)

M_B = 1.25\,M_{slab} + 0.90(M_{cant}+M_{C\&P}) + 1.5\,M_{FWS} + 1.75(1.33)\,M_{LL,+}

Substitute

= 1.25(0.872) + 0.90(-0.439-0.821) + 1.5(0.195) + 1.75(1.33)(5.110)

Result

M_B = +12.14\ \text{ft-kip/ft}

Moment at C (γp = 0.90 on cantilever + C&P since they reduce −M)

M_C = 1.25\,M_{slab} + 0.90(M_{cant}+M_{C\&P}) + 1.5\,M_{FWS} + 1.75(1.33)\,M_{LL,-}

Substitute

= 1.25(-1.211) + 0.90(0.258+0.483) + 1.5(-0.300) + 1.75(1.33)(-5.070)

Result

M_C = -13.10\ \text{ft-kip/ft}

Design demands (Strength I)

Positive-moment demand M_u⁺ = +12.14 ft-kip/ft (at B).
Negative-moment demand M_u⁻ = −15.02 ft-kip/ft (at A, overhang governs over −13.10 at C).
Reaction demand R_u = 17.24 kip/ft.

Step 6 — Flexural design for the governing negative moment

Try epoxy-coated #5 bars (d_b = 0.625 in, A_b = 0.31 in²) with 2.5 in cover. Maximum spacing s_max = min(1.5t, 18) = 1.5(8.5) = 12.75 in.

Fig. 2.62Deck slab section for reinforcement placement (1 ft-wide strip).

Effective depth d

d = t - t_{IWS} - c - \tfrac{d_b}{2}

Substitute

= 9.0 - 0.5 - 2.5 - \dfrac{0.625}{2}

Result

d = 5.68\ \text{in}

Equivalent stress-block depth (§5.7.3.2)

a = \dfrac{A_s\,f_y}{0.85\,f'_c\,b}

Substitute

= \dfrac{A_s(60)}{(0.85)(4.5)(12)}

Result

a = 1.307\,A_s\ \text{[in, with }A_s\text{ in in}^2\text{/ft]}

Factored flexural resistance (§5.5.4.2.1, φ = 0.9)

M_r = \varphi\,A_s\,f_y\!\left(d - \dfrac{a}{2}\right)

Set Mr = Mu and solve for As

(15.02)(12) = 0.9\,A_s(60)\!\left(5.68 - \dfrac{1.307}{2}A_s\right)

Result

A_s = 0.64\ \text{in}^2/\text{ft (required)}

Try #5 @ 5.5 in (provided As)

A_{s,prov} = A_b\!\left(\dfrac{12}{s}\right)

Substitute

= 0.31\!\left(\dfrac{12}{5.5}\right)

Result

A_{s,prov} = 0.68\ \text{in}^2/\text{ft} \geq 0.64\ \checkmark

Recompute a with As,prov

a = \dfrac{A_{s,prov}\,f_y}{0.85\,f'_c\,b}

Substitute

= \dfrac{(0.68)(60)}{(0.85)(4.5)(12)}

Result

a = 0.889\ \text{in}

Verify Mr

M_r = \varphi\,A_{s,prov}\,f_y\!\left(d - \dfrac{a}{2}\right)

Substitute

= (0.9)(0.68)(60)\!\left(5.68 - \dfrac{0.889}{2}\right)\!\dfrac{1}{12}

Result

M_r = 16.0\ \text{ft-kip/ft} \geq M_u = 15.02\ \checkmark

Step 7 — Minimum reinforcement (§5.7.3.3.2)

Modulus of rupture (§5.4.2.6)

f_r = 0.37\sqrt{f'_c}

Substitute

= 0.37\sqrt{4.5}

Result

f_r = 0.785\ \text{ksi}

Non-composite section modulus (structural 8.5 in)

S_{nc} = \dfrac{b\,t_s^2}{6}

Substitute

= \dfrac{(12)(8.5)^2}{6}

Result

S_{nc} = 144.5\ \text{in}^3

Cracking moment

M_{cr} = S_{nc}\,f_r

Substitute

= (144.5)(0.785)

Result

M_{cr} = 113.4\ \text{in-kip} = 9.45\ \text{ft-kip/ft}

Governing minimum (min of 1.2 Mcr and 1.33 Mu)

\min(1.2\,M_{cr},\ 1.33\,M_u)

Substitute

= \min(11.34,\ 19.98)\ \text{ft-kip/ft}

Result

= 11.34\ \text{ft-kip/ft} \leq M_r = 16.0\ \checkmark

Step 8 — Distribution reinforcement (§9.7.3.2)

Effective span (steel girder, tw = 0.39 in for W12×65)

S_e = S - t_w

Substitute

= 10\text{ ft} - (0.39\text{ in})(1/12)

