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A wood beam in a historic theater is reinforced with two angle sections at the outside lower corners (see figure). If the allowable stress in the wood is 12 M Pa and that in the steel is 140 M Pa, what is ratio of the maximum permissible moments for the beam before and after reinforcement with the angle sections? See Appendix F Table F-5(b) for angle section properties. Assume that e w = 12 GPa and E 3 =210 GPa.

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Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347

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BuyFindarrow_forward

Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 6, Problem 6.2.16P
Textbook Problem
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A wood beam in a historic theater is reinforced with two angle sections at the outside lower corners (see figure). If the allowable stress in the wood is 12 M Pa and that in the steel is 140 M Pa, what is ratio of the maximum permissible moments for the beam before and after reinforcement with the angle sections? See Appendix F Table F-5(b) for angle section properties. Assume that ew= 12 GPa and E3=210 GPa.

  Chapter 6, Problem 6.2.16P, A wood beam in a historic theater is reinforced with two angle sections at the outside lower corners

To determine

The ratio of maximum permissible moment of a beam before and after reinforcement.

Explanation of Solution

Given: .

  Area,AL=1090mm2.

Moment of Inertia,I11=1.14×106mm4..

Distance from the neutral axis, d=31mm..

  Es =12GPa..

  Ew =21GPa..

Width, b=240mm..

Height, h=480mm..

Calculation: .

The maximum moment if no angle section used is calculated as:.

  M1max=σwa(bh26).

  σwa - Allowable stress..

b- The width of section..

h- Height of section..

  M1max=12×(240×48026)M1max=110.592×106N.mm×(1KN1000N)×(1m1000mm)M1max=110.592kN.m.

The location of the neutral axis from the lower end is stated as:.

  h2=Ew(bh)(h2+6.4)+Es(2AL)dEw(bh)+Es(2AL)..

Young modulli of steel and wood areEw andEs.

SubstituteEs =12GPa,Ew =21GPa, b=240mm, d=31,Al =1090mm, h=480mm..

  h2={12×103(240×480)(4802+604)+{210×103(2×1090)31}{12×103(240×480)}+{210×103(2×1090)}}.

  h2={3.4062×1011}+{0.141918×1011}{13.824×108}+{4.578×108}.

  h2=0.192813×103mmh2=192.813mm.

The distance of the neural axis from the top..

  h1=h+6.4h2.

Substitute h=480mm andh2 =192.813mm.

  h1=480+6.4192.813h1=293.587mm.

Calculate the moment of Inertia for wood section..

  Iw=bh312+(bh)(h2h1)2.

Substitute b=240mm, h=480mm andh1 =293.587mm.

  Iw=240×480312+(240×480)(4802293.587)2.

  Iw=22.1184×108+3.3080×108Iw=25.43×108mm4×(1m41000mm4)Iw=2.543×103m4.

The moment of Inertia of the steel section..

  Is=2I11+2AL(h2d)2.

SubstituteI11 =1.14×106mm4 ,h2 =192.813mm,AL =1090mm , d=31mm..

  Is=2(1.14×106mm4)+2(1090)(192.81331)2.

  Is=0.0228×108+0.570799×108Is=0.5936×108mm4×(1m410004mm4)

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Chapter 6 Solutions

Mechanics of Materials (MindTap Course List)
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