Problem 30
Question
If the boundary portion \(A B\) of a structure under plane stress is stress-free as illustrated, the stress vector acting on the portion \(A B\) must be zero. Express the condition that the components of the stress vector must vanish in terms of the stress components with respect to the coordinate axes and the angle \(\alpha\).
Step-by-Step Solution
Verified Answer
The zero-stress condition, in terms of the stress components with respect to the coordinate axes and the angle \(\alpha\), is expressed as \(\sigma_x \cos^2(\alpha) + \sigma_y \sin^2(\alpha) + \tau_{xy}\sin(\alpha)\cos(\alpha) = 0\).
1Step 1: Define the stress vector
The stress vector on a plane at angle \(\alpha\) can be given by \(\sigma_n = \sigma_x \cos^2(\alpha) + \sigma_y \sin^2(\alpha) + \tau_{xy}\sin(\alpha)\cos(\alpha)\) where \(\sigma_x\) and \(\sigma_y\) are the stress components, \(\tau_{xy}\) is the shear stress and \(\alpha\) is the angle.
2Step 2: Define the zero-stress condition
As given, the stress vector acting on the portion \(A B\) must be zero, thus \(\sigma_n = 0\)
3Step 3: Express the zero-stress condition with respect to given parameters
Setting \(\sigma_n = 0\), we get the equation: \(\sigma_x \cos^2(\alpha) + \sigma_y \sin^2(\alpha) + \tau_{xy}\sin(\alpha)\cos(\alpha) = 0\) which implies that the relation of stress components \(\sigma_x, \sigma_y, \tau_{xy}\) and the angle \(\alpha\) must satisfy this equation for the condition of zero-stress vector.
Key Concepts
Plane StressStress ComponentsShear Stress
Plane Stress
Plane stress occurs when a scenario involves stress components acting parallel to a plane while perpendicular stress components are negligible. This simplifies the analysis to a two-dimensional problem.
In structures like thin plates or surfaces, forces predominantly lie within the plane, making plane stress assumptions applicable.
By neglecting the out-of-plane stress, the mathematical complexity reduces.
This makes it easier for engineers and students to analyze stress distributions.
In structures like thin plates or surfaces, forces predominantly lie within the plane, making plane stress assumptions applicable.
By neglecting the out-of-plane stress, the mathematical complexity reduces.
This makes it easier for engineers and students to analyze stress distributions.
- Plane stress is prominent in thin, wide structures.
- It focuses on in-plane directional forces.
- Out-of-plane (normal) stress is negligible.
Stress Components
Stress components are essential in determining how forces affect a material.
These components define the intensity of the force per unit area acting along specific directions within the material.
In the context of plane stress, the key components include
Analyzing these stresses is crucial for predicting how a structure will deform or fail under given loads.
These components define the intensity of the force per unit area acting along specific directions within the material.
In the context of plane stress, the key components include
- \(\sigma_x\)
- \(\sigma_y\)
- \(\tau_{xy}\)
Analyzing these stresses is crucial for predicting how a structure will deform or fail under given loads.
- Normal stresses affect stretching or compression.
- Shear stress contributes to sliding between material layers.
Shear Stress
Shear stress is a type of stress that lies parallel to the surface on which forces act.
It often occurs alongside normal stresses in practical scenarios and significantly impacts how materials behave.
Unlike normal stress that stretches or compresses, shear stress tends to cause material layers to slide over one another.
In plane stress situations, shear stress, denoted as \(\tau_{xy}\), plays a pivotal role in understanding material deformation.
When solving problems like the given exercise, analyzing shear stress helps in formulating the balance of forces.
It often occurs alongside normal stresses in practical scenarios and significantly impacts how materials behave.
Unlike normal stress that stretches or compresses, shear stress tends to cause material layers to slide over one another.
In plane stress situations, shear stress, denoted as \(\tau_{xy}\), plays a pivotal role in understanding material deformation.
When solving problems like the given exercise, analyzing shear stress helps in formulating the balance of forces.
- Shear stress arises from forces parallel to the material surface.
- It affects the potential slipping between different layers.
- Critical for stability analysis in structures.
Other exercises in this chapter
Problem 21
At a point in a body the principal strains are. $$ \begin{aligned} \epsilon_{1} &=700 \times 10^{-6} \\ \epsilon_{11} &=300 \times 10^{-6} \\ \epsilon_{111} &=-
View solution Problem 22
The readings of a \(45^{\circ}\) strain rosette (Fig. \(\left.4.40(b)\right)\) are (a) $$ \begin{aligned} &\epsilon_{a}=100 \times 10^{-6} \\ &\epsilon_{L}=200
View solution Problem 33
Recall from Chapter 3 that the cquilibrium cquations for a slender beam \((3.11)\) and \((3.12)\) are $$ \frac{d V}{d x}+q=0 \quad \frac{d M_{b}}{d x}+V=0 $$ Sh
View solution Problem 20
In a state of plane strain in the \(x y\) plane the strain components associated with the \(x y\) axes are $$ \begin{aligned} \epsilon_{x} &=800 \times 10^{-6}
View solution