Result

S_e \approx 9.97\ \text{ft}\ (\text{use }10\text{ ft})

Percent of positive-moment steel required in secondary direction

p = \min\!\left(\dfrac{220}{\sqrt{S_e}},\ 67\%\right)

Substitute

= \min(220/\sqrt{10},\ 67\%) = \min(69.6\%,\ 67\%)

Result

p = 67\%

Required distribution steel

A_{s,dist} = 0.67\,A_{s,prov}

Substitute

= 0.67(0.68)

Result

A_{s,dist} = 0.46\ \text{in}^2/\text{ft}\ \Rightarrow\ \text{#5 @ 8 in bottom longitudinal}

Step 9 — Shrinkage & temperature (§5.10.8)

Formula

A_{s,temp} \geq \dfrac{1.30\,b\,h}{2\,(b+h)\,f_y}

Substitute

= \dfrac{1.30(12)(8.5)}{2(12+8.5)(60)}

Result

A_{s,temp} = 0.054\ \text{in}^2/\text{ft (with } 0.11 \leq A_{s,temp} \leq 0.60 \text{)}

Adopt #4 @ 18 in top longitudinal (A_s = 0.13 in²/ft), satisfying the code floor of 0.11 in²/ft and the 3t / 18 in maximum spacing.

Design summary — Deck slab reinforcement

Primary (transverse) bottom & top: #5 @ 5.5 in (A_s = 0.68 in²/ft).
Distribution (longitudinal) bottom: #5 @ 8 in (0.46 in²/ft).
Shrinkage / temperature (longitudinal) top: #4 @ 18 in (0.13 in²/ft).
Total slab thickness: 9.0 in (8.5 in structural + 0.5 in integral wearing surface).
Governing checks: φMn = 16.0 ≥ Mu = 15.02 ft-kip/ft; 1.2 Mcr = 11.34 ≤ φMn.

6.8 — Detailing must-knows

Cover, splices, and cutoffs

AASHTO LRFD §5.10

Concrete cover (§5.10.1): 2 in top of deck (Class 2), 1 in bottom of deck, 1.5 in on stirrups, 3 in bottom of footings, 2 in for columns exposed to earth.
Development length (§5.10.8.2): ℓ_d = 1.25·A_b·f_y/√f'c ≥ 12 in. For #10 in 4 ksi concrete, ℓ_d ≈ 57 in.
Class B splice (§5.10.8.4): ℓ_s = 1.3·ℓ_d.
Bar cutoff (§5.10.8.1.2b): extend past theoretical cutoff by max(d, 15·d_b, L/20). At least ⅓ of positive-moment steel must extend into the support.
Skin reinforcement (§5.6.7): required in webs ≥ 3 ft deep; #4 @ 12 in on each face for ≥ h/2 from the tension face.

6.9 — Design challenge

Two-span continuous RC T-beam bridge

Design the two-span continuous RC T-beam bridge shown below. Each span is 70 ft, 44 ft roadway with four T-beam girders at 12 ft spacing, 9 in deck slab, 32 in barriers, and integral abutments plus an integral pier. Materials: f'c = 4.5 ksi, f_y = 60 ksi.

Required deliverables — every AASHTO check, both critical sections

Load calculation: DC, DW, LL+IM per interior and exterior girder. State formula first, then substitute.
LLDF (§4.6.2.2.2): Type (e) — one-lane and multi-lane distribution factors for moment and shear; lever rule check on exterior girder.
Envelopes: Strength I positive-moment envelope (midspan of each span) and negative-moment envelope (over the pier). Include Fatigue I.
Effective flange width (§4.6.2.6): interior and exterior. For negative-moment region, treat as rectangular b_w.
Positive-moment flexural design: longitudinal steel at midspan of Span 1.
Negative-moment flexural design: top steel over the pier (rectangular section, b_w only).
Minimum reinforcement (§5.6.3.3), ductility (§5.6.2.1), and crack control (§5.6.7).
Shear design (§5.7.3.4.1): stirrup schedule from support to midspan for both spans. Include §5.7.2.5 minimum and §5.7.2.6 maximum spacing.
Fatigue check (§5.5.3) on the bottom bars at midspan and on the top bars over the pier.
Deflection (§2.5.2.6.2): live-load deflection ≤ L/800 both spans.
Rebar drawings: longitudinal elevation with bar cutoffs (§5.10.8), transverse cross-sections at midspan, pier, and support, plus stirrup schedule.
Bill of materials & mass estimate.

