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 5.5.30P

Beam ABCDE has a moment release just right of joint B and has concentrated moment loads at D and E. In addition, a cable with tension P is attached at fand runs over a pulley at C (Fig, a). The beam is constructed using two steel plates, which arc welded to form a T cross section (see Fig. b). Consider ßexuralstresses only Find the maximum permissible value of load variable P if the allowable bending stress is 130 M Pa. Ignore the self-weight of the frame members and let length variable L = 0.75 m.

  Chapter 5, Problem 5.5.30P, Beam ABCDE has a moment release just right of joint B and has concentrated moment loads at D and E. , example  1

  Chapter 5, Problem 5.5.30P, Beam ABCDE has a moment release just right of joint B and has concentrated moment loads at D and E. , example  2

Expert Solution & Answer
Check Mark
To determine

The maximum permissible load.

Answer to Problem 5.5.30P

The maximum permissible load is 675.38N .

Explanation of Solution

Given information:

The allowable bending stress is 130MPa , the length of the beam is 0.75m .

Write the expression for the force equilibrium in x direction.

  Fx=0Ax=0 .....(I)

Here, the reaction at point A of the beam is Ax .

Write the expression for the moment at the point of release.

  Mrel=0Ay(4L)+4P(3L)5=0Ay=0.6P .....(II)

Here, the length of the beam is L , the load on the beam is P .

Write the expression for the moment about point E .

  ME=0Dy=14L(8PL2PL+P(8L)(Ay)(16L)).....(III)

Here, the vertical reaction at point D is Dy .

Write the expression for the force equilibrium in y direction.

  Fy=0Ey=PAyDy ….. (IV)

Here, the vertical reaction at point E is Ey .

Write the expression for the moment at 0x4L .

  M(x)=Ayx .....(V)

Here the moment is M(x) .

Write the expression for the moment at 4Lx8L .

  M(x)=[Ayx+45P(3L)35P(x4L)].....(VI)

Write the expression for the moment at 8Lx12L .

  M(x)=AyxP(x8L).....(VII)

Write the expression for the moment at other points.

  M(x)=Ey(16Lx)+8PL .....(VIII)

Write the expression for the area of the T shape.

  A=bftf+bwtw .....(IX)

Here, the width of the flange is bf , the thickness of the flange is tf , the width of the web is bw , the thickness of the flange is tw .

Write the expression for the centroid.

  c2=bftf(bw+tf2)+bwtw(bw2)A….. (X)

Here, the distance of the centroid from one end is c2 .

Write the expression for the other distance of the centroid.

  c1=bw+tfc2 .....(XI)

Here, the distance of the centroid from other end is c1 .

Write the expression for the moment of inertia about z axis.

  Iz=bft3f12+(bftf)(c1tf2)2+b3wtw12+(bwtw)(c2bw2)2 .....(XII)

Here, the moment of inertia about z axis is Iz .

Write the expression for the section modulus at the top section.

  Stop=Izc1 .....(XIII)

Here, the section modulus at top section is Stop

Here, the section modulus of top section is Stop .

Write the expression for the section modulus at the bottom section.

  Sbot=Izc2 .....(XIV)

Here, the section modulus of bottom section is Sbot .

Write the expression for the maximum permissible load.

  Pmax=σaSbotMmax.....(XV)

Here, the allowable stress is σa , the maximum permissible stress is Pmax .

Calculation:

Substitute 0.6P for Ay in Equation (III)

  Dy=14L(8PL2PL+P(8L)(0.6P)(16L))=14L(23.6PL)=5.9P

Substitute 0.6P for Ay , and 5.9Dy in Equation (IV)

  Ey=P(0.6P)5.9P=P+(0.6P)5.9P=4.3P

Substitute 150mm for bf , 18mm for tf , 100mm for bw , 12mm for tw in Equation (IX)

  A=((150mm×18mm)+(100mm×12mm))=2700mm2+1200mm2=3.9×103mm2

Substitute 150mm for bf , 18mm for tf , 100mm for bw , 12mm for tw , and 3.9×103mm2 for A in Equation (X)

  c2=[(150mm×18mm)(100mm+ 18 2mm)+(100mm×12mm)(100mm+ 100 2mm)3.9×103mm2]=[294300mm3+180000mm33.9×103mm2]=90.846mm

Substitute 18mm for tf , 100mm for bw , 90.846mm for c2 in Equation (XI)

  c1=100mm+18mm90.846mm=118mm-90.846mm=27.154mm

Substitute 150mm for bf , 18mm for tf , 100mm for bw , 12mm for tw in Equation (XII)

  Iz=[( 150mm× ( 18mm) 3 12)+(150mm×18mm)( 27.154mm 18mm 2 )2+( ( 100mm) 2 ×12mm 12)+(100mm×12mm)(90.846mm ( 100mm2 ) 2)]=[(72900 mm 4)+(150mm×18mm)( 27.154mm 18mm 2 )2+(10000 mm 4)+(100mm×12mm)(90.846mm ( 100mm2 ) 2)]=3.965×106mm4

Substitute 3.965×106mm4 for Iz , 27.154mm for c1 , in Equation (XIII)

  Stop=3.965×106mm427.154mm=1.46×105mm3

Substitute 3.965×106mm4 for Iz , 90.846mm for c2 , in Equation (XIV)

  Sbot=3.965×106mm490.846mm=4.364×104mm3

Substitute 130MPa for σa , 4.364×104mm3 for Sbot , 0.75mm for L , 11.2Nm for Mmax in Equation (XV)

  Pmax=130MPa×(4.364× 104 mm3)11.2×0.75=(130MPa)( 10 6 Pa 1MPa)×(4.364× 104 mm4)( 1m 1000mm )4(11.2Nm)×(0.75mm)( 1m 1000mm)=(130× 106Pa)( 1N/ m 2 1Pa)×(4.364× 104× 10 12m3)(11.2Nm)×(0.75× 10 3m)=675.38N

Conclusion:

The maximum permissible load is 675.38N .

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Chapter 5 Solutions

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