c
)
A
3
.
5
m high square storage platform is subject to a
4
kPa live load and a
1
.
5
kPa super imposed dead load which is distributed to the
2
supporting frames. The frames are made fr

Question

c
)
 ﻿A 
3
.
5
 ﻿m high square storage platform is subject to a 
4
 ﻿kPa live load and a 
1
.
5
 ﻿kPa super imposed dead load which is distributed to the 
2
 ﻿supporting frames. The frames are made from 
2
5
0
UB
2
5
.
7
 ﻿sections spaced at 
6
 ﻿m 
.
 ﻿The column supports are fixed, and the beams are pinned. Analysis yielded that ULS earthquake loading was most critical. Under this load case it was found that the structure is subject to a lateral force equivalent to 
2
0
%
 ﻿of the floor loading. Analyse a single frame to check if the column and beam section are adequate to sustain the load. Should the sections not satisfy ULS requirements, suggest methods and section size to satisfy these requirements. 
(
1
0
 ﻿marks
)

VIPSolutions · Accepted Answer

Let's break down the problem and solve step by step:

---

### **1. Given Data**

- **Platform height $ h $:** 3.5 m  
- **Live load ($ w_{live} $):** 4 kPa  
- **Superimposed Dead load ($ w_{dead} $):** 1.5 kPa  
- **Number of supporting frames:** 2  
- **Section:** 250UB25.7  
- **Frame spacing:** 6 m  
- **Frame width ($ L $):** (not specified, but let's denote as $ L $, will use in calculation or assume a reasonable value if needed)  
- **Column supports:** Fixed  
- **Beams:** Pinned  
- **Critical load case:** ULS earthquake loading  
- **Lateral force:** 20% of floor loading

---

### **2. Actions on a Single Frame**

#### **a. Floor Loading per Frame**

Total floor load (live + superimposed dead):

$$
w_{floor} = w_{live} + w_{dead} = 4 + 1.5 = 5.5 \,\text{kPa}
$$

Each frame supports half the floor (since 2 frames):

$$
\text{Load per frame} = w_{floor} \times \text{tributary width}
$$

Assuming each frame supports a tributary width = 6 m (the spacing):

$$
\text{Uniform load per frame} = 5.5\,\text{kPa} \times 6\,\text{m} = 33 \,\text{kN/m (along the beam)}
$$

#### **b. Lateral Force per Frame (Earthquake ULS)**

$$
\text{Lateral force per frame} = 0.2 \times 33 = 6.6\,\text{kN/m (horizontal)}
$$

Assuming this acts at the platform level (3.5 m high) and is distributed at the top of columns.

---

### **3. Section Properties for 250UB25.7**

From standard steel tables:

- **Depth ($ d $) =** 250 mm  
- **Weight =** 25.7 kg/m  
- **Section modulus ($ Z $) ≈** 194 cm³ = $ 1.94 \times 10^5 $ mm³  
- **Moment of inertia ($ I $) ≈** 2,420 cm⁴ = $ 2.42 \times 10^8 $ mm⁴  
- **Area ($ A $) =** 32.8 cm² = 3280 mm²

---

### **4. Beam Check (Pinned at ends)**

Assume the beam spans between columns, with a span = frame width (assume $ L = 6 $ m if not given).

#### **a. Moment under Uniform Load (ULS)**

$$
M_{max} = \frac{w L^2}{8}
$$

$$
w = 33\,\text{kN/m},\ L = 6\,\text{m}
$$
$$
M_{max} = \frac{33 \times 6^2}{8} = \frac{33 \times 36}{8} = \frac{1188}{8} = 148.5\,\text{kNm}
$$

#### **b. ULS Capacity Check**

Assume $ f_y = 300\,\text{MPa} $ (verify for actual steel grade).

