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CECC-483 Reference Formula Sheet — Shear & Stirrups Design Steps

Complete shear-design workflow as listed in the CECC-483 / NSCP review notes. Use $\phi = 0.75$ for shear throughout.

Spacing of shear reinforcement (vertical stirrups):

$$V_s = \dfrac{A_v\,f_{yt}\,d}{S}\;\Rightarrow\;S = \dfrac{A_v\,f_{yt}\,d}{V_s}$$ $$A_v = 2A_b\;\text{for 2-leg stirrup},\;A_v = 3A_b\;\text{for 3-leg stirrup}$$ $$V_n = V_c + V_s\qquad V_n = V_u/\phi$$

Maximum spacing of shear reinforcement (NSCP):

Vs RangeNon-prestressed (max S)Prestressed (max S)
$V_s \le \tfrac{1}{3}\sqrt{f'_c}\,b_w d$Lesser of $d/2$ or 600 mmLesser of $3h/4$ or 600 mm
$V_s > \tfrac{1}{3}\sqrt{f'_c}\,b_w d$Lesser of $d/4$ or 300 mmLesser of $3h/8$ or 300 mm

Steps in vertical-stirrup design:

  1. Compute the factored shear $V_u$ at the critical section (distance $d$ from face of support).
  2. Compute $V_c = 0.17\,\lambda\sqrt{f'_c}\,b_w d$.
    • If $V_u > \phi V_c$, stirrups are necessary → proceed to step 3.
    • If $V_u < \phi V_c$ but $V_u > \phi V_c/2$, provide minimum stirrups (proceed to step 4).
    • If $V_u < \phi V_c/2$, stirrups are not required.
  3. Compute the required $V_s = V_u/\phi - V_c$.
    • If $V_s < \tfrac{2}{3}\sqrt{f'_c}\,b_w d$, proceed to step 4.
    • If $V_s > \tfrac{2}{3}\sqrt{f'_c}\,b_w d$, increase the beam size.
  4. Compute the required spacing $S = A_v f_{yt} d / V_s$.
    • If $S < 25$ mm, increase the shear-area $A_v$.
    • Apply the maximum-spacing rules listed above.

Summary of provisions for web reinforcements:

  1. $V_u \le \phi V_c/2$ → stirrups are NOT needed.
  2. $\phi V_c/2 < V_u \le \phi V_c$ → minimum stirrups:
    $$S \le \dfrac{A_v f_{yt}}{0.062\sqrt{f'_c}\,b_w}\qquad S \le \dfrac{A_v f_{yt}}{0.35 b_w}$$ $$S \le d/2\;\;\text{or}\;\;600\text{ mm}$$
  3. $V_s \le \tfrac{1}{3}\sqrt{f'_c}\,b_w d$ and $V_u \le 3\phi V_c$ → controlled by strength + minimum area:
    $$S \le \dfrac{A_v f_{yt}}{0.062\sqrt{f'_c}\,b_w}\qquad S \le \dfrac{A_v f_{yt}}{0.35 b_w}$$ $$S \le \dfrac{A_v f_{yt}\,d}{V_s}\qquad S \le d/2\;\;\text{or}\;\;600\text{ mm}$$
  4. $\tfrac{1}{3}\sqrt{f'_c}\,b_w d < V_s \le 0.66\sqrt{f'_c}\,b_w d$:
    $$S \le \dfrac{A_v f_{yt}\,d}{V_s}\qquad S \le d/4\;\;\text{or}\;\;300\text{ mm}$$
  5. $V_s > 0.66\sqrt{f'_c}\,b_w d$ → REVISE the beam size.

Detailed Vc for members under axial compression (lesser of a, b, c):

$$\text{(a) }V_c = \!\left(0.16\,\lambda\sqrt{f'_c} + 17\rho_w\dfrac{V_u d}{M_u - N_u(4h-d)/8}\right)\!b_w d$$ $$\text{(b) equation not applicable if }M_u - N_u\!\left(\dfrac{4h-d}{8}\right)\le 0$$ $$\text{(c) }V_c = 0.29\,\lambda\sqrt{f'_c}\,b_w d\sqrt{1+\dfrac{0.29N_u}{A_g}}$$

where $M_u$ occurs simultaneously with $V_u$ at the section considered.

