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(a) A rolled steel joist RSJ of I section has top and bottom flanges 150mm * 25mm and web of the size 300mm * 12mm It is used as a simply supported beam over a span of 4 m to carry an uniformly distributed load of 80kN / m over its entre span. Draw bending and shearing stresses across the section at 1 / (4 deg) span.?
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(a) A rolled steel joist RSJ of I section has top and bottom flanges 1...
**Bending Stress**

To determine the bending stress in the rolled steel joist (RSJ), we need to calculate the moment of inertia (I) of the section and the bending moment (M) at the given point.

The moment of inertia (I) can be calculated using the formula:

I = (b * h^3) / 12

Where:
- b is the width of the section (flange width)
- h is the height of the section (web height)

In this case, the width of the section (b) is the sum of the top and bottom flange widths, so b = 2 * 150 mm = 300 mm.
The height of the section (h) is the sum of the web height and the flange thickness, so h = 300 mm + 2 * 25 mm = 350 mm.

Plugging these values into the formula, we get:

I = (300 mm * 350 mm^3) / 12 = 3,937,500 mm^4

The bending moment (M) at the given point can be calculated using the formula:

M = (w * L^2) / 8

Where:
- w is the uniformly distributed load
- L is the span length

In this case, the uniformly distributed load (w) is 80 kN/m, and the span length (L) is 4 m.
Plugging these values into the formula, we get:

M = (80 kN/m * 4 m^2) / 8 = 160 kNm

Now, we can calculate the bending stress (σ) using the formula:

σ = (M * y) / I

Where:
- σ is the bending stress
- y is the distance from the neutral axis to the point of interest

For the given point at 1/4 span, the distance from the neutral axis to the point of interest can be determined by:

y = h / 2

Plugging the values into the formula, we get:

y = 350 mm / 2 = 175 mm

Now we can calculate the bending stress at 1/4 span:

σ = (160 kNm * 175 mm) / 3,937,500 mm^4 = 7.12 MPa

**Shearing Stress**

To determine the shearing stress in the RSJ, we need to calculate the shear force (V) at the given point.

The shear force (V) can be calculated using the formula:

V = (w * L) / 2

Where:
- w is the uniformly distributed load
- L is the span length

In this case, the uniformly distributed load (w) is 80 kN/m, and the span length (L) is 4 m.
Plugging these values into the formula, we get:

V = (80 kN/m * 4 m) / 2 = 160 kN

Now we can calculate the shearing stress (τ) using the formula:

τ = (V * A) / (I * h)

Where:
- τ is the shearing stress
- A is the area of the section (flange area + web area)

The area of the section can be calculated as:

A = (2 * b * t) + (h * t)

Where:
- t is the thickness of the section (fl
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(a) A rolled steel joist RSJ of I section has top and bottom flanges 150mm * 25mm and web of the size 300mm * 12mm It is used as a simply supported beam over a span of 4 m to carry an uniformly distributed load of 80kN / m over its entre span. Draw bending and shearing stresses across the section at 1 / (4 deg) span.?
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(a) A rolled steel joist RSJ of I section has top and bottom flanges 150mm * 25mm and web of the size 300mm * 12mm It is used as a simply supported beam over a span of 4 m to carry an uniformly distributed load of 80kN / m over its entre span. Draw bending and shearing stresses across the section at 1 / (4 deg) span.? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about (a) A rolled steel joist RSJ of I section has top and bottom flanges 150mm * 25mm and web of the size 300mm * 12mm It is used as a simply supported beam over a span of 4 m to carry an uniformly distributed load of 80kN / m over its entre span. Draw bending and shearing stresses across the section at 1 / (4 deg) span.? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (a) A rolled steel joist RSJ of I section has top and bottom flanges 150mm * 25mm and web of the size 300mm * 12mm It is used as a simply supported beam over a span of 4 m to carry an uniformly distributed load of 80kN / m over its entre span. Draw bending and shearing stresses across the section at 1 / (4 deg) span.?.
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