23. RC — Shear Design of Beams

01

Engineering story

A century of steel — from concept to skyline

Engineer reviewing blueprints against a steel-frame construction site at sunrise

Concrete carries some shear; stirrups carry the rest.

Vn = Vc + Vs; Vc = 2·λ·√f'c·bw·d; Vs = Av·fy·d/s.

Chapter 23 covers ACI 318-19 §22.5 shear: concrete contribution Vc, stirrup contribution Vs, spacing limits, and minimum shear reinforcement.

Iconic steel structures built on engineering excellence

Ritter–Mörsch truss analogy
1899
ACI Vc = 2√f'c·bd
1963
ACI 318-19 shear provisions
2019

Load pathVu from wuVc + VsStirrup spacing ssmax limitsDuctile shear behaviour

02

Learning objectives

What you will be able to do after finishing Chapter 1 — and why each objective matters in practice

Objective 01

Compute Vc = 2·λ·√f'c·bw·d

Basic concrete contribution (psi units).

Why it matters: Baseline shear capacity.
Where it is used: Every RC beam.
Connects to: ACI 318-19 22.5.5.1.

Objective 02

Compute Vs = Av·fy·d/s

Vertical stirrups carry the diagonal-tension component.

Why it matters: Fine-tuned via s.
Where it is used: Everywhere Vu > φVc.
Connects to: ACI 22.5.10.

Objective 03

Enforce spacing limits smax

smax = d/2 (or d/4 if Vs > 4·√f'c·bw·d) and ≤ 24 in.

Why it matters: Ensures every 45° crack is crossed by a stirrup.
Where it is used: Detailing.
Connects to: ACI 9.7.6.2.

Objective 04

Recognise Av,min

Av,min = max(0.75·√f'c·bw·s/fyt, 50·bw·s/fyt) (psi).

Why it matters: Prevents brittle shear failure at low demands.
Where it is used: Everywhere Vu ≥ 0.5·φVc.
Connects to: ACI 9.6.3.4.

Objective 05

Understand diagonal-tension failure

Shear-critical beams crack diagonally from support toward load — brittle if unreinforced.

Why it matters: Consequence of shipping stirrups.
Where it is used: Design intent.
Connects to: ACI Commentary.

Objective 06

Punching-shear awareness (two-way)

Flat slabs punch through at columns; ACI 22.6 governs.

Why it matters: Related failure mode to beam shear.
Where it is used: Slab-column joints.
Connects to: ACI 22.6.

03

Engineering motivation

What each part of a steel-frame building actually does — and why it exists

Before you design any single member, you have to see the whole system. A steel-frame building is not a collection of independent shapes bolted together — it is a deliberate load path, engineered so that every kilonewton of gravity, wind, or seismic demand has a continuous route from where it starts to the ground where the earth can resist it.

The photograph below shows a typical steel framing detail. Drag each labelled chip onto the structural element it names — the drop is only accepted when it lands inside the correct element's outlined region. A correct answer locks in with a green outline and a short explanation; a wrong answer flashes the region red, tells you what you actually hit, and returns the chip so you can try again. Press Reveal expected placements to see the reference solution (that attempt is then marked as assisted).

Structural steel framing — identify each element by dragging the labels — Fig. 1.3 · Drop a chip inside the outlined element it names. Green = correct and locked; red flash = wrong element, chip returns.

04

Failure mechanisms

Why we design the way we do — six ways steel structures have failed, and what each disaster taught the profession

Every provision in AISC 360 is a scar. Behind each equation, load factor, and detailing rule is a bridge, a walkway, or a tower whose failure cost lives and rewrote the profession. The six case studies below trace the mechanisms that motivate the code you are about to learn.

Read each one as an engineer, not a spectator: identify the load, the limit state, the missing check, and the specific clause that exists today because that check was missed. When you meet those clauses again in Chapters 5–17, they will read as answers, not rules.

Failure mechanism

Diagonal-tension shear failure

Brittle crack from support toward load — no warning if stirrups are missing.

Root cause

Vu > φVc without adequate Vs.

Lesson learned

Design and detail stirrups.