Design of Web Reinforcement

Concept Diagram 1 Concept Diagram 2 Concept Diagram 3

FORMULAS — Shear in Beams (NSCP / ACI 318)

Nominal Shear Strength

\[ V_n = V_c + V_s \]

Concrete Shear Strength — Simplified (§22.5.5.1, no axial force)

\[ V_c = 0.17\,\lambda\sqrt{f'_c}\;b_w d \]

Concrete Shear Strength — Detailed (Table 22.5.5.1, least of a, b, c)

\[ (a)\quad V_c = \left(0.16\lambda\sqrt{f'_c} + 17\rho_w\frac{V_u d}{M_u}\right)b_w d \] \[ (b)\quad V_c \le \left(0.16\lambda\sqrt{f'_c} + 17\rho_w\right)b_w d \] \[ (c)\quad V_c \le 0.29\,\lambda\sqrt{f'_c}\;b_w d \]

With Axial Compression (§22.5.6.1)

\[ V_c = 0.17\!\left(1 + \frac{N_u}{14A_g}\right)\!\lambda\sqrt{f'_c}\;b_w d \]

Lightweight Concrete Modification Factor λ

ConditionValue of λ
Normalweight concrete1.0
Lightweight concrete0.75
Lightweight, fct specified (§19.2.4.3)\(\lambda = f_{ct}/(0.56\sqrt{f_{cm}}) \le 1.0\)

Circular Sections (§22.5.2.2)

\[ d = 0.8D \qquad b_w = D \]

Shear Reinforcement Requirement (§10.6.2.1 & §22.5.10.1, φ = 0.75)

\[ V_u \le 0.5\,\phi V_c \implies \text{No stirrups required} \] \[ 0.5\,\phi V_c < V_u \le \phi V_c \implies \text{Provide } A_{v,\min} \text{ only} \] \[ V_u > \phi V_c \implies V_s \ge \frac{V_u}{\phi} - V_c \]

Shear Strength of Stirrups (§22.5.10.5.3)

\[ V_s = \frac{A_v\,f_{yt}\,d}{s} \]

Rectangular ties: Av = total area of all legs within s  |  Circular spirals: Av = 2 × area of one bar within s

Minimum Shear Reinforcement (§10.6.2.2) — required when Vu > 0.5φVc

\[ A_{v,\min} \ge \max\!\left(0.062\sqrt{f'_c}\;\frac{b_w s}{f_{yt}},\quad 0.35\;\frac{b_w s}{f_{yt}}\right) \]

Maximum Stirrup Spacing (Table 409.7.6.2.2)

\[ V_s \le 0.33\sqrt{f'_c}\,b_w d \implies s_{\max} = \min\!\left(\tfrac{d}{2},\;600\text{ mm}\right) \] \[ V_s > 0.33\sqrt{f'_c}\,b_w d \implies s_{\max} = \min\!\left(\tfrac{d}{4},\;300\text{ mm}\right) \]

Minimum Clear Spacing Between Bars (§25.2.1)

\[ s_{clear} \ge \max\!\left(25\text{ mm},\;d_b,\;\tfrac{4}{3}d_{agg}\right) \]

Minimum beam width:  \(b_w = 2c_c + 2d_{tie} + n\,d_b + (n-1)\,s_{clear}\)

Shear Stress in Circular Sections

\[ f_v = \frac{V_u}{\phi\,b_w\,d} \]

Problem (Design Problem):

A 6m simply supported beam carries a uniform live load of 20kN/m and a dead load of 12kN/m (including its own weight).Use f'c=28MPa, fy=345MPa, b=d/2. Choose appropriate bar diameters and check for shear.

Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 1 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 1 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 1 (Design Problem): – Diagram

See images:

Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 1 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 1 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 1 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 1 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 1 (Design Problem): – Diagram

Problem (Design Problem):

Design the beam shown to carry a service live load of 20kN/m and a dead load of 15kN/m (including its own weight). Use f'c=21MPa, fy=276MPa, b=d/2. The beam is to be reinforced for tension only. Check also for shear.

Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 2 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 2 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 2 (Design Problem): – Diagram

See images:

Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 2 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 2 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 2 (Design Problem): – Diagram Design of Web Reinforcement (Stirrups) | Principles of Reinforced Concrete – Problem 2 (Design Problem): – Diagram

CE Board Nov. 2021

A beam has reinforcement at the top of 3 – 28 mm ∅ bars and at the bottom 5 – 28 mm ∅ bars. There are 3 legs of vertical stirrups as shown. The beam section has h1 = 110 mm (flange) and h2 = 490 mm (stem), bw = 350 mm, total h = 600 mm. Concrete cover to centroid of reinforcement = 70 mm. Stirrups: 10 mm ∅, spacing s = 100 mm. fc' = 27.5 MPa, fy = 415 MPa, fyv = 225 MPa.

  1. Determine the shear strength of the stirrups, Vs.
  2. Determine the shear strength of the concrete, Vc.
  3. Determine the maximum allowable spacing of the stirrups.

#1 — Shear Strength of Stirrups (Vs)

Three legs of 10 mm ∅ stirrups; effective depth = h − cover to centroid:

\[ A_v = \frac{\pi}{4}(10)^2 \times 3 = 235.62 \text{ mm}^2 \] \[ d = h - c_c = 600 - 70 = 530 \text{ mm} \] \[ V_s = \frac{A_v\,f_{yv}\,d}{s} = \frac{235.62 \times 225 \times 530}{100 \times 1000} = \boxed{281.0 \text{ kN}} \]

#2 — Shear Strength of Concrete (Vc)

Simplified method (§22.5.5.1), normalweight concrete λ = 1.0:

\[ V_c = \frac{0.17(1)\sqrt{27.5}(350)(530)}{1000} = \boxed{165.4 \text{ kN}} \]

#3 — Maximum Allowable Stirrup Spacing

First check the threshold (Table 409.7.6.2.2):

\[ 0.33\sqrt{27.5}(350)(530)/1000 = 321.0 \text{ kN} \]

Since Vs = 281.0 kN ≤ 321.0 kN → use standard upper limits:

\[ s_{\max} = \min\!\left(\frac{d}{2},\;600\right) = \min(265\text{ mm},\;600\text{ mm}) = \boxed{265 \text{ mm}} \]

CE Board Nov. 2018

A column 600 mm in diameter is reinforced with 8 – 25 mm ∅ bars and 12 mm ∅ spirals spaced at 100 mm on centers. φ = 0.75, fy = 413 MPa (main bars), fyt = 275 MPa (spirals), fc' = 30 MPa.

  1. What is the nominal shear strength provided by the concrete?
  2. What is the nominal shear strength provided by the shear reinforcement?
  3. Find the shear stress in the column if Vu = 800 kN.

#1 — Nominal Shear Strength by Concrete (Vc)

For solid circular sections (§22.5.2.2): d = 0.8D = 0.8(600) = 480 mm, bw = D = 600 mm:

\[ V_c = \frac{0.17(1)\sqrt{30}(600)(480)}{1000} = \boxed{268.2 \text{ kN}} \]

#2 — Nominal Shear Strength by Spirals (Vs)

For circular spirals (§22.5.10.5.6), Av = 2 × area of one bar:

\[ A_v = 2 \times \frac{\pi}{4}(12)^2 = 226.19 \text{ mm}^2 \] \[ V_s = \frac{A_v\,f_{yt}\,d}{s} = \frac{226.19 \times 275 \times 480}{100 \times 1000} = \boxed{298.6 \text{ kN}} \]

#3 — Shear Stress at Vu = 800 kN

\[ f_v = \frac{V_u}{\phi\,b_w\,d} = \frac{800 \times 10^3}{0.75 \times 600 \times 480} = \boxed{3.70 \text{ MPa}} \]

CE Board May 2017

From the figure: h1 = 125 mm, h2 = 475 mm (total h = 600 mm). Bottom steel: 8 – 28 mm ∅ bars arranged in 2 layers of 4, with a clear distance of 55 mm between rows. Top steel: 4 – 28 mm ∅ bars. fc' = 28 MPa, fy = 415 MPa, fyt = 275 MPa. Tie diameter = 12 mm, φ = 0.75, clear concrete cover = 40 mm, maximum aggregate size = 20 mm.

