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 5, Problem 5.5.25P

A beam with a T-section is supported and loaded as shown in the figure. The cross section has width b = 2 1/2 in., height c = 3 in., and thickness t = 3/8 in.

  1. Determine the maximum tensile and compressive stresses in the beam.

  • If the allowable stresses in tension and compression are 18 ksi and 12 ksi, respectively, what is the required depth h of the beam? Assume that thickness t remains at 3/8 in. and that flange width/) = 2.5 in.
  • Find the new values of loads P and q so that the allowable tension (18 ksi) and compression (12 ksi) stresses are reached simultaneously for the beam. Use the beam cross section in part (a) (see figure) and assume that Lh and L3are unchanged.
  •   Chapter 5, Problem 5.5.25P, A beam with a T-section is supported and loaded as shown in the figure. The cross section has width

    (a)

    Expert Solution
    Check Mark
    To determine

    The maximum tensile stress.

    The maximum compressive stress.

    Explanation of Solution

    Given information:

    The uniform load is 110lb/ft , the width of the beam is 212inthe height of the beam is 3inand the thickness of the beam is 38in .

    The following figure shows the free body diagram of the beam.

      Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.5.25P , additional homework tip  1

    Figure-(1)

    Write the expression for the distance of the neutral axis from the bottom layer.

      c2=b×t×(t2)+t×(ht)(ht2+t)(b×t)+t×(ht)  ......(I)

    Here, the width of the beam is b , the height of the t section is h , the thickness of the beam is b , and the distance of neutral axis is c2 .

    Write the expression for the distance of the neutral axis from the top layer.

      c1=hc2   ......(II)

    Here, the distance of neutral axis from the top layer.

      Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.5.25P , additional homework tip  2

    Figure-(2)

    Write the expression for the moment of inertia.

      I=bt312+bt(c2+t2)2+(ht)312t+t(ht)(c1( ht 2))2  ......(III)

    Here, the moment of inertia is I .

    Write the expression for the moment equilibrium about A .

      MA=0RB×L2=P×L1+qL3(L2+ L 32)   ......(IV)

    Here, the reaction at point Bis RB , the length of section ABis L2and the length of section BCis L3and the concentrated load is P , and the distance at which concentrated load is acting is L1 .

    Write the expression for the force equilibrium in vertical direction.

      Fy=0RA=P+qL3RB  ......(V)

    Here, the reaction at point Ais RA .

    Write the expression for the maximum sagging moment at load P .

      Mp=RA×L1  ......(VI)

    Here, the maximum sagging moment is MP .

    Write the expression for the maximum hogging moment at point B .

      MB=q×L3×L32  ......(VII)

    Here, the maximum hogging moment is MB .

    Write the expression for the maximum tensile stress at point B .

      σt=MBc1I   ......(VIII)

    Here, the maximum tensile stress is σt .

    Write the expression for the maximum compressive stress at point B .

      σc=MBc2I   ......(IX)

    Here the maximum compressive stress is σc .

    Calculation:

    Substitute 2.5infor b , 38infor t , and 3infor hin Equation (I).

      c2=2.5in×38in×( 38in 2)+38in×(3in 3 8in)( 3in38in 2+ 3 8in)(2.5in× 3 8in)+38in×(3in 3 8in)=0.1757in+1.6611in1.9219in=0.9558in

    Substitute 0.9558infor c2 , and 3infor hin Equation (II).

      c1=3in0.9558in=2.0442in

    Substitute 2.5infor b , 38infor t , 0.9558infor c2 , 2.442infor c1and 3infor hin Equation (III).

      I=[ (2.5in )( 38 in)3 12+( 2.5in×38 in)( (0.9558in)+( 38in)2)2+ ( (3in )(38in ))3 12( 38 in)+( 38 in)( (3in )(38in ))( (2.0442in)( 3in38in 2))2]=1.2411in4+0.619in4=1.8602in4

    Substitute 8ftfor L2 , 750lbfor P , 5ftfor L3 , 3ftfor L1and 110lb/ftfor q , in Equation (IV).

      (RB×8ft)=(750lb×3ft)+(110lb/ft)(5ft)(8ft+3ft2)RB=(750lb×3ft)+(110 lb/ ft)(5ft)(8ft+ 3ft 2)8ftRB=1003.125lb

    Substitute 8ftfor L2 , 750lbfor P , 5ftfor L3 , 1003.125lbfor RBand 110lb/ftfor q , in Equation (V).