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Chapter 6 challenge — Two-span continuous RC T-beam bridge

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Chapter summary

Key takeaways

RC slab, T-beam, and box girders cover the 20 – 120 ft span range and remain the workhorses of the U.S. short-span inventory.
AASHTO §4.6.2.6 effective flange width converts the true shear-lag stress distribution into a uniform design stress for T-beam flexure.
Whitney stress block (§5.6.2, §5.6.3) gives φM_n = φ·A_s·f_y·(d − a/2); tension-controlled requires c/d_t ≤ 0.375.
Simplified shear (§5.7.3.4.1) uses β = 2.0, θ = 45° — cleanly separates concrete and stirrup contributions.
Box girders add torsion (§5.7.3.6) and require closed hoops plus longitudinal steel around the entire perimeter.
Every calculation follows the formula-first, substitute-then-result pattern — no exceptions.

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

Flexural capacity of a T-beam

Basic

Problem

Compute φM_n and verify tension-controlled behavior.

Step-by-Step

a = A_{s}\cdot f_{y}/(0.85\cdot f'_{c}\cdot b_{f}) = 7.62\cdot 60/(0.85\cdot 4\cdot 84) = 1.60 in

1.60 < 7.0 \rightarrow rectangular behavior with b = b_{f}. \checkmark

Design Verification

Because a < t_s, all compression is in the flange — treat as a rectangular beam with width b_f.

Discussion

If a > t_s, use the flanged-section equations (compression in flange + web). Errors here are the most common cause of T-beam capacity over-estimates.

Worked Example 2

Shear design of an RC girder

Intermediate

Problem

Determine V_c, required V_s, and #4 stirrup spacing.

Step-by-Step

V_{c} = 0.0316\cdot \beta \cdot \sqrt{f}'_{c}\cdot b_{v}\cdot d_{v} = 0.0316\cdot 2.0\cdot 2\cdot 14\cdot 31

Result

V_{c} = 54.9 kip

0.90\cdot 54.9 = 49.4 kip; V_{u} > \phi V_{c} \rightarrow \ \text{stirrups required}

Design Verification

Check s_max: 0.8·d_v = 24.8 in and 24 in → 5-in spacing OK. ✓

Discussion

Non-prestressed members can adopt β = 2, θ = 45° per §5.7.3.4.2 without iteration — the price is slightly heavier stirrups vs full MCFT.

Worked Example 3

Serviceability crack-control check (Class 1 exposure)

Intermediate

Problem

Verify §5.6.7 rebar spacing limit for Class 1 exposure (γ_e = 1.00).

Step-by-Step

\beta _{s} = 1 + 2.5/(0.7\cdot (36-2.5)) = 1 + 0.106

Result

\beta _{s} = 1.11

s_{max} = 700\cdot \gamma _{e}/(\beta _{s}\cdot f_{ss}) - 2\cdot d_{c} = 700/(1.11\cdot 36) - 5 = 17.5 - 5

Result

s_{max} = 12.5 in

Design Verification

Class 2 (γ_e = 0.75) reduces s_max to 9.4 in — still OK. ✓

Discussion

Crack-control governs bar spacing before flexural capacity for many decks and shallow slabs. Always check §5.6.7 even when φM_n is fine.

Section 3

Guided Practice

Complete the missing steps. Use Hints for AASHTO article pointers and setup logic before revealing the full step. Submit at the end to send your work to your instructor.

Guided Problem 1

Rectangular RC beam — moment capacity

Interior girder rectangular RC beam: $b = 16\ \text{in}$ , $d = 32\ \text{in}$ , $A_{s} = 6.32\ \text{in}^{2}$ (4-#10), $f'_{c} = 4\ \text{ksi}$ , $f_{y} = 60\ \text{ksi}$ .

Step 1

Depth of the equivalent stress block $a$ (in). $a = \dfrac{A_{s} f_{y}}{0.85 f'_{c} b}$ .

Answer

Step 2

Neutral-axis depth $c$ (in). $\beta_{1} = 0.85$ for $f'_{c} = 4\ \text{ksi}$ .

Answer

Step 3

Tension-controlled check: strain in extreme steel. $\varepsilon_{t} = 0.003 \dfrac{d-c}{c}$ .

Answer

Step 4

Design moment $\phi M_{n}$ (k-ft). $\phi = 0.90$ .

Answer

Guided Problem 2

T-beam interior girder — effective flange width

Interior T-girder: web $b_{w} = 14\ \text{in}$ , deck $t_{s} = 8\ \text{in}$ , $S = 9\ \text{ft}$ c/c, span $L = 90\ \text{ft}$ .

Step 1

$L/4$ in inches.

Answer

Step 2

$12 t_{s} + b_{w}$ (in).

Answer

Step 3

Girder spacing in inches.

Answer

Step 4

Effective flange width $b_{\text{eff}}$ (in) = min of the three.

Answer

Guided Problem 3

Shear design — one-way concrete beam

$b_{w} = 14\ \text{in}$ , $d_{v} = 30\ \text{in}$ , $V_{u} = 220\ \text{kip}$ , $f'_{c} = 4\ \text{ksi}$ . Use AASHTO simplified $\beta = 2$ , $\theta = 45°$ .