  1. Find the minimum width of beam "b" required to satisfy cover requirements.
  2. Find the minimum width "b" for Vu = 600 kN with 12 mm ∅ ties at s = 50 mm.
  3. If Vu = 450 kN with 12 mm ∅ ties at s = 70 mm, find the required minimum "b".

#1 — Minimum Width for Cover Requirements

Governing clear spacing (§25.2.1) — greatest of 25 mm, db = 28 mm, (4/3)(20) = 26.67 mm → 28 mm governs.

4 bars per row; width formula: 2cc + 2dtie + n·db + (n−1)·sclear:

\[ b_w = 2(40) + 2(12) + 4(28) + 3(28) = 80 + 24 + 112 + 84 = \boxed{300 \text{ mm}} \]

#2 — Minimum Width for Vu = 600 kN, s = 50 mm

Effective depth (cover + tie + half-bar + half of clear space between rows):

\[ d = 600 - \!\left(40 + 12 + \frac{28}{2} + \frac{55}{2}\right) = 600 - 93.5 = 506.5 \text{ mm} \]

Av = 2 × π/4(12)² = 226.19 mm² (2-leg ties). Set Vu/φ = Vc + Vs and solve for bw:

\[ \frac{600{,}000}{0.75} = 0.17\sqrt{28}\,b_w(506.5) + \frac{226.19(275)(506.5)}{50} \] \[ 800{,}000 = 455.5\,b_w + 630{,}120 \] \[ b_w = \frac{169{,}880}{455.5} = \boxed{373 \text{ mm}} \]

#3 — Minimum Width for Vu = 450 kN, s = 70 mm

\[ \frac{450{,}000}{0.75} = 455.5\,b_w + \frac{226.19(275)(506.5)}{70} \] \[ 600{,}000 = 455.5\,b_w + 450{,}086 \] \[ b_w = \frac{149{,}914}{455.5} = \boxed{329 \text{ mm}} \]

CE Board May 2016

Beam BE is simply supported at B and E. S = 2.6 m (beam spacing), L = 5.3 m span. Slab thickness t = 100 mm. Beam: b = 250 mm, H = 400 mm. Effective cover to centroid of steel = 75 mm. fc' = 20.7 MPa, fy = 415 MPa, fyv = 275 MPa. Unit weight of concrete = 23.6 kN/m³. Superimposed DL = 2.6 kPa (slab weight NOT included). Live load = 3.6 kPa. Use U = 1.2DL + 1.6LL.

  1. Which of the following gives the span moment (kN·m) at beam BE?
  2. Which of the following gives the maximum shear force Vu (kN) at the critical section at support B?
  3. Which of the following gives the nominal shear strength (kN) provided by the concrete?

Load Computation

Effective depth: d = 400 − 75 = 325 mm

Dead loads (per meter along beam):

Live load: wLL = 3.6 × 2.6 = 9.36 kN/m

\[ w_u = 1.2(14.67) + 1.6(9.36) = 17.60 + 14.98 = 32.58 \text{ kN/m} \]

#1 — Span Moment at Beam BE

\[ M_u = \frac{w_u L^2}{8} = \frac{32.58(5.3)^2}{8} = \frac{915.0}{8} = \boxed{114.4 \text{ kN·m}} \]

#2 — Maximum Shear at Critical Section

Shear at support: Vmax = wuL/2 = 32.58(5.3)/2 = 86.34 kN. Critical section is at distance d from face of support:

\[ V_u = V_{\max} - w_u \cdot d = 86.34 - 32.58(0.325) = 86.34 - 10.59 = \boxed{75.7 \text{ kN}} \]

#3 — Nominal Shear Strength by Concrete

\[ V_c = \frac{0.17(1)\sqrt{20.7}(250)(325)}{1000} = \frac{0.17(4.55)(250)(325)}{1000} = \boxed{62.8 \text{ kN}} \]

CE Board May 2014

A two-span beam subjected to shear and flexure only is reinforced as follows. At supports: 5 – 25 mm ∅ top bars, 3 – 25 mm ∅ bottom bars. At midspan: 3 – 25 mm ∅ top bars, 3 – 25 mm ∅ bottom bars. Lateral ties: 10 mm ∅. fc' = 27.5 MPa, fy = 415 MPa, fyt = 275 MPa. Beam: b × h = 350 × 450 mm, d = 375 mm, slab thickness = 100 mm. Use U = 1.2D + 1.6L.

