## Step-by-Step Solution: Shear Design for Beam StirrupsLet's break down the solution to the shear design problem for the given beam. --- ### **1. Given Data** - **Concrete strength \( f'_c \):** 400 psi- **Yield of steel \( f \):** 40,000 psi - **Beam length \( L \):** 12 ft - **Width \( b \):** 16 in - **Effective depth \( d \):** 23.5 in - **Total height \( h \):** 26 in - **Concrete cover:** 2.5 in - **Uniform distributed load \( w \):** 11.6 k/ft - **Central point load \( P \):** 23.2 kips - **Design stirrups for half span (first 6 ft)** --- ### **2. Calculate Maximum Shear Force (\( V_u \))** For a simply supported beam: - **Shear due to UDL at support:** \( V_{u, w} = \frac{wL}{2} = \frac{11.6 \times 12}{2} = 69.6\, \text{kips} \) - **Shear due to point load at support:** Point load at center, reaction at support = \( \frac{P}{2} = \frac{23.2}{2} = 11.6\, \text{kips} \) - **Total shear at support:** \( V_{u, \text{max}} = V_{u, w} + V_{u, P} = 69.6 + 11.6 = 81.2\, \text{kips} \) --- ### **3. Calculate Shear Strength of Concrete (\( V_c \))** Using ACI 318 for non-prestressed beams: \[ V_c = 2 \sqrt{f'_c} b d \] - \( f'_c = 400 \) psi, \( b = 16 \) in, \( d = 23.5 \) in - \( \sqrt{400} = 63.25 \) - \( V_c = 2 \times 63.25 \times 16 \times 23.5 = 47,632 \) lbs \( = 47.6 \) kips --- ### **4. Shear Carried by Stirrups (\( V_s \))** \[ V_s = V_u - V_c = 81.2 - 47.6 = 33.6\, \text{kips} \] --- ### **5. Design Stirrups (Spacing Calculation)** - **Use #3 stirrups (area \( A_v \) for 2 legs):** \( A_v = 2 \times A_{bar} \) From table: \( A_{bar} \) for #3 = .11 in\(^2\) \( A_v = 2 \times .11 = .22\, \text{in}^2 \) - **Stirrup spacing \( s \):** \[ V_s = \frac{A_v f_y d}{s} \] \[ s = \frac{A_v f_y d}{V_s} \] Plug in numbers (convert kips to lbs): - \( A_v = .22\, \text{in}^2 \) - \( f_y = 40,000\, \text{psi} \) - \( d = 23.5\, \text{in} \) - \( V_s = 33,600\, \text{lbs} \) \[ s = \frac{.22 \times 40,000 \times 23.5}{33,600} = \frac{206,800}{33,600} \approx 6.16\, \text{in} \] --- ### **6. Maximum Allowable Spacing per ACI** - **Maximum spacing (\( s_{max} \)):** - Not more than \( d/2 = 23.5/2 = 11.75\, \text{in} \) - Not more than 24 in **Use the smaller of calculated and maximum:** So, **use \( s = 6\, \text{in} \) (rounded down)** --- ### **7. Specify Location on Shear Force Diagram** - **Place #3 stirrups at 6 in. spacing from the support up to midspan (6 ft).** - **Symmetry:** Other half of beam will have same stirrups. --- ### **8. Bend Diameter and Length (per ACI Table)** - For #3 bar, 135° hook: - **Minimum inside diameter = 6 × bar diameter** - #3 bar diameter = 3/8 in - **Inside diameter = 6 × 3/8 = 2.25 in** --- ## **Final Answers** ### **Stirrup Design Summary** - **Stirrup size:** #3 (two-legged) - **Spacing:** 6 in (from support to midspan, 6 ft) - **Bend:** 135° hook, minimum inside diameter = 2.25 in (per ACI) - **Location:** First 6 ft from each end (symmetrical) --- ### **Shear Force Diagram Placement** - Mark stirrups at 6 in intervals from support to midspan ( to 6 ft), both sides. --- **If you need a hand sketch of the shear force diagram with stirrup locations, let me know!**

## Hand Sketch of Shear Force Diagram with Stirrup Locations ### **Shear Force Diagram Overview** The shear force diagram illustrates the variation of shear force along the length of the beam. For this beam, we analyze the shear forces due to a uniform distributed load (UDL) and a point load at the center. ### **Shear Force Diagram** ```plaintext ↑ | | | | | | 81.2 |-------| | | | | | | | | | | | | | | | | | | | | | | | | | |-------|----------------|-----------→ 0 6 ft 12 ft ``` ### **Stirrup Locations** - **Stirrup spacing:** 6 in - **Locations:** - From the left support (0 ft) to midspan (6 ft) - Place stirrups at intervals of 6 inches ### **Stirrup Placement Markings** ```plaintext |---6 in---|---6 in---|---6 in---|---6 in---|---6 in---|---6 in---| 0 ft 1 ft 2 ft 3 ft 4 ft 5 ft 6 ft ``` ### **Final Notes** - **Stirrups are placed up to midspan (6 ft).** - **Spacing between stirrups is maintained at 6 in.** - **The same design applies symmetrically to the other half of the beam.** ### **Image Alt Text** *Shear force diagram with markings showing stirrup placements at 6 in intervals from the left support to midspan.* If you have any further requests or need additional details, feel free to ask!

## Hand Sketch of Shear Force Diagram with Stirrup Locations ### **Shear Force Diagram Overview** The shear force diagram illustrates the variation of shear force along the length of the beam, accounting for both a uniform distributed load (UDL) and a point load at the center. ### **Shear Force Diagram** ```plaintext Shear Force (kips) ↑ | 81.2 |-------------------- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |--------------------|-------------------→ Length of Beam (ft) 0 ft 12 ft ``` ### **Stirrup Locations** - **Stirrup spacing:** 6 in - **Locations:** - From the left support (0 ft) to midspan (6 ft) - Place stirrups at intervals of 6 inches ### **Stirrup Placement Markings** ```plaintext |---6 in---|---6 in---|---6 in---|---6 in---|---6 in---|---6 in---| 0 ft 1 ft 2 ft 3 ft 4 ft 5 ft 6 ft ``` ### **Final Notes** - **Stirrups are placed up to midspan (6 ft).** - **Spacing between stirrups is maintained at 6 in.** - **The same design applies symmetrically to the other half of the beam.** ### **Image Alt Text** *Neatly drawn shear force diagram with markings showing stirrup placements at 6 in intervals from the left support to midspan.* If you have any further requests or need additional details, feel free to ask!