§ACI 22.5

Failure mechanism

Punching-shear failure

Truncated-cone breaks through the slab at a column.

Root cause

vu > φvc on the critical perimeter b0.

Lesson learned

Add shear caps / stud rails or drop panels.

§ACI 22.6

How failure propagates

The five-stage failure progression

1Vu computed

2Compare to φVc

3Add Vs if needed

4Check smax and Av,min

5Ductile mode locked-in

Design codes intervene at the transition from yield to instability. Everything before yield is elastic and reversible; everything after instability is a race to collapse. LRFD keeps the demand well below the first transition.

05

Lecture notes

The full textbook chapter — figures, equations, and engineering narrative

Reflection · Think before you read

Diagonal tension cracks look nothing like flexural cracks, yet stirrups are placed vertically to resist them. How do vertical stirrups actually engage a 45° crack, and why does spacing s = d/2 matter so much?

1. Why Shear Governs — Brittleness of Diagonal Tension

The code deliberately makes shear capacity larger than flexural capacity so a beam warns you by sagging, cracking, and deflecting before it ever fails. A shear (diagonal-tension) failure is sudden and brittle: little deflection, little cracking, then collapse. That is why φ_shear = 0.75 — much lower than φ_flexure = 0.90.

2. Shear Stress in Concrete Beams (Elastic Mechanics)

Concrete does not fail from pure shear — it fails from the diagonal tension that shear induces. Start from elementary beam theory:

Flexural stress

(1)

f = M \cdot c / I

Shear stress

(2)

v = V \cdot Q / (I \cdot b)

An element off the neutral axis feels both f and v. The maximum principal (tensile) stress and its inclination α to the beam axis are:

Principal tensile stress

(3)

f_{p} = f /2 + ((f /2)^{2} + v^{2})

Angle of principal stress

(4)

t an (2 α) = 2 v / f

At the neutral axis f = 0, so f_p = v and α = 45° — diagonal cracks form at 45°.

3. How Diagonal Cracks Actually Form

Two crack patterns govern:

Flexure–shear crack — the common case. A vertical flexural crack starts at the tension face, then bends toward ~45° as it climbs into the web where shear stress is largest.
Web–shear crack — thin webs (I-beams, prestressed girders) crack directly in the web where combined f + v is critical, before any flexural crack forms.

Cracks propagate at ≈ 45° once they enter the web, matching the principal-tension direction.

4. Shear Strength of Concrete Alone — V_c

Test data show that concrete without web reinforcing carries an average diagonal-tension stress ≈ 2λ√f'_c (psi). Multiplied by the effective web area b_w·d, this is the simplified ACI expression:

Simplified Vc (ACI Eq. 11-3)

(5)

V_{c} = 2 \cdot λ \cdot f_{c}^{'} \cdot b_{w} \cdot d (l b)

SI form (f'c in MPa)

(6)

V_{c} = (λ \cdot f_{c}^{'} /6) \cdot b_{w} \cdot d (N)

A more refined expression (ACI Eq. 11-5) captures the beneficial effect of longitudinal steel ρ_w and the demand ratio V_ud / M_u:

Detailed Vc (ACI Eq. 11-5)

(7)

V_{c} = (1.9 \cdot λ \cdot f_{c}^{'} + 2500 \cdot ρ_{w} \cdot V_{u} d / M_{u}) \cdot b_{w} \cdot d \leq 3.5 \cdot λ \cdot f_{c}^{'} \cdot b_{w} \cdot d

where ρ_w = A_s/(b_w·d) and V_ud/M_u is capped at 1.0. Use the simplified form for hand design and the detailed one only when you need the extra capacity (thin webs, near supports where M_u is small).

5. Design Equation — V_n, φV_n, and V_s

Nominal strength

(8)

V_{n} = V_{c} + V_{s}

Design strength

(9)

ϕ V_{n} = ϕ V_{c} + ϕ V_{s}

Safety check (φ = 0.75)

(10)

V_{u} \leq ϕ V_{n}

Required stirrup capacity

(11)

V_{s, r e q} = (V_{u} - ϕ V_{c}) / ϕ

V_n vs φV_n — do not confuse these.

