Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card
Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card
9th Edition
ISBN: 9781337594318
Author: Barry J. Goodno; James M. Gere
Publisher: Cengage Learning
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Textbook Question
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Chapter 8, Problem 8.5.24P

Repeat Problem 8.5-22 but replace the square tube column with a circular tube having a wall thickness r = 5 mm and the same cross-sectional area (3900 mm2) as that of the square tube in figure b in Problem 8.5-22. Also, add force P. = 120 N at B

(a) Find the state of plane stress at C. (b) Find maximum normal stresses and show them on a sketch of a properly oriented element.

(c) Find maximum shear stresses and show them on a sketch of a properly oriented element.

  Chapter 8, Problem 8.5.24P, Repeat Problem 8.5-22 but replace the square tube column with a circular tube having a wall

(a)

Expert Solution
Check Mark
To determine

The state of plane stress on an element C.

Answer to Problem 8.5.24P

The state of stress on element C are

Tensile stress σt=0.499MPa

Compressive stress σc=0.5694MPa

Explanation of Solution

Given information:

The cross-section of the bicycle rack tubing is circular.

The weight of the bicycle = 14 kg is represented as point load applied at B on a plane frame model of the rack.

A force in z direction has also been added Pz= 120 N

Weight density of the steel γ=77kN/m3

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 8, Problem 8.5.24P , additional homework tip  1

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 8, Problem 8.5.24P , additional homework tip  2

Calculation:

From the figure, let us consider horizontal portion of the cycle rack. The weight of the cycle is considered as the point load and this load produces a moment and a direct compressive load in the beam along Y-direction.

Also, a load Pzhas been added in z direction. This load will produce bending moment in the horizontal beam and twisting moment in the vertical portion of the beam.

The cross-sectional area of the beam is given by

  A=3900mm2=3.9×103m2

  A=π4(d2(dt)2)3900=π4(d2(d5)2)

  d=499.06mm

The moment in the vertical beam =

  M=14×9.81×1M=137.34Nm

The moment in the horizontal beam due to load Pz= Mz=Pz×1m

  Mz=120×1=120Nm

Total moment Me=137.34+120=257.34Nm

The load Pzwill produce twisting moment equal to Mz. This twisting moment will produce the shear stress at the base of rack.

The direct compressive loading = P=14×9.81=137.34N

From the properties of the cross-section, the moment of inertia is

  I=πd464π(dt)464

  I=π×(499.06)464π×(499.065)464

  I=1.202×108mm4I=1.202×104m4

Now, plane stress can be calculated as follows

  σt=PA+M×(d2)I

  σt=14×9.8139×104+257.34×(499.06×1032)1.202×104

  σt=0.499MPa

And maximum compressive stress is

  σc=PAM×(a2)I

  σc=14×9.8139×104257.34×(499.06×1032)1.202×104

  σc=0.5694MPa

Conclusion: Thus, the state of stress on element C are

Tensile stress σt=0.499MPa

Compressive stress σc=0.5694MPa

(b)

Expert Solution
Check Mark
To determine

The maximum normal stress and the sketch of properly oriented element.

Answer to Problem 8.5.24P

The normal stress values on element C are, σ1=0.094166MPa , σ2=0.1646MPa .

Explanation of Solution

Given information:

The cross-section of the bicycle rack tubing is circular.

The weight of the bicycle = 14 kg is represented as point load applied at B on a plane frame model of the rack.

A force in z direction has also been added Pz= 120 N

Weight density of the steel γ=77kN/m3

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 8, Problem 8.5.24P , additional homework tip  3

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 8, Problem 8.5.24P , additional homework tip  4

Calculation:

From the figure, let us consider horizontal portion of the cycle rack. The weight of the cycle is considered as the point load and this load produces a moment and a direct compressive load in the beam along Y-direction.

Also, a load Pzhas been added in z direction. This load will produce bending moment in the horizontal beam and twisting moment in the vertical portion of the beam.

The cross-sectional area of the beam is given by

  A=3900mm2=3.9×103m2

  A=π4(d2(dt)2)3900=π4(d2(d5)2)

  d=499.06mm

The moment in the vertical beam =

  M=14×9.81×1M=137.34Nm

The moment in the horizontal beam due to load Pz= Mz=Pz×1m

  Mz=120×1=120Nm

Total moment Me=137.34+120=257.34Nm

The load Pzwill produce twisting moment equal to Mz. This twisting moment will produce the shear stress at the base of rack.