      RA=(750lb)+(110lb/ft)(5ft)(1003.125lb)=(750lb)+(550lb)(1003.125lb)=296.875lb

    Substitute 296.875lbfor RA , 3ftfor L1in Equation (VI).

      MP=(296.875lb)×(3ft)=890.625lbft

    Substitute 5ftfor L3 , 110lb/ftfor qin Equation (VII).

      MB=(110lb/ft)×5ft×5ft2=(110lb/ft)×12.5ft2=1375lbft

    Substitute 1375lbftfor MB , 1.8602in4for Iand 2.0442infor c1 , in Equation (VIII).

      σt=(1375lbft)×(2.0442in)1.8602in4=(1375lbft)( 12lbin 1lbft)×(2.0442in)1.8602in4=(18132.08lb/ in2)(1psi1 lb/ in2 )=18132.08psi

    Substitute 1375lbftfor MB , 1.8602in4for Iand 0.9558infor c2 , in Equation (IX).

      σc=(1375lbft)×(0.9558in)1.8602in4=(1375lbft)( 12lbin 1lbft)×(0.9558in)1.8602in4=(8477.95lb/ in2)(1psi1 lb/ in2 )=8477.95psi

    Conclusion:

    The maximum tensile stress is 7810psi .

    The maximum compressive stress is 13885psi .

    (b)

    Expert Solution
    Check Mark
    To determine

    The required depth of the beam.

    Answer to Problem 5.5.25P

    The required depth of the beam is 3.26in .

    Explanation of Solution

    Given Information:

    The allowable stress in tension is 18ksi , the allowable stress in compression is 12ksi .

    Write the expression for the maximum tensile stress at point B .

      σt=MBc1I   ......(X)

    Write the expression for the distance of the neutral axis from the bottom layer.

      c1=b×t×(t2)+t×(ht)(ht2+t)(b×t)+t×(ht)  ......(XI)

    Write the expression for the moment of inertia.

      I=bt312+bt(c2+t2)2+(ht)312t+t(ht)(c1( ht 2))2  ......(XII)

    Calculation:

    Substitute 2.5infor b , and 38infor tin Equation (XI).

      c1=2.5in×38in×( 38in 2)+38in×(h 3 8in)( h38in 2+ 3 8in)(2.5in× 3 8in)+38in×(h 3 8in)c1=( 3 16 h 2+ 51 64h+ 153 1024)( 51 64+ 3 89h)

    Substitute 2.5infor b , 38infor t , 0.9558infor c2 , 2.442infor c1and 3infor hin Equation (XII).

      I=[ (2.5in )( 38 in)3 12+( 2.5in×38 in)( (0.9558in)+( 38in)2)2+ ( h(38in ))3 12( 38 in)+( 38 in)( (h )(38in ))( (2.0442in)( h38in 2))2]=[454906+1516( hc1 316 )2+396( h38 )3+38(h38)( c1 (h23 16 ))2]

    Substitute 18ksifor σt , 1375lbftfor MB , (316h2+5164h+1531024)(5164+389h)for c1and [454906+1516(hc13 16)2+396(h38)3+38(h38)(c1( h2 316 ))2]for Iin Equation (X).

    .   ......(XIII)

    Substitute 3infor hin Equation (XIII).

      (12in11)=( (( 316in) (3in) 2+( 5164in)( 3in)+( 1531024in)) ( 51 64in+( 389in)( 3in)))[ 454906 in2 +1516 in(( 3in) 3 16)2 +396 in(( 3in) 3 8)3 +38 in(( 3in)38in )( c 1( ( 3in)2316in))2 ]12in11=2.1997in21.7031in21.091=1.2915

    Substitute 3.26infor hin Equation (XIII).

      (12in11)=( (( 316in) (3.26in) 2+( 5164in)( 3.26in)+( 1531024in)) ( 51 64in+( 389in)( 3.26in)))[ 454906 in2 +1516 in(( 3.26in) 3 16in)2 +396 in(( 3.26in) 3 8)3 +38 in(( 3.26in)38in )(( ( 3.26in)2316in))2 ]12in11=2.3472in22.1400in1.091=1.091

    Conclusion:

    The required depth of the beam is 3.26in .

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