V_n (nominal) = the strength the section can theoretically carry using specified material properties (f'_c, f_y). Computed from strain compatibility. It is not the design capacity.
φV_n (design) = the strength the code allows you to count on. Always < V_n. This is what you compare with the factored demand V_u.
Design check: φV_n ≥ V_u. If you write V_n ≥ V_u, you have skipped safety.

Same rule applies to shear (φV_n ≥ V_u), axial (φP_n ≥ P_u), and torsion — always insert φ before comparing.

6. Stirrup Mechanics — Where V_s Comes From

Assume the diagonal crack projects horizontally by d (a 45° crack). The number of vertical stirrups it crosses is n = d/s. Each stirrup yields at f_yt with cross-area A_v (= 2 × area of one bar for a U-stirrup — two legs cross the plane):

Stirrups crossing the crack

(12)

n = d / s

Vs from vertical stirrups (ACI Eq. 11-15)

(13)

V_{s} = A_{v} \cdot f_{y t} \cdot n = A_{v} \cdot f_{y t} \cdot d / s

Solve for spacing

(14)

s = A_{v} \cdot f_{y t} \cdot d / V_{s, r e q}

Inclined stirrups at angle α (ACI Eq. 11-16)

(15)

V_{s} = A_{v} \cdot f_{y t} \cdot (s in α + cos α) \cdot d / s

Every stirrup that crosses the crack yields → V_s = A_v·f_yt·d/s.

7. Critical Section — Where to Compute V_u

For beams supported on top with loads applied on top and no concentrated load within a distance d of the support face, ACI §9.4.22.3 lets you take V_u at a section a distance d from the face of the support, not at the support itself. Cracks in that end zone are forced to pass through the support and can't form.

8. ACI Code Requirements — Spacing Decision Tree

Zone	Condition on V_u	Requirement
1	V_u ≤ ½·φV_c	No stirrups required
2	½·φV_c < V_u ≤ φV_c	Minimum stirrups: s ≤ min[A_v·f_yt/(0.75·√f'_c·b_w), A_v·f_yt/(50·b_w)]
3	V_u > φV_c	Design stirrups: V_s,req = (V_u − φV_c)/φ, then s = A_v·f_yt·d/V_s,req
4	V_s ≤ 4·√f'_c·b_w·d	s_max = min(d/2, 24 in)
5	V_s > 4·√f'_c·b_w·d	s_max = min(d/4, 12 in) — tight-spacing zone
Cap	V_s ≤ 8·√f'_c·b_w·d	If violated: resize the section. Stirrups can't rescue an under-sized web.

📐 Cross-Section with Stirrups

A_v = 2 × (area of one leg) for a U-stirrup — two legs cross the failure plane.

9. Worked Example 23.1 — Theoretical Stirrup Spacing

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Beam: b_w = 14 in, d = 24 in, f'_c = 3000 psi, f_yt = 60,000 psi, #3 U-stirrups (A_v = 0.11 in²), normal-weight (λ = 1.0). Find theoretical s for each V_u.

Concrete capacity

(16)

ϕ V_{c} = ϕ \cdot 2 \cdot λ \cdot f_{c}^{'} \cdot b_{w} \cdot d ϕ V_{c} = 0.75 \cdot (2) (1.0) (3000) (14) (24) ϕ V_{c} = 27, 605 l b

(a) V_u = 12,000 lb: ½·φV_c = 13,803 lb > V_u ⇒ no stirrups required.

(b) V_u = 40,000 lb:

(17)

V_{s, r e q} = (V_{u} - ϕ V_{c}) / ϕ = (40, 000 - 27, 605) /0.75 = 16, 527 l b s = A_{v} \cdot f_{y t} \cdot d / V_{s, r e q} = (0.11) (60, 000) (24) /16, 527 = 19.17 in s_{ma x} = d /2 = 12 in \leftarrow g o v er n s U ses = 12.0 in .

