14.3 Shear and Diagonal Tension, Stirrup Design, and Development Length

14.3 Shear and Diagonal Tension, Stirrup Design, and Development Length

Unlike flexural failures—which are preceded by ductile warnings of extensive cracking and excessive deflection—shear failures in reinforced concrete beams are typically sudden, brittle, and catastrophic. Shear stresses interact with flexural normal stresses to produce principal tensile stresses inclined at roughly $45^\circ$ to the longitudinal beam axis. Because plain concrete has low tensile capacity, web cracks propagate diagonally along the principal compression trajectories. To protect against diagonal tension failure, structural engineers design transverse reinforcement—typically vertical closed or U-shaped stirrups—and enforce strict development length criteria to anchor longitudinal bars.

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1. Shear Behavior and Diagonal Tension Cracking

Consider an infinitesimal element in a beam web subjected to transverse shear stress $\tau$ and bending normal stress $\sigma$. From Mohr's circle of stress, the principal tensile stress $\sigma_1$ is:

$\sigma_1 = \frac{\sigma}{2} + \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau^2}$

• Near mid-span where bending moment is high and shear is low, $\sigma$ dominates, producing nearly vertical flexural cracks.
• Near supports where shear is high and bending moment is low (near the neutral axis where $\sigma \approx 0$), $\sigma_1 \approx \tau$ acts at $45^\circ$. Once $\sigma_1$ exceeds the tensile splitting strength of concrete ($f_{ct} \approx 0.33\sqrt{f'_c}\text{ to } 0.62\sqrt{f'_c}\text{ MPa}$), an inclined web-shear crack opens suddenly.
• Between supports and mid-span, existing flexural cracks bend diagonally into the web, forming flexure-shear cracks.

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2. Critical Section for Shear Design

Under NSCP 2015 Section 409.4.3.2, for beams supported by bearing reactions that introduce vertical compression into the end regions of the member (e.g., resting on top of columns or walls):

• The maximum factored shear force $V_u$ used for design is computed at a critical section located at a distance $d$ from the face of the support.
• All sections between the support face and distance $d$ are designed for this same critical shear force $V_u$.

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3. Basic USD Shear Design Equations

Under NSCP 2015 Section 422.5, nominal shear strength $V_n$ is the sum of the shear resisted by concrete ($V_c$) and the shear resisted by transverse reinforcement ($V_s$):

$\phi V_n \ge V_u \implies \phi (V_c + V_s) \ge V_u$

Where:

• $\phi = 0.75$ is the strength reduction factor for shear and torsion.

Concrete Shear Strength ($V_c$)

For members subjected to flexure and shear only (normal-weight concrete, $\lambda = 1.0$):

$V_c = 0.17 \lambda \sqrt{f'_c} b_w d \quad (\text{in N, with } f'_c \text{ in MPa and dimensions in mm})$

Where:

• $\lambda = 1.0$ for normal-weight concrete.
• $\lambda = 0.85$ for sand-lightweight concrete; $\lambda = 0.75$ for all-lightweight concrete.
• $b_w$ = web width of beam (mm).
• $d$ = effective depth to tension reinforcement (mm).

Required Stirrup Shear Strength ($V_s$)

When factored shear $V_u$ exceeds the design concrete capacity $\phi V_c$:

$V_s \ge \frac{V_u}{\phi} - V_c = \frac{V_u}{0.75} - V_c$

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4. Stirrup Design Formulation

For vertical stirrups perpendicular to the longitudinal axis of the beam, the shear force resisted across a potential $45^\circ$ crack traversing $(d / s)$ stirrups is:

$V_s = \frac{A_v f_{yt} d}{s}$

Solving for required stirrup spacing $s$:

$s_{\text{req}} = \frac{A_v f_{yt} d}{V_s}$

Where:

• $A_v$ = total cross-sectional area of all stirrup legs crossing the shear plane within spacing $s$. For a standard two-legged closed or U-stirrup, $A_v = 2 A_b = 2 \left(\frac{\pi d_v^2}{4}\right)$.
• $f_{yt}$ = specified yield strength of transverse reinforcement (limited to $f_{yt} \le 420\text{ MPa}$ for crack control per NSCP).
• $s$ = center-to-center longitudinal spacing of stirrups.

