Mechanics of Materials - MindTap Access
Mechanics of Materials - MindTap Access
9th Edition
ISBN: 9781337093620
Author: GOODNO
Publisher: Cengage
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Chapter 5, Problem 5.11.11P

The three beams shown have approximately the same cross-sectional area. Beam 1 is a W 14 X 82 with flange plates; beam 2 consists of a web plate with four angles; and beam 3 is constructed of 2 C shapes with flange plates.

  1. Which design has the largest moment capacity?
  2. Which has the largest shear capacity?
  3. Which is the most economical in bending?
  4. Which is the most economical in shear?

Assume allowable stress values are: = 18 ksi and ra=11 ksi. The most economical beam is that having the largest capacity-to-weight ratio. Neglect fabrication costs in answering parts (c) and (d) above. Note: Obtain the dimensions and properties of all rolled shapes from tables in Appendix F.

  Chapter 5, Problem 5.11.11P, The three beams shown have approximately the same cross-sectional area. Beam 1 is a W 14 X 82 with

Expert Solution & Answer
Check Mark
To determine

The beam of the largest momentum capacity.

Answer to Problem 5.11.11P

The beam of the largest momentum capacity is beam-1.

Explanation of Solution

Given information:

The allowable normal stress is 18ksi and the allowable shear stress is 11ksi .

Write the expression for the total moment of inertia of the beam-1.

  I1=(Iw)1+2[b1t1312+b1t1( ( h w ) 1 2+ t 1 2)2]   ......(I)

Here, the total moment of inertia of the beam-1.is I1 , moment of inertia of the web-1 is (Iw)1 , height of the web-1 is (hw)1 , thickness of the cover plate is t1 , and width of the cover plate is b1 .

Write the expression for the total height of the beam-1.

  h1=(hw)1+2t1   ......(II)

Here, the total height of the beam-1 is h1 .

Write the expression for the total area of the beam-1.

  A1=(Aw)1+2b1t1   ......(III)

Here, the total area of the beam-1.is A1 and area of the web-1 is (Aw)1

Write the expression for the section of the modulus of the beam-1.

  S1=2I1h1   ......(IV)

Here, the section of the modulus of the beam-1 S1 .

Write the expression for the total moment of inertia of the beam-2.

  I2=b2t2312+4[Ia+Aa( h 2 2 C a)2]   ......(V)

Here, the total moment of inertia of the beam-2 is I2 , width of the plate is b2 , thickness of the plate is t2 , moment of the inertia of the angle is Ia , area of the angle is Aa , height of the plate is h2 , and centroid of the angle is Ca .

Write the expression for the total area of the beam-2.

  A2=4Aa+b2t2   ......(VI)

Here, the total area of the beam-2 is A2 .

Write the expression for the section of the modulus of the beam-2.

  S2=2I2h2   ......(VII)

Here, the section of the modulus of the beam-2 is S2 .

Write the expression for the total height of beam-3.

  h3=hc+2t3   ......(VIII)

Here, the total height of beam-3 is h3 , height of the channel is hc , thickness of the flange plate is t3 .

Write the expression for the total moment of inertia of the beam-3.

  I3=2Ic+2[b3t3312+b3t3( h 2+ t 3 2)2]   ......(IX)

Here, the total moment of inertia of the beam-2 is I3 , width of the flange plate is b3 , moment of the inertia of the channel is Ic .

Write the expression for the total area of the beam-3.

  A3=4Ac+2b3t3   ......(X)

Here, the total area of the beam-3 is A3 .

Write the expression for the section of the modulus of the beam-3.

  S3=2I3h3   ......(XI)

Here, the section of the modulus of the beam-3 S2 .

Calculation:

Refer Table-F-1(a) “Properties of wide flange section” to obtain (Aw)1 as 24in , (Iw)1 as 881in4 , b1 as 8in , t1 as 0.52in , tf1 as 0.855in , bf1 as 10.1in .

Substitute 881in4 for (Iw)1 , 8in for b1 , 0.52in for t1 , 14.3in for (hw)1 in the Equation (I)

  I1=881in4+2×[8in ( 0.52in ) 312+(8in)(0.52in)( 14.3in 2 + 0.52in 2 )2]=881in4+2(0.937 in4+228.42 in4)=1339.71in4

Substitute 0.52in for t1 and 14.3in for (hw)1 in the Equation (II).

  h1=14.3in+2×0.52in=14.3in+1.04in=15.34in

Substitute 24in for (Aw)1

  8in for b1 , and 0.52in for t1 in the Equation (III).

  A1=24in+2×(8in×0.52in)=24in2+8.32in2=32.32in2

Substitute 1339.71in4 for I1 , 15.34in for h1 in the Equation (IV).

  S1=2×1339.71 in415.34in=2×87.33in3=174.66in3

Refer Table-F-4(a) “Properties of wide equal legs L-shape channel” L6×6×12 to obtain

  Aa as 5.77in2 , Ia as 19.9in4 , b2 as 0.675in , t2

  19.9in4 , and Ca as 1.67in .

Substitute 0.675in for b2 , 14in for t2 , 19.9in4 for Ia , 5.77in2 for Aa and 1.67in for Ca in the Equation (V).

  I2=( 0.675in) ( 14in )312+4[19.9in4+5.77in2( 14in 2 1.67in)2]=154.35in4+4(19.9 in4+163.92 in4)=889.62in4

Substitute 0.675in for b2 , 14in for t2 , and 5.77in2 for Aa in the Equation (VI).

  A2=4(5.77 in2)+(14in)(0.675in)=23.08in2+9.45in2=32.53in2

Substitute 889.62in4 for I2 and 14in for h2 in the Equation (VII).

  S2=2×889.62 in414in=2×63.54in3=127.09in3

Refer Table-F-3(a) “Properties of C-shape channel” for C15×150 to obtain hc as 15in , t3 as 0.375in , Ic as 404in4 , b3 as 4in , Ac as 14.7in2 .

Substitute 15in for hc and 0.375in for t3 . in the Equation (VIII).

  h3=15in+2×(0.375in)=15in+0.75in=15.75in

Substitute 404in4 for Ic , 4in for b3

  4in for t3 and 15in for h in the Equation (IX).

  I3=2×404in4+2[4in ( 0.375in ) 3112+(4in×0.375in)( 15in 2 + 0.375in 2 )2]=808in4+2[0.0176in4+88.65in4]=808in4+177.33in4=985.33in4

Substitute 14.7in2 for Ac , 4in for b3 and 4in for t3 in the Equation (X).

  A3=2(14.7 in2)+2(4in)(0.375in)=29.4in2+3in2=32.4in2

Substitute 985.33in4 for I3 and 15.75in for h3 in the Equation (XI).

  S3=2×985.33 in415.75in=2×62.56in3=125.12in3

Since stress due to moment is inversely proportional to section modulus, therefore the beam which has the highest section modulus will have the highest capacity.

The value of the section modulus of the beam-1 is highest i.e., (S1=174.66in3)

Conclusion:

The beam of the largest momentum capacity. is beam-1.

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Chapter 5 Solutions

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