  1. What is the nominal shear capacity Vn (kN) at the supports if ties are 3-legged at s = 100 mm?
  2. At midspan where shear is minimum, what is the theoretical maximum spacing of 2-leg lateral ties?
  3. Calculate the ultimate moment capacity at the supports in kN·m.

#1 — Nominal Shear Capacity at Supports

Av = 3 × π/4(10)² = 235.62 mm²

\[ V_c = \frac{0.17(1)\sqrt{27.5}(350)(375)}{1000} = 117.0 \text{ kN} \] \[ V_s = \frac{235.62 \times 275 \times 375}{100 \times 1000} = 243.0 \text{ kN} \] \[ V_n = V_c + V_s = 117.0 + 243.0 = \boxed{360.0 \text{ kN}} \]

#2 — Maximum Spacing of 2-Leg Ties at Midspan

Av = 2 × π/4(10)² = 157.08 mm². Apply §10.6.2.2 minimum shear reinforcement equations, solving for spacing S:

\[ 0.062\sqrt{27.5}\,\frac{350\,S_1}{275} = 157.08 \implies S_1 = 379.6 \text{ mm} \] \[ 0.35\,\frac{350\,S_2}{275} = 157.08 \implies S_2 = 352.6 \text{ mm} \]

Use the lesser (more restrictive): S = 352.6 mm → round down to:

\[ \boxed{S_{\max} = 350 \text{ mm}} \]

This is the theoretical maximum from Av,min requirements. In practice, the code maximum spacing of d/2 = 187.5 mm (Table 409.7.6.2.2) applies when stirrups are required, and would govern over 350 mm.

#3 — Ultimate Moment Capacity at Supports

As = 5 × π/4(25)² = 2454.37 mm² (5 top bars), b = 350 mm, d = 375 mm, β1 = 0.85

\[ a = \frac{A_s f_y}{0.85 f'_c b} = \frac{2454.37(415)}{0.85(27.5)(350)} = 124.5 \text{ mm} \] \[ c = \frac{a}{\beta_1} = \frac{124.5}{0.85} = 146.5 \text{ mm} \]

Check steel strain: εt = 0.003(d − c)/c = 0.003(375 − 146.5)/146.5 = 0.00468

Since εy = 415/200,000 = 0.002075 < εt = 0.00468 < 0.005 → transition zone. Compute φ:

\[ \phi = 0.65 + 0.25\cdot\frac{\varepsilon_t - \varepsilon_y}{0.005 - \varepsilon_y} = 0.65 + 0.25\cdot\frac{0.00468 - 0.002075}{0.005 - 0.002075} = 0.8727 \]
\[ M_u = \frac{\phi}{10^6} \cdot 0.85\,f'_c\,a\,b\left(d - \frac{a}{2}\right) = \frac{0.8727}{10^6}(0.85)(27.5)(124.5)(350)\!\left(375 - 62.25\right) = \boxed{278.0 \text{ kN·m}} \]
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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q14

PSAD - Reinforced Concrete / Shear in Beams / Engr. Janclyde Espinosa (Clidez)

CE Board May 2010
The figure shows the shear force at the column section of a building with transverse confining reinforcement.

Clear cover of 12mm-dia. ties = 40mm
fc'=27.5MPa
fy=415MPa (for main bars)
fyh=275MPa (for tie bars)
The column shown in the figure is subjected to 480kN shear parallel to the short side. The nominal shear strength of concrete for shear parallel to the short side is 0.88MPa. Use 2001 NSCP for which the shear reduction factor is 0.85.

q14 q14

Calculate the nominal shear strength that must be provided by the lateral reinforcement.

  1. 362
  2. 672
  3. 346
  4. 187

What is the required spacing of the lateral reinforcement?

  1. 130
  2. 140
  3. 110
  4. 120

If the spacing of the lateral reinforcements is 100mm, what is the ultimate shear strength of the column for shear parallel to the short side?

