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A simply supported beam spans 8 m. It carries the uniformly distributed load throughout of its length.

Dead Load, D= 15 kN/m (beam weight not included)
Live Load, L = 10.8 kN/m
Given:
Beam section: W460 x 97 kg/m
Area = 12, 320 mm²
Depth, d = 465 mm
Flange width, bf = 193 mm
Web thickness, tw = 11 mm
Flange thickness, t = 19 mm
Zx = 1.89 x 10⁶ mm³
Zy = 2.18 x 10⁶ mm³
E= 200,000 MPa
Fy = 345 MPa
Use LRFD design.
What is the nominal shear strength Vn (kN) at the stage of yielding factored shear stress; Vu = 0.6Fy, web shear coefficient = 1.0.____

1 Answer

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Final answer:

The nominal shear strength (V_n) is calculated using the formula V_n = 0.6 F_y A_w. Given the web thickness and depth, the area of the web (A_w) is 5115 mm² and with the yield strength (F_y) of 345 MPa, the shear strength is found to be 1059.199 kN.

Step-by-step explanation:

The student asks for the nominal shear strength of a simply supported beam when subject to a factored shear stress. The provided characteristics allow calculation of the nominal shear strength at the stage of yielding in accordance with the LRFD (Load and Resistance Factor Design) design method. This involves the application of the shear formula:


V_n = 0.6 F_y A_w

where:

  • V_n is the nominal shear strength,
  • F_y is the yield strength of the steel,
  • A_w is the area of the web.

Given that the web thickness, tw, is 11 mm and the depth, d, is 465 mm, we can calculate A_w as:

A_w = tw × d = 11 mm × 465 mm = 5115 mm²

Using the yield strength of the steel F_y = 345 MPa, we find:

V_n = 0.6 × 345 MPa × 5115 mm² = 1,059,199 N = 1059.199 kN

The nominal shear strength, V_n, is therefore 1059.199 kN when considering a web shear coefficient of 1.0.

It is important to note that for design purposes, additional factors such as safety factors and material properties would be taken into consideration to ensure the actual design strength is within acceptable limits.

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