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-15 A composite beam is constructed froma wood beam (3 in. x 6 in.) and a steel plate (3 in, wide). The wood and the steel are securely fastened to act as a single beam. The beam is subjected to a positive bending moment M. = 75 kip-in. Calculate the required thickness of the steel plate based on the following limit states: Allowable compressive stress in the wood = 2 ksi Allowable tensile stress in the wood = 2 ksi Allowable tensile stress in the steel plate = 16 ksi Assume that E w = 1,500 ksi and e s = 30,000 ksi.

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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 6.2.15P
Textbook Problem
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-15 A composite beam is constructed froma wood beam (3 in. x 6 in.) and a steel plate (3 in, wide). The wood and the steel are securely fastened to act as a single beam. The beam is subjected to a positive bending moment M. = 75 kip-in. Calculate the required thickness of the steel plate based on the following limit states:

  1. Allowable compressive stress in the wood = 2 ksi
  2. Allowable tensile stress in the wood = 2 ksi
  3. Allowable tensile stress in the steel plate = 16 ksi
  4. Assume that Ew= 1,500 ksi and es= 30,000 ksi.

  Chapter 6, Problem 6.2.15P, -15 A composite beam is constructed froma wood beam (3 in. x 6 in.) and a steel plate (3 in, wide).

a.

To determine

The thickness required for the steel plate.

Explanation of Solution

Given:

The given figure

  

The wooden beam of 3in.*6in. and steel plate with wide 3in. forms the beam. The positive bending moment is given as Mz=75kip.in . Compressive stress that is maximum is 2 ksi.

Concept Used:

Normal stress that is maximum for steel,

  σsa=M2h2(Es)EsIs+EwIw

Where,

  M2=bending momentIw=wood Inertia momentIs=steel Inertia momentEW=wood modulus ElasticityEs=Steel modulus Elasticityh2=height

Calculation:

Neutral axis location at the lower end is given as,

  h2=Es(bts)ts2+Ew(bhw)(ts+ h w 2)Es(bts)+Ew(bhw)

Substituting the values we have,

  =30000ts2+1500(6)(2ts+6)2(30000ts)+1500(6)

  =10ts2+6ts+1820ts+6=5ts2+3ts+910ts+3

From the top the neutral axis distance is given as,

  h1=hw+tsh2

  =6+ts5ts2+3ts+910ts+3=( 6+ t s )( 3+10 t s )( 5 t s 2 +3 t s +9)10ts+3( 60 t s )+10ts+3ts+185ts23ts910ts+3=( 60 t s )+5ts2+910ts+3

The wooden section moment of inertia is given as,

  Iw=bhw312+bhw(h1 h w 2)2

  =3*6312+3*6( ( 60 t s )+5

b.

To determine

The thickness required for the steel plate.

c.

To determine

The thickness required for the steel plate.

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Chapter 6 Solutions

Mechanics of Materials (MindTap Course List)
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Ch. 6 - A bimetallic beam used in a temperature-control...Ch. 6 - A simply supported composite beam 3 m long carries...Ch. 6 - A simply supported wooden I-beam with a 12-ft span...Ch. 6 - -14 A simply supported composite beam with a 3.6 m...Ch. 6 - -15 A composite beam is constructed froma wood...Ch. 6 - A wood beam in a historic theater is reinforced...Ch. 6 - Repeat Problem 6.2-1 but now assume that the steel...Ch. 6 - Repeat Problem 6.2-17 but now use a...Ch. 6 - A sandwich beam having steel faces enclosing a...Ch. 6 - A wood beam 8 in. wide and 12 in. deep (nominal...Ch. 6 - A simple beam of span length 3.2 m carries a...Ch. 6 - A simple beam that is 18 ft long supports a...Ch. 6 - The composite beam shown in the figure is simply...Ch. 6 - The cross section of a beam made of thin strips of...Ch. 6 - Consider the preceding problem if the beam has...Ch. 6 - A simple beam thai is IS ft long supports a...Ch. 6 - The cross section of a composite beam made of...Ch. 6 - A beam is constructed of two angle sections, each...Ch. 6 - 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