Upper Web Crushing Limit on $V_s$

To prevent catastrophic diagonal compression failure (crushing of the concrete web between diagonal tension cracks before stirrups reach yield), the code imposes a strict ceiling on stirrup resistance:

$V_s \le 0.66 \sqrt{f'_c} b_w d$

If $V_s > 0.66 \sqrt{f'_c} b_w d$, adding more stirrups is useless; the cross-section must be enlarged ($b_w$ or $d$ increased) or concrete strength $f'_c$ increased.

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5. NSCP 2015 Stirrup Spacing Zones and Rules

NSCP 2015 Section 409.7.6.2.2 establishes four operational shear regimes:

Minimum Area of Shear Reinforcement ($A_{v,\min}$)

In Zones 2 and 3, when stirrups are provided, the minimum area must satisfy:

$A_{v,\min} = 0.062 \sqrt{f'_c} \frac{b_w s}{f_{yt}} \ge \frac{0.35 b_w s}{f_{yt}}$

$\implies s_{\max} \le \frac{A_v f_{yt}}{0.35 b_w} \quad \text{and} \quad s_{\max} \le \frac{A_v f_{yt}}{0.062 \sqrt{f'_c} b_w}$

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6. Torsion Design of Beams (ACI 318-14 / NSCP 2015)

The PSAD TOS asks for design of "flexural, shear and torsion reinforcements." Torsion in beams usually comes from spandrel beams supporting slabs on one side, or from eccentric loads.

Section properties.

• $A_{cp}$ and $p_{cp}$ are the area and perimeter of the outer concrete outline.
• $A_{oh}$ and $p_h$ are the area and perimeter enclosed by the centerline of the outermost closed stirrup.
• $A_o = 0.85 A_{oh}$.

When torsion can be neglected. For non-prestressed members, the threshold torsion is:

$T_{th} = 0.083 \lambda \sqrt{f'_c} \left( \frac{A_{cp}^2}{p_{cp}} \right)$

Torsion may be neglected if $T_u < \phi T_{th}$, with $\phi = 0.75$. Above that value, closed stirrups and longitudinal bars must resist the full torsion.

Equilibrium versus compatibility torsion. If the torsion is needed for equilibrium (a canopy beam), design for the full $T_u$. If it arises only from compatibility (a spandrel beam restrained by a slab), $T_u$ may be reduced to the cracking torque, $\phi T_{cr} = \phi(0.33 \lambda \sqrt{f'_c}\, A_{cp}^2 / p_{cp})$, and the moments redistributed.

Transverse reinforcement (one leg of a closed stirrup), with $\theta = 45^\circ$ for non-prestressed members:

$\frac{A_t}{s} = \frac{T_u}{\phi\, 2 A_o f_{yt} \cot\theta}$

Longitudinal reinforcement:

$A_\ell = \frac{A_t}{s}\, p_h \left( \frac{f_{yt}}{f_y} \right) \cot^2\theta$

Detailing rules:

• Combine with shear stirrups as $(A_v + 2A_t)/s$, at least $\max(0.062\sqrt{f'_c},\; 0.35)\, b_w / f_{yt}$.
• Torsion stirrup spacing must not exceed the smaller of $p_h/8$ and $300\text{ mm}$.
• Longitudinal bars are distributed around the perimeter at no more than $300\text{ mm}$, with a bar in each corner.

Example. A $300 \times 500\text{ mm}$ beam has $f'_c = 28\text{ MPa}$: $A_{cp} = 150{,}000\text{ mm}^2$ and $p_{cp} = 1{,}600\text{ mm}$. Then $T_{th} = 0.083(5.2915)(150{,}000^2/1{,}600) = 6.18\text{ kN}\cdot\text{m}$ and $\phi T_{th} = 4.63\text{ kN}\cdot\text{m}$. A factored torsion of $10\text{ kN}\cdot\text{m}$ therefore needs torsion reinforcement.