(c) V_u = 60,000 lb:

(18)

V_{s, r e q} = (60, 000 - 27, 605) /0.75 = 43, 193 l b s = (0.11) (60, 000) (24) /43, 193 = 7.33 in C h ec k 4 f_{c}^{'} \cdot b_{w} \cdot d = 73, 614 l b > V_{s}, so s_{ma x} = d /2 = 12 in . U ses = 7.33 in .

(d) V_u = 150,000 lb:

(19)

V_{s, r e q} = (150, 000 - 27, 605) /0.75 = 163, 193 l b C a p = 8 f_{c}^{'} \cdot b_{w} \cdot d = 147, 228 l b S ec t i o n t oos ma l l — r es i z e t h e b e am or r ai se f_{c}^{'} .

10. Worked Example 23.2 — Uniform-Load Stirrup Design

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Simple beam, clear span 14 ft, w_D = 4 klf, w_L = 6 klf, b_w = 15 in, d = 22.5 in, f'_c = 4000 psi, f_yt = 60,000 psi, #3 stirrups.

Factored load & end shear

(20)

w_{u} = 1.2 (4) + 1.6 (6) = 14.4 k l f V_{u} (f a ce) = 7 f t \cdot 14.4 k l f = 100, 800 l b V_{u} (a t d) = 100, 800 \cdot (84 - 22.5) /84 = 73, 800 l b

Concrete capacity

(21)

ϕ V_{c} = 0.75 \cdot (2) (1.0) (4000) (15) (22.5) = 32, 018 l b

Vs required at critical section

(22)

V_{s} = (73, 800 - 32, 018) /0.75 = 55, 709 l b

Theoretical spacing at d

(23)

s = (0.11) (60, 000) (22.5) /55, 709 = 5.33 in

Step A — Write Vu(x) as a straight line from the support face

(24)

B ec a u se w_{u} i s u ni f or m, V_{u} d r o p s l in e a r l y f r o m t h e f a ce (x = 0) t oz er o a t mi d s p an (x = 7 f t) : V_{u} (x) = V_{u} (f a ce) - w_{u} \cdot x = 100, 800 - 14, 400 \cdot x (l b, w i t h x in f t) Q u i c k c h ec k : V_{u} (0) = 100, 800 l b ✓ V_{u} (7) = 0 ✓

Step B — Where can we relax to s = 9 in? (this gives x = 3.058 ft)

(25)

S o l v es t i r r u p e q u a t i o n f or V_{s} a t s = 9 in : V_{s} = A_{v} \cdot f_{y t} \cdot d / s = (0.11) (60, 000) (22.5) /9 = 16, 500 \cdot (9) /9 \dots l e t^{'} s j u s t co m p u t e : V_{s} = (0.11 \cdot 60, 000 \cdot 22.5) /9 = 148, 500/9 = 16, 500 l bW ai t — t ha t^{'} s t h e V_{s} t h e 9 - in s p a c in g c an s u ppl y . M a t c h t hi s t o w ha t^{'} sr e q u i r e d : V_{s, r e q} = (V_{u} - ϕ V_{c}) / ϕ . S e tt in g V_{s, r e q} = 33, 000 l b (t h e v a l u e t ha t g i v es a t h eor e t i c a l 9 - in s p a c in g, s in ces = 148, 500/33, 000 \approx 4.5 in

w e^{'} r e u s in g 33, 000 l bb ec a u se w e p ai r t h e 9 - in d ec i s i o n w i t h t h er o u n d p o in t o n t h es h e a r c u r v e — see t ab l er o w) : V_{u} = ϕ V_{c} + ϕ \cdot V_{s, r e q} = 32, 018 + 0.75 (33, 000) = 56, 768 l b N o w so l v e V_{u} (x) = 56, 768 : 100, 800 - 14, 400 \cdot x = 56, 768 \Rightarrow 14, 400 \cdot x = 44, 032 \Rightarrow x = 3.058 f t (\approx 36.7 in f r o m f a ce) . M e anin g : p a s t 3.058 f tt h e b e amn ee d s l esss h e a r s t ee l, so w ec an w i d e n t h es p a c in g t o ab o u t 9 in .

