Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 6, Problem 6.2.16P

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

Expert Solution & Answer
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To determine

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

Answer to Problem 6.2.16P

The ratio of the maximum permissible moment before and after reinforcement is0.755 .

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).

  Is=5.936×105m4.

The maximum moment based on allowable stress in steel..

  Mmax=σsa[EsIs+EwIwh2Es].

σsa - the allowable stress in steel

Substituteσsa -140MPa,Es -210GPa ,Is -5.936×105.

Ew -12GPa ,Iw -2.543×103.

  Mmax=140×106((210×109×5.936×105)+(12×109×2.543×103)192.813×103×210×109).

  Mmax=148.597×103N.m×(1kN1000N)Mmax=148.597KN.m.

.

The maximum moment based on the allowable stress on top..

  MwmaxTop=σwa[EsIs+EwIwh1Ew].

  MwmaxTop=12×106((210×109×5.936×105)+(12×109×2.543×103)293.587×103×12×109).

  MwmaxTop=146.387×103N.m×(1KN1000N)MwmaxTop=146.387KN.m.

The maximum moment based on the allowable stress in wood on bottom..

  MwmaxBot=σwa[EsIs+EwIw(h26.4)Ew].

Substitute the value

Substituteσsa -140MPa,Es -210GPa,Is -5.936×105.

Ew -12GPa ,Iw -2.543×103.

  MwmaxBop=12×106((210×109×5.936×105)+(12×109×2.543×103)(192.8136.4)×103×12×109).

  MwmaxBot=230.548×103N.m×(1KN1000N)Mwmaxbot=230.548KN.m.

The lower magnitude of the moment among the three values for the cross section is to be safe. Hence, the maximum allowable moment isMwmaxTop=146.387KN.m..

Therefore the ratio of the maximum permissible moment before and after reinforcement is

  ratio=MimaxMwmaxTop.

Substitute the valueMimax =110.592KN andMwmaxTop =146.387KN

  ratio=110.592146.387ratio=0.755.

Therefore, the ratio of the maximum permissible moment before and after reinforcement is0.755 .

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Chapter 6 Solutions

Mechanics of Materials (MindTap Course List)

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