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 6.2.14P

-14 A simply supported composite beam with a 3.6 m span supports a triangularly distributed load of peak intensity q0at mid-span (see figure part a). The beam is constructed of two wood joists, each 50 mm x 280 mm, fastened to two steel plates, one of dimensions 6 mm × 80 mm and the lower plate of dimensions 6 mm x 120mm (see figure part b). The modulus of elasticity for the wood is 11 GPa and for the steel is 210 GPa.

If the allowable stresses are 7 MPa for the wood and 120 MPa for the steel, find the allowable peak load intensity q0maxwhen the beam is bent about the z axis. Neglect the weight of the beam.

  Chapter 6, Problem 6.2.14P, -14 A simply supported composite beam with a 3.6 m span supports a triangularly distributed load of

Expert Solution & Answer
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To determine

The peak maximum load intensity q0,max with the bent beam at the z axis.

Answer to Problem 6.2.14P

The peak maximum load intensity q015.6kN/m

Explanation of Solution

Given Information:

The given figure,

  Mechanics of Materials (MindTap Course List), Chapter 6, Problem 6.2.14P

A supported beam with the span of 3.6m supports a load that is triangularly distributed with the peak intensity at the mid span. The two wood joists of the beam each measuring 50mm*280mm that is fastened to steel plates of two that has dimensions of 6mm*80mm and 6mm*120mm. The wood elasticity modulus is 11GPa and for steel it is 210GPa.

By the given equation the neutral axis location is determined,

  EsxydA+EwwydA=0Es(y s1¯As1)+Es(y s2¯As2)+Ew(y w3¯Aw3)+Ew(y w4¯Aw4)=0Es[(y s1¯As1)+(y s2¯As2)]+Ew(y w3¯Aw3)+Ew(y w4¯Aw4)=0(1)

Where,

  EsSteel elasticity modulusEwwood elasticity modulus

Using the figure we have,

The distance between the section 1centroid and composite section neutral axis is,

  ys1¯=(h140)

The distance between the section 2 centroid and composite section neutral axis is,

  ys2¯=(280h160)

The distance between the section 3 centroid and composite section neutral axis is,

  yw3¯=(h12802)

The distance between the section 4 centroid and composite section neutral axis is,

  yw4¯=(h12802)

Section 1 area As1=80*6=480mm2

Section 2 area As2=120*6=720mm2

Section 3 area Aw3=280*50=14000mm2

Section 4 area Aw4=280*50=14000mm2

Substituting the given values in (1)

  Es[(y s1¯As1)+(y s2¯As2)]+Ew(y w3¯Aw3)+Ew(y w4¯Aw4)=0210*|103|[(h140)480+[(280h160)]720]+11*|103|[(h140)14000+[h1140]14000]=0

  h1=143.6mm

Assume the load is symmetric at the span centre. So at point A the reaction is half the total load

  RA=12*1.8*q0

  =0.9q0

The bending moment maximum is seen at the mid span as the symmetric load is at the span centre.

In the diagram the beam at the mid span is cut and the equilibrium cut at the left side is considered. Here M is considered as the bending moment maximum.

The right section moment is given as,

  M mid=0RA*1.8+12*1.8*q0(13*1.8)+M=00.9q0*1.8+12*1.8*q0(13*1.8)+M=00.9q0*2.3*1.8*q0(0.6)+M=0M=1.08q0Nm

  =1.08q0Nm*|103mm1m|1080q0Nmm

With the help of parallel axis theorem the steel section’s second moment of inertia at the neutral axis is determined.

  Is=Is1+Is2

  =( b 1 d 1 3 12+A s1 ( y s1 ¯ )2)+( b 2 d 2 3 12+A s2 ( y s2 ¯ )2)=( 6* 80 3 12+(6*80) ( h 1 40 )2)+( 6* 120 3 12+(6*120) ( 280 h 1 60 )2)

  =( 6* 80 3 12+(6*80) ( 143.640 )2)+( 6* 120 3 12+(6*120) ( 280143.660 )2)=(3072000+(480) ( 103.6 )2)+(864000+(720) ( 280143.660 )2)=10474432mm4

With the help of parallel axis theorem the wooden section’s second moment of inertia at the neutral axis is determined.

  Iw=Iw3+Iw4

  =( b 3 d 3 3 12+A w3 ( y w3 ¯ )2)+( b 4 d 4 3 12+A w4 ( y s4 ¯ )2)=( 50* 280 3 12+(50*2800) ( h 1 280 2 )2)+( 50* 280 3 12+(5*2800) ( h 1 280 2 )2)

  =( 50* 280 3 12+(50*2800) ( 143.6 280 2 )2)+( 50* 280 3 12+(5*2800) ( 143.6 280 2 )2)=186562133mm4

The equation of the flexural stress at the maximum of the steel that has the top layer,

  σts=Mh1EsEsIs+EwIw

Substituting the values we have,

  120=1080q0*143.6*210*|103|210*|103|*10474432+11*|103|*186562133(2)

  q0=15666N/m

As the top layer of the steel is compressive the negative sign is given in the equation (2)

The equation of the flexural stress at the maximum of the steel that has the bottom layer,

  σbs=M(h2)EsEsIs+EwIw

  120=1080q0EsEsIs+EwIw120=1080q0*( (280143.6))*210*| 10 3|210*| 10 3|*10474432+11*| 10 3|*186562133(3)

  q0=16493N/m

The equation of the flexural stress at the maximum of the wood that has the top layer,

  σtw=M(h1)EwEsIs+EwIw

  7=1080q0*143.6*11*|103|210*|103|*10474432+11*|103|*186562133(4)

  q0=17446N/m

The equation of the flexural stress at the maximum of the wood that has the bottom layer,

  σbw=M((280h1))EwEsIs+EwIw

  7=1080q0*((280143.6))*11*|103|210*|103|*10474432+11*|103|*186562133(5)

  q0=18367N/m

The load with the lowest magnitude is selected from the equations (2), (3), (4), (5)

  q0=15666N/m15.6kN/m

Conclusion:

Thus, the peak maximum load intensity is calculated by substituting section areas in equation 1.

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Chapter 6 Solutions

Mechanics of Materials (MindTap Course List)

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