Concept explainers
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.
(a)
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
Compressive stress
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
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
The moment in the vertical beam =
The moment in the horizontal beam due to load Pz=
Total moment
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 =
From the properties of the cross-section, the moment of inertia is
Now, plane stress can be calculated as follows
And maximum compressive stress is
Conclusion: Thus, the state of stress on element C are
Tensile stress
Compressive stress
(b)
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,
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
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
The moment in the vertical beam =
The moment in the horizontal beam due to load Pz=
Total moment
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 =
From the properties of the cross-section, the moment of inertia is
Now, plane stress can be calculated as follows
And maximum compressive stress is
The shear stress on the element is given by the torque and shear stress relation.
The normal stresses are given by
And,
The state of stress on the element can be shown as follows
Conclusion: Thus, the normal stress values on element C are,
(c)
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
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
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
The moment in the vertical beam =
The moment in the horizontal beam due to load Pz=
Total moment
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 =
From the properties of the cross-section, the moment of inertia is
Now, plane stress can be calculated as follows
And maximum compressive stress is
The shear stress on the element is given by the torque and shear stress relation.
The normal stresses are given by
And,
The in plane maximum shear stress value is given by
The state of stress on the element can be shown as follows
Conclusion: Thus, the maximum stress values on element C is
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Chapter 8 Solutions
Mechanics of Materials (MindTap Course List)
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