VIPSolutions4. Find the distance from point A on a shear force diagram of simply supported beam shown i below, after that no shear reinforcement is required as per ACI 318-19. (0.5 Marks) Data Given: @ Section of beam (b x h) = 12-inch x 24 inch [@ Bottom cover to reinforcement is 3.0 inch @ fc'=4000 psi and fy = 60000 psi. 70k . 26K 26K c A B | 46K 70k f— tft fb ef ——]
Question:
4. Find the distance from point A on a shear force diagram of simply supported beam shown i
below, after that no shear reinforcement is required as per ACI 318-19. (0.5 Marks)
Data Given:
@ Section of beam (b x h) = 12-inch x 24 inch
[@ Bottom cover to reinforcement is 3.0 inch
@ fc'=4000 psi and fy = 60000 psi.
70k
. 26K
26K
c
A B |
46K
70k
f— tft fb ef ——]
This is AI generated Response.
Asked by: Test User 100005
Created at: 2025-05-12 05:54:39
Asked by: Test User 100005
Created at: 2025-05-12 05:54:39
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Let's solve the problem step by step as per **ACI 318-19**.
### **Step 1: Calculate the Shear Strength Provided by Concrete (\( \phi V_c \))**
According to ACI 318-19, for non-prestressed members:
\[
\phi V_c = \phi \cdot 2 \sqrt{f'_c} \cdot b \cdot d
\]
Where:
- \( \phi = 0.75 \) (Strength reduction factor for shear)
- \( f'_c = 4000 \) psi
- \( b = 12 \) in
- \( d = \) effective depth (h - cover - 0.5 × bar dia, if not given use h - cover)
Given:
- \( h = 24 \) in
- Bottom cover = 3.0 in
Assuming bar diameter ≈ 1 in (typical for this section, but you can ignore or use 0.5 in if not specified):
\[
d \approx 24 - 3 = 21 \text{ in}
\]
Now, calculate \( V_c \):
\[
V_c = 2 \sqrt{4000} \cdot 12 \cdot 21
\]
\[
\sqrt{4000} \approx 63.25
\]
\[
V_c = 2 \times 63.25 \times 12 \times 21
\]
\[
V_c = 2 \times 63.25 \times 252
\]
\[
V_c = 2 \times 15,939 = 31,878 \text{ lbs}
\]
\[
V_c = 31.88 \text{ kips}
\]
Apply phi:
\[
\phi V_c = 0.75 \times 31.88 = 23.91 \text{ kips}
\]
### **Step 2: Find the Distance Where Shear Equals \( \phi V_c \)**
Shear force at A = 70 kips. Shear decreases linearly to B (46 kips at 6 ft), then to C.
Let's use the left side of the diagram (from A):
- Shear at A = 70 kips
- Shear at 6 ft (B) = 46 kips
Shear varies linearly:
\[
\text{Shear at distance } x \text{ from A:}
\]
\[
V(x) = 70 - \left(\frac{70-46}{6}\right)x
\]
\[
V(x) = 70 - \left(\frac{24}{6}\right)x = 70 - 4x
\]
Set \( V(x) = 23.91 \) kips:
\[
70 - 4x = 23.91
\]
\[
4x = 70 - 23.91 = 46.09
\]
\[
x = \frac{46.09}{4} = 11.52 \text{ feet}
\]
### **Step 3: Final Answer**
**Distance from point A after which no shear reinforcement is required:**
\[
\boxed{11.5~\text{ft}}
\]
(after rounding to 1 decimal place)
---
**Summary of Steps:**
1. Calculate effective depth, \( d \).
2. Compute \( \phi V_c \) using ACI 318-19.
3. Find where the shear force diagram drops to \( \phi V_c \).
4. That's the distance from A after which no shear reinforcement is required.
Let me know if you'd like the calculation for the right side or need the detailed steps for the rest!
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