7. Development Length ($l_d$) of Deformed Bars in Tension

For reinforcing bars to develop their design yield strength $f_y$ through flexure, they must be embedded into concrete by a sufficient length called the development length ($l_d$). Bond transfer occurs via mechanical bearing of rebar deformations (ribs) against surrounding concrete, chemical adhesion, and frictional resistance.

General Code Equation (NSCP 2015 Section 425.4.2)

$l_d = \left[\frac{f_y}{1.1 \lambda \sqrt{f'_c}} \frac{\psi_t \psi_e \psi_s}{\left(\frac{c_b + K_{tr}}{d_b}\right)}\right] d_b \ge 300\text{ mm} \quad (\text{SI units: MPa, mm})$

Where the confinement term $\frac{c_b + K_{tr}}{d_b} \le 2.5$.

Simplified Design Equations (NSCP 2015 Table 425.4.2.2)

For beams with clear spacing $\ge d_b$, clear cover $\ge d_b$, and stirrups meeting code minimums:

Modification Factors:

1. $\psi_t$ (Casting Position Factor):
   • $\psi_t = 1.3$ for top horizontal bars where more than $300\text{ mm}$ of fresh concrete is cast below the reinforcement. Bleed water and rising air bubbles settle beneath top bars, degrading bond strength.
   • $\psi_t = 1.0$ for other bars (bottom bars and vertical bars).
2. $\psi_e$ (Coating Factor):
   • $\psi_e = 1.5$ for epoxy-coated bars with cover $< 3 d_b$ or clear spacing $< 6 d_b$.
   • $\psi_e = 1.2$ for other epoxy-coated bars.
   • $\psi_e = 1.0$ for uncoated (black) bars.
   • (Note: The product $\psi_t \psi_e$ need not exceed $1.7$.)
3. $\psi_s$ (Size Factor): $\psi_s = 0.8$ for 20 mm (No. 19) and smaller bars; $\psi_s = 1.0$ for 22 mm and larger bars.
4. $\lambda$ (Lightweight Factor): $1.0$ for normal-weight; $0.75$ for lightweight.

Standard Hooks in Tension ($l_{dh}$)

When straight development length cannot be accommodated within member dimensions (e.g., at exterior beam-column joints), $90^\circ$ or $180^\circ$ standard hooks provide anchorage:

$l_{dh} = \left(\frac{0.24 f_y \psi_e}{\lambda \sqrt{f'_c}}\right) d_b \ge \max(8 d_b, \quad 150\text{ mm})$

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8. Comprehensive Worked Examples

Worked Example 1: Full Beam Stirrup Design

Problem: A simply supported rectangular beam spans $L = 6.0\text{ m}$ between column supports ($300\text{ mm} \times 300\text{ mm}$ columns). It supports a factored uniformly distributed load $w_u = 80\text{ kN/m}$ (inclusive of self-weight). Beam cross-section: $b_w = 300\text{ mm}$, effective depth $d = 500\text{ mm}$. Materials: $f'_c = 28\text{ MPa}$, $f_{yt} = 280\text{ MPa}$ (Grade 280 stirrups). Using $10\text{ mm}$ two-legged stirrups ($A_v = 157.08\text{ mm}^2$): (a) Determine the factored shear $V_u$ at the critical section. (b) Compute concrete shear capacity $\phi V_c$. (c) Calculate the required stirrup spacing at the critical section and check all code limits.

Solution:

• Step 1: Factored Shear at Critical Section: Centerline reaction: $R = \frac{w_u L}{2} = \frac{80 \times 6.0}{2} = 240.0\text{ kN}$. Distance from support centerline to face of support: $\frac{0.30}{2} = 0.15\text{ m}$. Critical section is at distance $d = 500\text{ mm} = 0.50\text{ m}$ from face of support: $x_{\text{crit}} = 0.15 + 0.50 = 0.65\text{ m}$ $V_u = R - w_u x_{\text{crit}} = 240.0 - 80(0.65) = 240.0 - 52.0 = 188.0\text{ kN}$

• Step 2: Concrete Shear Strength: $V_c = 0.17 \lambda \sqrt{f'_c} b_w d = 0.17(1.0)\sqrt{28} \times 300 \times 500 \times 10^{-3} = 134.93\text{ kN}$ $\phi V_c = 0.75 \times 134.93 = 101.20\text{ kN}$ $0.5 \phi V_c = 0.5 \times 101.20 = 50.60\text{ kN}$ Since $V_u = 188.0\text{ kN} > \phi V_c = 101.20\text{ kN}$, web shear reinforcement is required.