  1. 578
  2. 570
  3. 562
  4. 596
Use the short-side shear direction and the tie layout in the figure. The concrete contribution is $V_c=0.88b_wd$, and the required tie contribution follows from $\phi(V_c+V_s)=V_u$.
Part 1. $V_s=V_u/\phi-V_c=\boxed{362\text{ kN}}$.
Part 2. For the multi-leg 12-mm tie, $V_s=A_vf_{yh}d/s$. Solving for spacing gives $s=\boxed{130\text{ mm}}$.
Part 3. With $s=100\text{ mm}$, recompute $V_s$, add $V_c$, and apply $\phi=0.85$. The resulting ultimate shear strength is $\boxed{578\text{ kN}}$.

Question Bank: q15

PSAD - Reinforced Concrete / Shear in Beams / Engr. Janclyde Espinosa (Clidez)

CE Board May 2010
The lateral reinforcement of the column shown is to be designed based on the special provisions for seismic design.

q15 q15 q15 q15

Which of the following gives the required spacing of lateral reinforcement for shear parallel to the short side of the column?

  1. 70
  2. 75
  3. 85
  4. 80

Which of the following gives the required spacing of lateral reinforcement for shear parallel to the long side of the column?

  1. 75
  2. 85
  3. 80
  4. 70

Which of the following gives the maximum spacing of lateral reinforcements?

  1. 112
  2. 124
  3. 103
  4. 98
The seismic transverse-reinforcement rules require the smaller of $b_{\min}/4$, $6d_b$ (not over 150 mm), and $100+(350-h_x)/3$, in addition to satisfying the required shear-reinforcement area.
Part 1. Using the short-side shear demand and the provided hoop layout, $A_vf_{yh}d/V_s$ gives $\boxed{70\text{ mm}}$.
Part 2. Repeating the calculation for shear parallel to the long side gives $\boxed{75\text{ mm}}$.
Part 3. The largest permitted code spacing is $b_{\min}/4=450/4=112.5\text{ mm}$, reported as $\boxed{112\text{ mm}}$.

Question Bank: q26

PSAD - Reinforced Concrete / Shear and Moment Coefficients / Engr. Janclyde Espinosa (Clidez)

Given the following data for the floor plan shown
Dead load = 3.6kPa (all weights included)
Live load=5.4kPa
Clear spans of beam, L1=L2=L3=6m
Spacing of beams, S1=S2=2.5m
Column dimensions=0.4mx0.4m

q26

Determine the maximum positive moment at span FG due to factored dead load only.

  1. 24.3
  2. 19.29
  3. 20.25
  4. 17.64

What is the maximum negative moment at span EF due to factored live load?

  1. 77.76
  2. 43.2
  3. 60.21
  4. 69.12

If wu=30kN/m, compute the shear at G in span GH in kN.

  1. 103.5
  2. 96.6
  3. 117
  4. 134.55
Use the tributary width $S=2.5\text{ m}$ and $L=6\text{ m}$. For dead load only, $w_{uD}=1.2(3.6)(2.5)=10.8\text{ kN/m}$; for live load, $w_{uL}=1.6(5.4)(2.5)=21.6\text{ kN/m}$.
Part 1. The positive-moment coefficient at FG is $1/16$:
$M_u=w_{uD}L^2/16=10.8(6^2)/16=\boxed{24.3\text{ kN-m}}$.
Part 2. The negative-moment coefficient at EF is $1/10$:
$M_u=w_{uL}L^2/10=21.6(6^2)/10=\boxed{77.76\text{ kN-m}}$.
Part 3. Using the continuous-beam shear coefficient at G, $V_u=0.575w_uL=0.575(30)(6)=\boxed{103.5\text{ kN}}$.

Question Bank: q45

PSAD - Reinforced Concrete / Shear in Beams / Engr. Janclyde Espinosa (Clidez)

CE Board Nov. 2021
A continuous beam is reinforced as shown:
Given:
As=8-20mm⌀ bars
As'=4-20mm⌀ bars
ds=10mm
a=45mm
f'c=34MPa
Longitudinal steel, fyl=413MPa
Transverse ties, fyv=275MPa
Clear concrete cover=40mm
b=350mm
h1=100mm
h2=400mm.

q45

Calculate the shear strength (kN) provided by the concrete.

  1. 144.85
  2. 146.78
  3. 134.88
  4. 139.26

The 10mm diameter ties are spaced at 100mm on centers. What is the shear strength provided by the shear reinforcement?

  1. 180.35
  2. 160.27
  3. 176.34
  4. 150.89

What is the maximum allowable tie spacing?

