Let's break down the problem step by step: --- ## **Step 1: Understanding the Unit Hydrograph (UH)** - The table gives a **20-minute Unit Hydrograph (UH₁/₃)**, meaning the values are the stream flows (Q) produced by **1/3 inch of runoff** over the entire 310-acre watershed. - The hydrograph's time increment is **10 minutes**. --- ## **Step 2: Rainfall Event (Runoff Hyetograph)** **Runoff Depths:** - First 20 min: **1.5 in** - Next 20 min: **0.5 in** - **Total runoff = 2.0 in** (1.5 in + 0.5 in) But the storm is: - 1.5 in in first **20 min** - 0.5 in in next **20 min** --- ## **Step 3: Construct the Direct Runoff Hydrograph (TRH)** The direct runoff hydrograph is the **sum of scaled and lagged unit hydrographs** for each runoff pulse. ### **a. Number of Unit Hydrographs Needed** Since UH is for **1/3 inch**, we need to see how many "units" of 1/3-inch are in each period: - **First 20 min:** 1.5 in / (1/3 in) = **4.5 UHs** - **Next 20 min:** 0.5 in / (1/3 in) = **1.5 UHs** ### **b. Lag the UHs** - The first 20 min pulse (4.5 UHs) starts at time 0. - The next 20 min pulse (1.5 UHs) starts at 20 min (lagged by 20 min). ### **c. Multiply and Add the Hydrographs** Let's tabulate: | Time (min) | UH₁/₃ (cfs) | 4.5 × UH₁/₃ (Pulse 1) | 1.5 × UH₁/₃ (Pulse 2, lag 20 min) | Sum | +Base Flow (20 cfs) | **TRH (cfs)** | |:----------:|:-----------:|:---------------------:|:----------------------------------:|:---:|:------------------:|:-------------:| | 0 | 0 | 0 | 0 | 0 | 20 | 20 | | 10 | 140 | 630 | 0 |630 | 20 | 650 | | 20 | 320 | 1440 | 0 |1440 | 20 | 1460 | | 30 | 500 | 2250 | 210 |2460 | 20 | 2480 | | 40 | 420 | 1890 | 480 |2370 | 20 | 2390 | | 50 | 280 | 1260 | 750 |2010 | 20 | 2030 | | 60 | 160 | 720 | 630 |1350 | 20 | 1370 | | 70 | 60 | 270 | 420 |690 | 20 | 710 | | 80 | 0 | 0 | 240 |240 | 20 | 260 | | 90 | 0 | 0 | 60 | 60 | 20 | 80 | | 100 | 0 | 0 | 0 | 0 | 20 | 20 | **Explanation:** - For the first 20 min, only the first pulse (4.5 UHs) contributes. - Starting at time = 20 min, the second pulse (1.5 UHs) starts contributing (use UH values for 0, 10, ..., lagged by 20 min). --- ## **Step 4: Determine Runoff Volume (Acre-feet)** Total runoff = **2.0 inches** over **310 acres**. Convert inches to feet: 2.0 in × (1 ft / 12 in) = 0.1667 ft \[ \text{Runoff Volume} = \text{Area} \times \text{Depth} = 310\, \text{acres} \times 0.1667\, \text{ft} = 51.1\, \text{acre-feet} \] --- ## **Step 5: Final Answers** ### **a. Stream Flow Hydrograph (TRH):** Time (min) | Flow (cfs) -----------|----------- 0 | 20 10 | 650 20 | 1460 30 | 2480 40 | 2390 50 | 2030 60 | 1370 70 | 710 80 | 260 90 | 80 100 | 20 ### **b. Runoff Volume:** \[ \boxed{51.1\, \text{acre-feet}} \] --- **Let me know if you need the hydrograph plotted or further explanations!**

Let's carefully work through the problem step-by-step, considering all the details: --- ### **Given Data:** - **Column section:** W12 x 152 - **F_y (yield strength of steel):** 50 ksi - **Factored axial load (P_u):** 960 kips - **Steel:** A36 (assumed to be grade A36 steel) - **Concrete pedestal:** fc = 3 ksi --- ### **Step 1: Determine the required steel area (A_s)** Since the column is W12 x 152, the **nominal steel area (A_c)** is: \[ A_c = 12\, \text{in} \times 0.25\, \text{in} \text{ (approximate web thickness)} + roughly the flange area; but since the designation is W12 x 152, the **area is given directly**: \[ A_{column} = 152\, \text{in}^2 \] *(Alternatively, the "152" in W12 x 152 indicates the weight per foot in lb/ft; but for this problem, it's likely referring to the section weight, so the **area is 12 in × web thickness**; for simplicity, we take the **section area as 12 in x 12 in** - but to be precise, we need to check standard tables.)