• Step 3: Required Stirrup Resistance ($V_s$): $V_s = \frac{V_u}{\phi} - V_c = \frac{188.0}{0.75} - 134.93 = 250.67 - 134.93 = 115.74\text{ kN}$

  Check Upper Web Crushing Limit: $V_{s,\max} = 0.66 \sqrt{f'_c} b_w d = 0.66 \sqrt{28} \times 300 \times 500 \times 10^{-3} = 523.85\text{ kN}$ Since $V_s = 115.74\text{ kN} \ll 523.85\text{ kN}$, web crushing is avoided.

• Step 4: Determine Maximum Permissible Spacing ($s_{\max}$): Check threshold: $0.33 \sqrt{f'_c} b_w d = 0.33 \sqrt{28} \times 300 \times 500 \times 10^{-3} = 261.9\text{ kN}$. Since $V_s = 115.74\text{ kN} \le 261.9\text{ kN}$, the spacing is in Zone 3: $s_{\max} = \min\left(\frac{d}{2}, \quad 600\text{ mm}\right) = \min(250\text{ mm}, \quad 600\text{ mm}) = 250\text{ mm}$

  Check Minimum Stirrup Area Spacing Limit: $s \le \frac{A_v f_{yt}}{0.35 b_w} = \frac{157.08 \times 280}{0.35 \times 300} = \frac{43,982.4}{105} = 418.9\text{ mm}$

• Step 5: Theoretical Spacing at Critical Section: $s_{\text{req}} = \frac{A_v f_{yt} d}{V_s} = \frac{157.08 \times 280 \times 500}{115,740} = \frac{21,991,200}{115,740} = 189.99\text{ mm}$ Selection: Provide $10\text{ mm}$ stirrups at $s = 180\text{ mm}$ on centers (since $180\text{ mm} \le s_{\text{req}} = 190\text{ mm}$ and $180\text{ mm} \le s_{\max} = 250\text{ mm}$).

Worked Example 2: Tension Development Length of Top and Bottom Bars

Problem: A continuous beam has $\phi 25\text{ mm}$ longitudinal bars ($d_b = 25\text{ mm}$) with $f_y = 420\text{ MPa}$ and $f'_c = 28\text{ MPa}$ in normal-weight concrete. The bars are uncoated (black) and satisfy code minimum cover and clear spacing. Calculate the required development length $l_d$ for: (a) bottom bars, and (b) top negative moment bars having $400\text{ mm}$ of fresh concrete placed beneath them.

Solution:

• For $\phi 25\text{ mm}$ bars (22 mm and larger group), the simplified code equation is: $l_d = \left(\frac{f_y \psi_t \psi_e}{1.7 \lambda \sqrt{f'_c}}\right) d_b$

• Material terms: $\lambda = 1.0$, $\psi_e = 1.0$, $\sqrt{f'_c} = \sqrt{28} = 5.2915\text{ MPa}$. $\frac{f_y}{1.7 \lambda \sqrt{f'_c}} = \frac{420}{1.7 \times 1.0 \times 5.2915} = \frac{420}{8.9956} = 46.689$

• (a) Bottom Bars: $\psi_t = 1.0$: $l_d = 46.689 \times 1.0 \times 25\text{ mm} = 1167.2\text{ mm} \approx 1170\text{ mm}$

• (b) Top Bars: Since $> 300\text{ mm}$ of fresh concrete is cast beneath the bars, $\psi_t = 1.3$: $l_d = 46.689 \times 1.3 \times 25\text{ mm} = 1517.4\text{ mm} \approx 1520\text{ mm}$ (Top bars require a 30% longer embedment length!)

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9. Licensure Exam Pitfalls & Review Notes