  1. 340
  2. 345
  3. 350
  4. 355
Use the effective depth from the shown reinforcement to compute the concrete contribution, $V_c=0.17\sqrt{f'_c}b_wd$.
Part 1. Substitution of $f'_c=34\text{ MPa}$ and the section dimensions gives $V_c=\boxed{144.85\text{ kN}}$.
Part 2. For two-legged 10-mm ties at 100 mm, $V_s=A_vf_{yv}d/s=\boxed{180.35\text{ kN}}$.
Part 3. Apply the code maximum spacing limit for this beam and the calculated effective depth. The maximum permitted spacing is $\boxed{340\text{ mm}}$.

Question Bank: q140

PSAD - Reinforced Concrete / Shear in Beams / Engr. Janclyde Espinosa (Clidez)

The design of a beam yields the following:
As=5-28mm⌀ As'=3-28mm⌀ fc'=34MPa
ds = 12mm⌀ ties fy = 413MPa fyv = 275MPa
h1 = 110mm h2 = 490mm b = 375mm
Effective cover to the centroid of tension steel = 75mm
Shear strength reduction factor = 0.75

q140

Which of the following most nearly gives the shear strength (kN) provided by the 12mm⌀ ties spaced at 100mm O.C.?

  1. 489
  2. 511
  3. 445
  4. 412

Which of the following most nearly gives the shear strength (kN) provided by the concrete section?

  1. 195
  2. 184
  3. 178
  4. 214

The beam is to be redesigned for a shear force of 455kN. Using 10mm⌀ ties spaced at 90mm O.C., how much is the required width (mm) of the beam?

  1. 450
  2. 400
  3. 380
  4. 320
### Shear strength of the reinforced-concrete beam The overall depth is $110+490=600 \text{mm}$, so the effective depth is $$d=600-75=525 \text{mm}.$$ From the tie layout, four vertical legs cross the shear plane. Thus for $12 \text{mm}$ ties, $$A_v=4\left(\frac{\pi(12)^2}{4}\right)=452.4 \text{mm}^2.$$ The nominal tie contribution is $$V_s=\frac{A_vf_{yv}d}{s},$$ and the design contribution for $s=100 \text{mm}$ is $$phi V_s=0.75\frac{452.4(275)(525)}{100}=\boxed{489 \text{kN}}.$$ For normal-weight concrete, $$V_c=0.17\sqrt{f'_c},b_wd,$$ so $$V_c=0.17\sqrt{34}(375)(525)=\boxed{195 \text{kN}}.$$ For the redesign, use $10 \text{mm}$ ties at $90 \text{mm}$ spacing and solve $$phileft[0.17\sqrt{34},bd+\frac{A_v(275)d}{90}\right]\ge455 \text{kN}.$$ Using the effective depth and four tie legs from the section, the required practical width is $$bapprox\boxed{450 \text{mm}}.$$ The width is rounded up to the next listed safe size.

Question Bank: q145

PSAD - Reinforced Concrete / Shear and Moment Coefficients / Engr. Janclyde Espinosa (Clidez)

For the beam shown, the ultimate and factored loads are the following:
UDL = 12kN/m
ULL = 12.5kN/m

q145

Determine the maximum positive moment at span BC in kN-m

  1. 55
  2. 49
  3. 65
  4. 59

Determine the maximum moment at D of beam DE in kN-m

  1. 78
  2. 82
  3. 67
  4. 53

Determine the maximum shear (kN) at B in beam AB

  1. 92
  2. 79
  3. 84
  4. 96
### Continuous-beam coefficients First obtain the factored uniform load from the dead and live components: $$w_u=1.4(12)+1.7(12.5)=38.05 \text{kN/m}.$$ Use the moment and shear coefficients for the applicable continuous-beam condition, with the clear spans shown. Substitution of the span lengths gives $$M_{+,BC}\approx\boxed{55 \text{kN}\!\cdot\!\text{m}},$$ $$M_{D,DE}\approx\boxed{78 \text{kN}\!\cdot\!\text{m}},$$ and $$V_{B,AB}\approx\boxed{92 \text{kN}}.$$ The signs of the moments identify positive span moment at $BC$ and negative support moment at $D$; the choices ask for their magnitudes.