* **However,** since the problem states **supporting a factored load P_u = 960 kip**, and the question asks to determine the **plate dimensions and concrete pedestal**, **the column's cross-sectional area is less critical**; instead, the focus is on the **plate and pedestal design**. --- ### **Step 2: Calculate the required steel reinforcement (A_s)** The **design axial capacity** for the steel plate and concrete pedestal must resist P_u. --- ### **Step 3: Design of the steel plate** The plate is designed to **transfer the load to the concrete pedestal**. The **plate's area** is related to the **bearing strength** of steel and the **area of contact**. --- ### **Step 4: Design of the concrete pedestal** The **concrete pedestal** must resist the axial load transferred through the plate, considering the **concrete's bearing capacity**. --- ### **Assumptions and Approach:** - The **plate covers the entire concrete area**, so the **area of the plate (A_plate)** determines the load transfer. - The **bearing strength of concrete (f_c)** is given as 3 ksi. - The **bearing pressure (p_b)** is: \[ p_b = \frac{P_{u}}{A_{plate}} \] which must **not exceed** the concrete's **bearing capacity**: \[ p_b \leq f_c \quad \Rightarrow \quad A_{plate} \geq \frac{P_u}{f_c} \] --- ### **Step 5: Calculate the required **area of the plate** (A_plate):** \[ A_{plate} = \frac{P_u}{f_c} = \frac{960\, \text{kips}}{3\, \text{ksi}} = \frac{960,000\, \text{lb}}{3,000\, \text{lb/in}^2} = 320\, \text{in}^2 \] --- ### **Step 6: Determine the **plate dimensions** Assuming a **square plate** for simplicity: \[ \text{Side length} = \sqrt{A_{plate}} = \sqrt{320} \approx 17.89\, \text{in} \] **Round up** for practical purposes: \[ \boxed{ \text{Plate dimensions} \approx 18\, \text{in} \times 18\, \text{in} } \] --- ### **Step 7: Check the steel reinforcement (if needed)** - The **plate is A36 steel** with \(F_y = 50\, \text{ksi}\). - The **plate thickness** **t** can be estimated based on the **load transfer capacity**. Assuming the **bearing pressure** is the main concern, and the **plate thickness** is sufficient: \[ \text{Steel area (A_s)} = \frac{P_u}{F_y} \quad \text{(for tension)} \] But since the plate primarily acts as bearing surface, the **plate's cross-sectional area** is the **area of the plate**, which we've established as **about 320 in²**. --- ### **Step 8: Final answer in the correct format** **- Plate dimensions:** **Approximately 18 inches by 18 inches** **- Concrete pedestal area:** same as the plate area, **about 320 in²** **- Converting to ft²:** \[ 320\, \text{in}^2 = \frac{320}{144} \approx 2.22\, \text{ft}^2 \] --- ### **Summary:** | Item | Dimension / Area | |------------------------------|----------------------------| | **Plate dimensions** | **Approximately 18 in x 18 in** | | **Concrete pedestal area** | **Approximately 320 in² (about 2.22 ft²)** | --- ### **Final note:** In actual practice, detailed reinforcement calculations, plate thickness, and safety factors would be considered, but based on the given data and typical assumptions, these are the approximate dimensions. --- **Let me know if you'd like a more detailed reinforcement calculation or clarification!**