The direct compressive loading = P=14×9.81=137.34N

From the properties of the cross-section, the moment of inertia is

  I=πd464π(dt)464

  I=π×(499.06)464π×(499.065)464

  I=1.202×108mm4I=1.202×104m4

Now, plane stress can be calculated as follows

  σt=PA+M×(d2)I

  σt=14×9.8139×104+257.34×(499.06×1032)1.202×104

  σt=0.499MPa

And maximum compressive stress is

  σc=PAM×(a2)I

  σc=14×9.8139×104257.34×(499.06×1032)1.202×104

  σc=0.5694MPa

The shear stress on the element is given by the torque and shear stress relation.

  τ=16×T×dπ(d4(dt)4)

  τ=16×120×0.499π((0.499)4(0.499(5×103))4)

  τ=0.124581MPa

The normal stresses are given by

  σ1,2=σx+σy2±12(σxσy)2+4τxy2

  σ1,2=0.4990.569442±12(0.4990.56944)2+4×(0.1245)2

  σ1=0.094166MPa

And,

  σ2=0.1646MPa

The state of stress on the element can be shown as follows

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 8, Problem 8.5.24P , additional homework tip  5

Conclusion: Thus, the normal stress values on element C are, σ1=0.094166MPa , σ2=0.1646MPa .

(c)

Expert Solution
Check Mark
To determine

The maximum shear stress and the sketch of properly oriented element.

Answer to Problem 8.5.24P

the maximum stress values on element C is τ=0.129383MPa .

Explanation of Solution

Given information:

The cross-section of the bicycle rack tubing is circular.

The weight of the bicycle = 14 kg is represented as point load applied at B on a plane frame model of the rack.

A force in z direction has also been added Pz= 120 N

Weight density of the steel γ=77kN/m3

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 8, Problem 8.5.24P , additional homework tip  6

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 8, Problem 8.5.24P , additional homework tip  7

Calculation:

From the figure, let us consider horizontal portion of the cycle rack. The weight of the cycle is considered as the point load and this load produces a moment and a direct compressive load in the beam along Y-direction.

Also, a load Pzhas been added in z direction. This load will produce bending moment in the horizontal beam and twisting moment in the vertical portion of the beam.

The cross-sectional area of the beam is given by

  A=3900mm2=3.9×103m2

  A=π4(d2(dt)2)3900=π4(d2(d5)2)

  d=499.06mm

The moment in the vertical beam =

  M=14×9.81×1M=137.34Nm

The moment in the horizontal beam due to load Pz= Mz=Pz×1m

  Mz=120×1=120Nm

Total moment Me=137.34+120=257.34Nm

The load Pzwill produce twisting moment equal to Mz. This twisting moment will produce the shear stress at the base of rack.

The direct compressive loading = P=14×9.81=137.34N

From the properties of the cross-section, the moment of inertia is

  I=πd464π(dt)464

  I=π×(499.06)464π×(499.065)464

  I=1.202×108mm4I=1.202×104m4

Now, plane stress can be calculated as follows

  σt=PA+M×(d2)I

  σt=14×9.8139×104+257.34×(499.06×1032)1.202×104

  σt=0.499MPa

And maximum compressive stress is

  σc=PAM×(a2)I

  σc=14×9.8139×104257.34×(499.06×1032)1.202×104

  σc=0.5694MPa

The shear stress on the element is given by the torque and shear stress relation.

  τ=16×T×dπ(d4(dt)4)

  τ=16×120×0.499π((0.499)4(0.499(5×103))4)

  τ=0.124581MPa

The normal stresses are given by

  σ1,2=σx+σy2±12(σxσy)2+4τxy2

  σ1,2=0.4990.569442±12(0.4990.56944)2+4×(0.1245)2

  σ1=0.094166MPa

And,

  σ2=0.1646MPa

The in plane maximum shear stress value is given by

  τ=|σ1σ22|

  τ=0.094166(0.1646)2τ=0.129383MPa

The state of stress on the element can be shown as follows

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card, Chapter 8, Problem 8.5.24P , additional homework tip  8

Conclusion: Thus, the maximum stress values on element C is τ=0.129383MPa .

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Chapter 8 Solutions

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 1 Term (6 Months) Printed Access Card

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