$$
\phi M_{b,Rd} = 0.9 \times Z \times f_y
$$
$$
Z = 1.94 \times 10^5 \,\text{mm}^3,\ f_y = 300\,\text{MPa}
$$
$$
\phi M_{b,Rd} = 0.9 \times 1.94 \times 10^5 \times 300 = 0.9 \times 58,200,000 = 52,380,000\,\text{Nmm} = 52.38\,\text{kNm}
$$

**This value does not make sense** (should be much higher). Let's double-check:

$$
M_{b,Rd} = Z \times f_y = 1.94 \times 10^5 \times 300 = 58,200,000\,\text{Nmm} = 58.2\,\text{kNm}
$$

This is **much less** than the required 148.5 kNm.  
(Either the Z value is off, or perhaps a larger section modulus is needed.)

**But per OneSteel tables, 250UB25.7 has:**
- Zx (major axis) = 194 × 10³ mm³
- So, $ M_{b,Rd} = 194 \times 10^3 \times 300 = 58,200,000 $ Nmm = 58.2 kNm.

**Therefore, the beam section is NOT adequate**.

---

### **5. Column Check (Fixed at base, free at top)**

Columns carry:
- **Axial load** from floor
- **Moment** from lateral force (earthquake)

#### **a. Axial Load**

Total vertical load on each column:

$$
\text{Total load per frame} = w_{floor} \times \text{frame width} = 33\,\text{kN/m} \times 6\,\text{m} = 198\,\text{kN}
$$
Each column supports half: $ 99\,\text{kN} $.

#### **b. Moment from Lateral Loading**

Lateral force per frame at top = 6.6 kN/m × 6 m = 39.6 kN

Assuming **single sway frame** (two columns), moment at base of each column:

$$
M = \frac{F \times h}{2} = \frac{39.6\,\text{kN} \times 3.5\,\text{m}}{2} = \frac{138.6}{2} = 69.3\,\text{kNm}
$$

#### **c. Interaction Check (Axial + Moment)**

Check using interaction formula:

$$
\frac{N^*}{\phi N_c} + \frac{M^*}{\phi M_{c}} \leq 1.0
$$

- $ N^* = 99\,\text{kN} $
- $ M^* = 69.3\,\text{kNm} $
- $ \phi N_c = 0.9 \times f_y \times A = 0.9 \times 300 \times 3280 = 885,600\,\text{N} = 885.6\,\text{kN} $
- $ \phi M_{c} = 0.9 \times 194,000 \times 300 = 52,380,000\,\text{Nmm} = 52.38\,\text{kNm} $

$$
\frac{99}{885.6} + \frac{69.3}{52.38} = 0.112 + 1.323 = 1.435 > 1.0
$$

**Column section is NOT adequate**.

---

### **6. Recommendations**

#### **a. Use Larger Section**

Let's try a **310UB40.4** section:

- Z ≈ 292,000 mm³
- Area = 5160 mm²

- $ \phi N_c = 0.9 \times 300 \times 5160 = 1,393,200 $ N = 1,393 kN
- $ \phi M_{c} = 0.9 \times 292,000 \times 300 = 78,840,000\,\text{Nmm} = 78.8\,\text{kNm} $

Interaction check:

$$
\frac{99}{1,393} + \frac{69.3}{78.8} = 0.071 + 0.88 = 0.951 < 1.0
$$

**This section is adequate for the column.**

#### **b. For the Beam**

Required $ M_{b,Rd} > 148.5\,\text{kNm} $

Try **410UB53.7** (Z ≈ 402,000 mm³):

$$
\phi M_{b,Rd} = 0.9 \times 402,000 \times 300 = 108,540,000\,\text{Nmm} = 108.5\,\text{kNm}
$$

Still not enough. Try **460UB67.1** (Z ≈ 516,000 mm³):

$$
\phi M_{b,Rd} = 0.9 \times 516,000 \times 300 = 139,320,000\,\text{Nmm} = 139.3\,\text{kNm}
$$

Still less than 148.5 kNm.

Try **530UB82.0** (Z ≈ 645,000 mm³):

\[
\phi M_{b,R

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