SITUATION 2: Refer to FigureS1GHGiven:S1 = S2 = S3 = 2.8 mL1 = L2 = L3 = 6.0 mb = 300 mmh = 400 mmSzKPL1L2L3t = 100 mmColumn = 0.3 m x 0.3 mTotal Ultimate load = 12 kN/mPLAN2. What is the moment for the interior span bounded by F-G-K-J?hTYP. SECTION

L = 6m S = 2.8m W = 12KN/m h = 400mm = 0.4m b = 300mm = 0.3m

SITUATION 2: Refer to Figure G H Given: Sz S1 = S2 = S3 = 2.8 m L1 = L2 = L3 = 6.0 m b = 300 mm h = 400 mm K L P L1 L2 L3 t= 100 mm PLAN Column = 0.3 m x 0.3 m Total Ultimate load = 12 kN/m 2. What is the moment for the interior span bounded by F-G-K-J? h

SITUATION 2: Refer to Figure G H Given: Sz S1 = S2 = S3 = 2.8 m L1 = L2 = L3 = 6.0 m b = 300 mm h = 400 mm K L P L1 L2 L3 t= 100 mm PLAN Column = 0.3 m x 0.3 m Total Ultimate load = 12 kN/m 2. What is the moment for the interior span bounded by F-G-K-J? h

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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/. u/ es ug up w21E A W21 x 50 spans 11.0 m on a simple span. The compression flange is laterally supported at the third points. A36 Steel is used. A = 9,484 mm² b₁ = 165.9 mm Properties of W21 x 50 section tw = 9.7 mm Ix=409.57 x 10 mm Iy = 10.36 x 10 mm Sx = 1548.58 x 10³ mm³ Sy = 125.20 x 10³ mm³ d=529.1 mm tf = 13.6 mm w = 0.73 kN/m a) Determine the allowable bending stress of this beam. b) Determine the allowable superimposed uniformly distributed load be place on this beam.
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A W section is used as a beam without lateral support. A36 steel is used with Fy3D248 MPa. Compute the allowable bending stress(MPa) if it has a simple span of 9 m. Properties of W Section A=24709.6 mm2 d 840.74 mm tw3D14.73 mm I 2792.9 x 106 mm I, 90.74 x 106 mm S= 6653.15 x 103 mm b= 292.35 mm ty%3D21.72mm We Ar Sy 621.07 x 103 mm %3D De ry=335.28 mm Fla Ty = 60.45 mm Flar Wel Mor Sect Ime tw mm d. by
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