Step C — Where do stirrups stop? (this gives x = 5.89 ft)

(26)

A C I 22.5.6.5 s a y ss t i r r u p s a r e n ee d e d u n t i l V_{u} d r o p s t o ½ \cdot ϕ V_{c} : ½ \cdot ϕ V_{c} = ½ \cdot 32, 018 = 16, 009 l b S o l v e V_{u} (x) = 16, 009 : 100, 800 - 14, 400 \cdot x = 16, 009 \Rightarrow 14, 400 \cdot x = 84, 791 \Rightarrow x = 5.888 f t \approx 5.89 f t I nin c h es : 5.89 f t \cdot 12 = 70.7 in \approx 71 in f r o m f a ce . B ey o n d t hi s p o in t n os t i r r u p s a r er e q u i r e d b y co d e .

Summary of the two distances: x = 3.058 ft is where the required spacing loosens to 9 in; x = 5.89 ft ≈ 71 in is the last point that still needs a stirrup (V_u just reaches ½·φV_c). Everything past 71 in is code-exempt. Rounding to whole inches and typical spacings-of-multiples-of-3:

x from face (ft)	V_u (lb)	V_s (lb)	Theoretical s (in)
0 → d = 1.875	73,800	55,709	5.33
2	72,000	53,309	5.57
3	57,600	34,109	8.71
3.058	56,768	33,000	9
4	43,200	14,909	> d/2 = 11.25

Selected spacing (symmetric about centerline): 1 @ 2 in (2 in), then 7 @ 5 in (35 in), then 4 @ 9 in (36 in) — total 73 in from face.

Stirrups extend from the support face to where V_u drops to ½·φV_c.

🧭 Master Recipe — Shear Design of a Beam

Factor the load → w_u = 1.2·w_D + 1.6·w_L; build the V_u diagram.
Locate the critical section at distance d from the support face; read V_u(d).
Compute φV_c = φ·2·λ·√f'_c·b_w·d (φ = 0.75).
Classify the section using the decision table (no stirrups / minimum / designed).
Check the 8√f'_c·b_w·d cap. If exceeded, resize.
Solve for s = A_v·f_yt·d / V_s,req; enforce s_max (d/2 or d/4) and 12/24-in caps.
Extend stirrups to where V_u drops to ½·φV_c; round to practical multiples (2 in, 3 in, 4 in, 5 in, 9 in, 12 in).

⚠ Common Mistakes in RC Shear

Using φ = 0.90. Shear uses φ = 0.75 — diagonal-tension failure is brittle.
Using A_v = area of one bar. A U-stirrup crosses the plane twice: A_v = 2·A_bar.
Reading V_u at the support face instead of at distance d (ACI §9.4.22.3).
Forgetting s_max = d/4 (not d/2) when V_s exceeds 4√f'_c·b_w·d.
Adding more stirrups when V_s exceeds the 8√f'_c·b_w·d cap. That cap means the concrete crushes before the stirrups yield — resize the section.
Stopping stirrups at φV_c instead of ½·φV_c. Between those two, minimum stirrups are still required.

06

Professional practice, safety & ethics

RC shear practice

Professional practice

Show stirrup size, spacing zones, and hook details explicitly; spacing changes are where field errors happen.
State whether ACI 318-19 size effect (λs) and the detailed Vc equations were used.
Coordinate stirrup spacing with congested joints and openings.

Safety in design & construction

Shear failure is brittle and sudden — minimum shear reinforcement exists precisely to force ductile behavior.
The 135° seismic hook is not optional; a 90° hook opens under load.
Web openings cut for MEP after the pour can eliminate shear capacity — require EOR approval always.

Engineering ethics

Do not relax s ≤ d/2 (or d/4) limits for placement convenience.
Report unauthorized penetrations; the 2021 Surfside collapse investigation highlighted the cost of ignoring deterioration and detailing.
Never certify a beam based on stirrups you have not verified were placed.

Inspector verifying reinforcing bar size, spacing, and cover before a concrete pour — Field inspection: rebar size, spacing, and cover verified before placement.

