In Steel Design please answer completely with step by step solutions. Problem No. 2A simply supported beam having a simple span o 8 m must support a uniformly distributed service load of 10kN/m including its own weight. A36 steel is used.Properties of the steel beam:2A = 10452 mmd = 598.68 mmb₁ = 177.93 mmt= 12.33 mmtw = 10.03 mm.F, = 248 MPa1₂ = 562 x 106 mm4ly = 12 x 10 mmZ, 2196 x 10³ mm³r = 231.39 mma.Determine the allowable bending stress using elastic design (ASD).C.b. Determine the safe uniform live load it could support using ASD method due to its flexural strength.Determine the safe uniform live load it could support using plastic design method due to its flexuralstrength. U = 1.2 DL + 1.6 LL

Span(L)= 8 mdead load (uniformaly distributed load) = 10 KN/mProperties of steel beam Need to finda.…

Answered: a. Determine the allowable bending…

a. Determine the allowable bending stress using elastic design (ASD). 09 3.98333TRE b. Determine the safe uniform live load it could support using ASD method due to its flexural strength.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Composite Construction

Composite construction is the construction of structures in which two or more materials are used. Regarding structural engineering, composite construction is a construction where two materials are bound together in a technical manner such that both the materials act as a single unit to carry and transfer the load. The motives of composite construction are to increase the structural strength, make structures more aesthetic, and provide a sustainable environment.

Question

In Steel Design please answer completely with step by step solutions.

Problem No. 2
A simply supported beam having a simple span o 8 m must support a uniformly distributed service load of 10
kN/m including its own weight. A36 steel is used.
Properties of the steel beam:
2
A = 10452 mm
d = 598.68 mm
b₁ = 177.93 mm
t= 12.33 mm
tw = 10.03 mm.
F, = 248 MPa
1₂ = 562 x 106 mm4
ly = 12 x 10 mm
Z, 2196 x 10³ mm³
r = 231.39 mm
a.
Determine the allowable bending stress using elastic design (ASD).
C.
b. Determine the safe uniform live load it could support using ASD method due to its flexural strength.
Determine the safe uniform live load it could support using plastic design method due to its flexural
strength. U = 1.2 DL + 1.6 LL

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