Step 1

$V_{c} = 0.0316 \beta \sqrt{f'_{c}}\, b_{w} d_{v}$ (kip). $f'_{c}$ in ksi.

Answer

Step 2

Required $V_{s} = V_{u}/\phi - V_{c}$ (kip). $\phi = 0.90$ .

Answer

Step 3

Stirrup spacing $s = \dfrac{A_{v} f_{y} d_{v}}{V_{s}}$ (in) for #4 double-leg ( $A_{v} = 0.40\ \text{in}^{2}$ ).

Answer

Step 4

Max stirrup spacing per §5.7.2.6 if $v_{u} < 0.125 f'_{c}$ (in).

Answer

Guided Problem 4

Crack control — Class 2 exposure spacing

Bottom of deck slab, $\gamma_{e} = 0.75$ (Class 2). $f_{ss} = 36\ \text{ksi}$ (Service I), $d_{c} = 1.25\ \text{in}$ , $h = 8\ \text{in}$ .

Step 1

$\beta_{s} = 1 + \dfrac{d_{c}}{0.7(h-d_{c})}$ .

Answer

Step 2

Denominator $\beta_{s} f_{ss}$ (ksi).

Answer

Step 3

$s_{\max} = \dfrac{700 \gamma_{e}}{\beta_{s} f_{ss}} - 2 d_{c}$ (in).

Answer

Step 4

Actual bar spacing (in) using #5 @ 8-in c/c satisfying $s_{\max}$ ?

Answer

Section 4

Independent Practice

Every problem randomizes its inputs. Work each step, submit for immediate feedback, request new values to practice again.

Practice 1

Whitney stress-block depth a

A_s

As = 6.5 in²

f_y

fy = 60 ksi

f′_c

fc = 7 ksi

b = 22 in

Step 1a = A_s·f_y / (0.85·f′_c·b).

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 2

Nominal moment capacity M_n

A_s

As = 10 in²

f_y

fy = 70 ksi

d = 18 in

a = 3.5 in

Step 1A_s·f_y·(d − a/2)/12 in k-ft.

Answer (k-ft)

Randomized inputs, symbolic grading (±2%).

Practice 3

Tension-controlled strain check

d = 18 in

c = 5.1000000000000005 in

Step 1ε_t = 0.003(d−c)/c.

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 4

Minimum reinforcement ratio

f′_c

fc = 3 ksi

f_y

fy = 75 ksi

Step 13·√f′_c/f_y (ACI/AASHTO lower bound).

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 5

V_c simplified

f′_c

fc = 3 ksi

b_w

bw = 19 in

d_v

dv = 59 in

Step 10.0316·2·√f′_c·b_w·d_v (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 6

Stirrup spacing

V_s

Vs = 195 kip

A_v

Av = 0.4 in²

f_y

fy = 80 ksi

d_v

dv = 49 in

Step 1s = A_v·f_y·d_v/V_s (in).

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 7

Development length (tension #6 bar)

d_b

db = 1.375 in

f_y

fy = 65 ksi

f′_c

fc = 5.5 ksi

Step 1l_d ≈ 2.4·d_b·f_y/√f′_c.

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 8

Crack-control spacing

γ_e

ge = 1 -

d_c

dc = 1.25 in

h = 11.5 in

f_ss

fss = 47 ksi

Step 1β_s = 1 + d_c/(0.7(h−d_c)).

Answer (-)

Step 2s_max = 700·γ_e/(β_s·f_ss) − 2·d_c.

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 9

Modulus of elasticity of concrete

f′_c

fc = 4.5 ksi

Step 1E_c = 33·(150)^1.5·√f′_c ≈ 1820·√f′_c (ksi).

Answer (ksi)

Randomized inputs, symbolic grading (±2%).

Practice 10

Deck slab dead-load moment

t = 9 in

S = 11 ft

Step 1w = 0.150·(t/12) klf/ft.

Answer (klf/ft)

Step 2M = w·S²/10 (continuous).

Answer (k-ft/ft)

Randomized inputs, symbolic grading (±2%).

Practice 11

Effective flange width (min rule)

Span

L = 105 ft

t_s

ts = 9 in

b_w

bw = 18 in

S = 11.5 ft

Step 1L/4 (in).

Answer (in)

Step 212 t_s + b_w.

Answer (in)

Step 3S in inches.

Answer (in)

Step 4b_eff = min of three.

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 12

Modulus of rupture and cracking moment

f′_c

fc = 3 ksi

I_g (in⁴)

Ig = 10500 in^4

y_t (in)

yt = 11 in

Step 1f_r = 0.24·√f′_c (ksi).

Answer (ksi)

Step 2M_cr = f_r·I_g/y_t /12 in k-ft.

Answer (k-ft)

Randomized inputs, symbolic grading (±2%).