Engineers reviewing sealed structural drawings across a conference table — Design review: documenting assumptions before the drawings are sealed.

ABET / licensure link. These points map to ABET Student Outcomes 2 and 4 — engineering design within realistic constraints, and recognition of ethical and professional responsibilities. Expect NCEES FE and PE exam questions on the NSPE Code of Ethics, OSHA construction requirements, and the engineer's standard of care.

07

Cost analysis

Stirrup cost

Approach

Stirrups are labor-intensive: each one is cut, bent, tied, and inspected. Spacing, not bar size, drives cost.
Increasing bw or d to reduce Vs demand is often cheaper than tight stirrup spacing over a long region.
Standardize spacing into 2–3 zones rather than a continuously varying schedule.

Worked cost example — s = 6″ vs s = 10″ over a 20-ft beam

Basis: #3 two-leg stirrups

Line item	Qty	Rate	Cost
Stirrups @ 6″ (41 total)	41 ea installed	$11	$451
Stirrups @ 10″ (25 total)	25 ea installed	$11	$275
Beam depth increase to allow 10″ spacing	1 ls	$180	$180
Estimated total			$906

Takeaway. Widening spacing saves ~$20 net here — the real gain is placement speed and less congestion at the supports.

Unit rates are representative US averages for teaching purposes. On a real project, price with current local rates (RSMeans, fabricator quotes, or contractor pricing) and state the estimate date.

08

Animated concepts

Key mechanics visualised — watch the strain profile, stress block, or buckled shape evolve

Truss analogy for RC shear

Diagonal concrete strut carries Vc; vertical stirrups engage as diagonal cracks form. Vn = Vc + Vs (ACI 22.5).

09

Engineering figures

Full-page reference diagrams — the visual vocabulary you will use for the rest of the course

§23.6.1

Shear cracks

Fig. 23.1Diagonal crack in RC beam
Diagonal tension = principal tension in web.

§23.6.2

Stirrup detail

Fig. 23.2Stirrups at close spacing
Every crack must be crossed by a stirrup.

§23.6.3

Brittle shear

Fig. 23.3Diagonal-tension failure
No warning without Vs.

§23.6.4

Punching shear

Fig. 23.4Two-way slab failure
Cone punches through flat slab.

10

Worked examples

Full textbook solutions — problem, theory, step-by-step, verification, interpretation

Example 23.1

Stirrup design for Vu = 40 kips

Full ACI 22.5 shear design line-by-line.

Problem statement

Beam bw = 12 in, d = 20 in, f'c = 4,000 psi (normal wt λ=1.0), fyt = 60 ksi. At d from support, Vu = 40 kips. Design stirrups.

RC beam shear crack — FIG. 23.1 — diagonal shear crack in beam.

DIMStirrups cross the 45° crack — that is Vs.

Given

bw=12, d=20 in
f'c=4,000 psi
fyt=60 ksi
Vu = 40 k

Find

Stirrup size and spacing s

Assumptions

λ = 1.0 (normal wt)
Simply supported beam

Code references

ACI 318-19 22.5.5.1
ACI 22.5.10
ACI 9.7.6.2

Theory & approach

Concrete resists 2√f'c·bw·d; stirrups make up the rest via Av·fy·d/s.

Step-by-step solution

1
Vc (concrete contribution)
FormulaACI 22.5.5.1
V_c = 2·λ·√f'_c·b_w·d (psi)
V_c = 2·1.0·√4,000·12·20 = 2·63.25·240 = 30,360 lb ≈ 30.4 k
φV_c = 0.75·30.4 = 22.8 k
2
Required Vs
Formula
V_s,req = V_u / φ − V_c
V_s,req = 40 / 0.75 − 30.4 = 53.33 − 30.4 = 22.9 k
3
Trial #3 stirrup
Formula
s = A_v·f_yt·d / V_s,req
s = 0.22·60·20 / 22.9 = 264 / 22.9 = 11.53 in
Try s = 11 in.
4
smax check
FormulaACI 9.7.6.2
s_max = d/2 if V_s ≤ 4·√f'_c·b_w·d
4·√f'_c·b_w·d = 4·63.25·240 = 60,720 lb ≈ 60.7 k
V_s = 22.9 k < 60.7 k → s_max = d/2 = 10 in
Use s = 10 in (governs).
5
Av,min
FormulaACI 9.6.3.4
A_v,min = max(0.75·√f'_c·b_w·s / f_yt, 50·b_w·s/f_yt)
= max(0.75·63.25·12·10 / 60,000, 50·12·10 / 60,000)
= max(0.095, 0.100) = 0.10 in²
Av (#3, 2 legs) = 0.22 in² > 0.10 in² ✓

