Problem 3
Question
Imagine that the horizontal bar in the sketch is rigid and is constrained to remain horizontal as load \(P\) and displacement \(v\) increase. The three vertical bars are elasticperfectly plastic, with \(A=1, E=1\), and \(L=2\). Bars have the respective yieldpoint loads \(F_{1}=2, F_{2}=4\), and \(F_{3}=6\). Use three steps of a tangent stiffness method to generate the \(P\) versus \(v\) relation. Scale the result of each step so that one bar begins to yield at the beginning of the next step.
Step-by-Step Solution
Verified Answer
In each step of the tangent stiffness method, the load and displacement increase respectively as follows: In the first step, the load is 12 units and displacement is 8 units. In the second step, the load is 16 units and displacement is 10.67 units. In the third step, the load is 24 units and displacement is 16 units.
1Step 1: Determine load on each bar
Initially all bars are elastic and share the load in proportion to their stiffness. The stiffness of elastic perfectly plastic bar is \( \frac{AE}{L} = \frac{1 \times 1}{2} = 0.5 \). So, each bar will take a load equal to 0.5P. Since total load P is shared among 3 bars, \( P = F_{1} + F_{2} + F_{3} = 2 + 4 + 6 = 12 \) units.
2Step 2: Calculate displacement on each bar
The displacement \(v\) can be calculated using Hooke's law \( F = AE(v/L) \). Solving for \(v\), we have \(v = FL/AE = 12 \times 2/(3 \times 1 \times 1) = 8\) units.
3Step 3: Perform the tangent stiffness method
In the second step of tangent stiffness method, the load increases until the second bar yields. The load per bar could be calculated by \( P_{2} = F_{1} + 2 \times F_{2} + F_{3} = 2 + 2 \times 4 + 6 = 16 \) units and displacement would be \( v=FL/AE = 16 \times 2/(3 \times 1 \times 1) = 10.67\) units. In the third step, the load further increases until the third bar yields. The load per bar could be calculated by \( P_{3} = F_{1} + F_{2} + 3 \times F_{3} = 2 + 4 + 3 \times 6 = 24 \) units and displacement would be \( v=FL/AE = 24 \times 2/(3 \times 1 \times 1) = 16\) units.
Key Concepts
Finite Element AnalysisElastic Perfectly Plastic MaterialHooke's Law
Finite Element Analysis
Finite Element Analysis, often abbreviated as FEA, is a computational method used to predict how structures will respond to external forces, such as heat, vibration, or other physical effects.
FEA works by breaking down a real-life physical structure into a finite number of discrete elements, which are then analyzed to determine the distribution of stresses and deformations across the structure. This process requires solving many simultaneous equations, typically done using powerful computers.
Essentially, it's like making a complex puzzle out of a structure where each piece (element) affects its neighbors and is affected by them. By solving for each piece, we can predict how the entire structure will behave under certain conditions, allowing engineers and scientists to test designs virtually before physical prototypes are made.
FEA works by breaking down a real-life physical structure into a finite number of discrete elements, which are then analyzed to determine the distribution of stresses and deformations across the structure. This process requires solving many simultaneous equations, typically done using powerful computers.
Essentially, it's like making a complex puzzle out of a structure where each piece (element) affects its neighbors and is affected by them. By solving for each piece, we can predict how the entire structure will behave under certain conditions, allowing engineers and scientists to test designs virtually before physical prototypes are made.
Elastic Perfectly Plastic Material
An elastic perfectly plastic material is a simplified model used to describe the behavior of materials under load. It follows two basic phases:
This model is crucial for the tangent stiffness method, which involves calculating the response of structures with such materials. Knowing when a material yields allows engineers to design structures that maintain integrity under load without incurring unnecessary material costs.
- In the elastic phase, the material deforms under stress but returns to its original shape when the stress is removed, much like a spring. This behavior is governed by Hooke's law.
- In the perfectly plastic phase, when the stress exceeds a certain threshold (the yield point), the material deforms permanently, without any additional stress increase, even if the load increases.
This model is crucial for the tangent stiffness method, which involves calculating the response of structures with such materials. Knowing when a material yields allows engineers to design structures that maintain integrity under load without incurring unnecessary material costs.
Hooke's Law
Hooke's Law is a principle that states that the force needed to extend or compress a spring by some distance is proportional to that distance. In other words, if you pull on a spring twice as hard, it will stretch twice as much, up to the point of elastic limit.
In the context of materials, Hooke's Law is used to describe the initial linear relationship between stress (force per unit area) and strain (deformation) in materials. Mathematically, it's expressed as: \( F = kx \) where \( F \) is the force applied, \( k \) is the spring constant (stiffness), and \( x \) is the extension or compression of the spring (deformation).
This law is foundational in understanding material behavior and is used extensively in finite element analysis to predict how a structure will behave when subjected to forces during its elastic phase.
In the context of materials, Hooke's Law is used to describe the initial linear relationship between stress (force per unit area) and strain (deformation) in materials. Mathematically, it's expressed as: \( F = kx \) where \( F \) is the force applied, \( k \) is the spring constant (stiffness), and \( x \) is the extension or compression of the spring (deformation).
This law is foundational in understanding material behavior and is used extensively in finite element analysis to predict how a structure will behave when subjected to forces during its elastic phase.
Other exercises in this chapter
Problem 3
Describe steps of a tangent-stiffness solution algorithm in which each load incre. ment causes a single additional sampling point to be brought to the initiatio
View solution Problem 4
Consider an FE model of a plane structure. The material is isotropic, linearly elastic, and brittle: it cracks when tensile stress in any direction exceeds the
View solution Problem 4
The bar shown is of length \(L\) when unstressed, where \(L^{2}=a^{2}+c^{2}\). Let \(a \gg>c\). When load \(P\) is zero, so is displacement \(D\). The bar has a
View solution Problem 5
Assume that members of a truss carry only uniaxial stress, and that members in compression will buckle elastically at their critical loads without ever yielding
View solution