Verification

All three checks (Vs ≥ Vs,req, s ≤ smax, Av ≥ Av,min) satisfied.

Final answer

Use #3 stirrups @ 10 in (smax controls).

Common mistakes

Using Av of one leg instead of two.
Forgetting the smax = d/4 branch when Vs is high.
Skipping Av,min.

References

· ACI 318-19

11

Guided practice

Compute the governing variables — hints unlock as you need them

For bw=10, d=18, f'c=4000, fyt=60, Vu=25 k, find φVc and required Vs.

Your turn

V_c =kipphi*V_c =kipV_s,req =kip

Hints

1.V_c (simplified, ACI Eq. 22.5.5.1a) = 2·λ·√f'c·b_w·d with f'c in psi → V_c = 2·1·√4000·10·18 = 22,772 lb ≈ 22.8 kip.

12

Independent practice

Solve the chapter's design task — compute each governing variable

Design task

A rectangular RC beam (bw = 12 in, d = 20 in, NWC f'c = 4000 psi, fyt = 60,000 psi) uses #3 two-legged stirrups (Av = 0.22 in²) at s = 8 in. Compute Vc, Vs, and the design shear strength φVn per ACI 318.

Given

bw = 12 in, d = 20 in
f'c = 4000 psi, λ = 1.0 (NWC)
#3 stirrups: Av = 0.22 in², s = 8 in, fyt = 60,000 psi

Approach

Simplified concrete: Vc = 2·λ·√f'c·bw·d (lb) — ACI 318 Table 22.5.5.1.
Vs = Av·fyt·d / s (kips when Av, d in in and fyt in ksi).
φVn = 0.75·(Vc + Vs).

Submit your answer

V_c =kipV_s =kipphi*V_n =kip

13

Mini design challenge

Select the option that satisfies every code and serviceability requirement in the brief

Brief

Select transverse reinforcement (#3 or #4 U-stirrups) and spacing for the beam above so that V_s ≥ V_s,req = 10.5 k, s ≤ s_max per ACI §9.7.6.2.2, and A_v ≥ A_v,min per ACI §9.6.3.4. Include cost/constructability judgment.

Requirements

V_s = A_v·f_yt·d/s ≥ V_s,req = 10.5 kip
s ≤ s_max = min(d/2, 24 in) = 9 in when V_s ≤ 4√f'c·b_w·d
A_v,min/s ≥ max[0.75√f'c·b_w/f_yt, 50·b_w/f_yt] (ACI §9.6.3.4)
Round spacing to a practical 1-inch (or 2-in) increment

Section

Wt (lb/ft)

Δ (in)

Ru/Rn

Cost

Pick

14

Chapter summary

A mind map of how every concept connects

Graded Chapter Quiz(15 FE-style questions · ACI 318-19 required)

These questions reference ACI 318-19 — sections, equations, and tables are cited explicitly. Use a calculator. Each question offers a clue you may reveal before answering. Submissions are recorded to your account once signed in.

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Include all diagrams, FBDs, shear/moment diagrams and section sketches — label them.
Number each question the same way the assignment does, and note anything you skipped.
Show units on every line and box your final answers.
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16

FE exam preparation

NCEES-style practice with timer, equation sheet, and mastery tracking

Exam mode

30:00 Calculator

Question 1 / 10

bw=12, d=20, f'c=4000. Vc is closest to:

◆◆ MediumACI 318-19 §22.5.5.1 · ASCE 7-22 §2.3.1

A. 15 k
B. 23 k
C. 30 k
D. 45 k