/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Two identical bars \(A B\) and \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two identical bars \(A B\) and \(B C\) support a vertical load \(P\) (see figure). The bars are made of steel having a stress-strain curve that may be idealized as elastoplastic with yield stress \(\sigma_{Y}\) Each bar has cross-sectional area \(A\) Determine the yield load \(P_{Y}\) and the plastic load \(P_{P^{-}}\)

Short Answer

Expert verified
The yield load \( P_Y = 2 \sigma_Y A \) and the plastic load \( P_{P^-} = 4 \sigma_Y A \).

Step by step solution

01

Understand Yield Load

The yield load \(P_Y\) is the maximum load the structure can support without permanent deformation. It occurs when the stress in the bar reaches the yield stress \( \sigma_Y \).
02

Apply Force on a Single Bar

Consider the force in one bar. When the bar yields, the stress \( \sigma = \frac{P}{2A} \) (since there are two bars) needs to be equal to the yield stress \( \sigma_Y \). Set this up as: \( \frac{P_Y}{2A} = \sigma_Y \).
03

Solve for Yield Load

From the equation \( \frac{P_Y}{2A} = \sigma_Y \), solve for \( P_Y \):\[ P_Y = 2 \sigma_Y A \]
04

Understand Plastic Load

The plastic load \(P_{P^-}\) occurs when the entire structure undergoes plastic deformation. For two identical bars, this is essentially when both bars are fully plastically deformed.
05

Determine Plastic Load

Since after yielding, the load continues to be supported on an elastoplastic basis, the maximum plastic force before complete failure is given by the sum of the loads supported by both bars. Because both bars are fully yielded but still supporting loads, the plastic capacity is twice the yield capacity of one bar: \[ P_{P^-} = 2P_Y = 4 \sigma_Y A \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Yield Load
The yield load is a crucial concept in understanding the behavior of materials under stress. When we talk about yield load, we're discussing the point at which a material, like steel, begins to deform permanently. Up to this load, any stress on the material can be removed without leaving a permanent bend or change. It's like stretching a rubber band - as long as you don't stretch it too far, it snaps back to its original shape.

Yield load is reached when the stress in the bars equals the yield stress of the material. For instance, in our problem with the steel bars, we calculate the yield load using the formula: \[P_Y = 2 \, \sigma_Y \, A\]Here, \(\sigma_Y\) is the yield stress, and \(A\) is the cross-sectional area of the bars. This formula helps us determine how much load each bar can handle before starting to change form permanently.

Understanding yield load is essential because it tells us the safe limit of loading for structures to avoid unexpected failure. It acts as a critical safety benchmark in engineering designs.
Plastic Load
After the yield load, materials enter a phase where they can undergo what's known as plastic deformation. This is when the material is permanently deformed, and it doesn't return to its original shape once the force is removed. The load at which this occurs in a complete manner for our structure is called the plastic load.

In the exercise, once both bars have yielded, they both continue to support a load, but in a completely deformed state. The plastic load is essentially the point when the structure as a whole is fully engaged in a plastic flow. For two identical bars, our equation becomes: \[P_{P^-} = 4 \, \sigma_Y \, A\]This shows that the plastic load is twice the yield load. When planning for construction or analyzing existing structures, understanding the plastic load helps us design systems that can endure beyond their yield point without failing immediately.

It's important to note that while structures can support loads even beyond the yield point up to their plastic load, this is not sustainable for long-term use. It gives warning of impending failure and should only be considered in exceptional circumstances.
Stress-Strain Curve
The stress-strain curve is a graphical representation that shows how a material reacts under loading. It plots stress on one axis and strain (or deformation) on the other. This curve is invaluable for engineers when designing materials and structures, as it highlights key points such as yield and maximum strength.

For steel, which is often conceptualized with an idealized elastoplastic behavior, the stress-strain curve starts with a linear region. In this portion, stress and strain are proportional, meaning the material behaves elastically like a perfect spring. This linear portion ends at the yield stress (\(\sigma_Y\)), beyond which the material behaves plastically.

Once the yield point is passed, the curve flattens out, indicating that the material will continue to deform under constant stress without a significant increase in load. This is the hallmark of plastic deformation.
  • The initial linear section represents the elastic region.
  • The plateau beyond this is where plastic deformation occurs.
  • Understanding this curve enables precise predictions about when and how a material will fail.
This curve not only depicts at what load materials like steel can pass from being elastic to plastic but also helps in understanding their full range of mechanical properties, letting engineers make informed decisions about the safe limits of structural materials.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A rigid bar of weight \(W=3560 \mathrm{N}\) hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is \(32 \mathrm{mm}\). Before they were loaded, all three wires had the same length. What temperature increase \(\Delta T\) in all three wires will result in the entire load being carried by the steel wires? (Assume \(E_{s}=205\) GPa, \(\alpha_{s}=12 \times 10^{-6 /^{\circ} \mathrm{C}},\) and \\[ \alpha_{a}=24 \times 10^{-6 / 9} \mathrm{C} \\]

A brass wire of diameter \(d=2.42 \mathrm{mm}\) is stretched tightly between rigid supports so that the tensile force is \(T=98 \mathrm{N}\) (see figure). The coefficient of thermal expansion for the wire is \(19.5 \times 10^{-6 / 8} \mathrm{C}\) and the modulus of elasticity is \(E=110 \mathrm{GPa}\) (a) What is the maximum permissible temperature drop \(\Delta T\) if the allowable shear stress in the wire is 60 MPa? (b) At what temperature change does the wire go slack?

A post \(A B\) supporting equipment in a laboratory is tapered uniformly throughout its height \(H\) (see figure). The cross sections of the post are square, with dimensions \(b \times b\) at the top and \(1.5 b \times 1.5 b\) at the base. Derive a formula for the shortening \(\delta\) of the post due to the compressive load \(P\) acting at the top. (Assume that the angle of taper is small and disregard the weight of the post itself.

A rectangular bar of length \(L\) has a slot in the middle half of its length (see figure). The bar has width \(b\) thickness \(t,\) and modulus of elasticity \(E .\) The slot has width \(b / 4\) (a) Obtain a formula for the elongation \(\delta\) of the bar due to the axial loads \(P\) (b) Calculate the elongation of the bar if the material is high-strength steel, the axial stress in the middle region is \(160 \mathrm{MPa}\), the length is \(750 \mathrm{mm}\), and the modulus of elasticity is \(210 \mathrm{GPa}\) (c) If the total elongation of the bar is limited to \(\delta_{\max }=0.475 \mathrm{mm},\) what is the maximum length of the slotted region? Assume that the axial stress in the middle region remains at \(160 \mathrm{MPa}\)

A small rubber ball (weight \(W=450 \mathrm{mN}\) ) is attached by a rubber cord to a wood paddle (see figure). The natural length of the cord is \(L_{0}=200 \mathrm{mm},\) its crosssectional area is \(A=1.6 \mathrm{mm}^{2},\) and its modulus of elasticity is \(E=2.0\) MPa. After being struck by the paddle, the ball stretches the cord to a total length \(L_{1}=900 \mathrm{mm}\) What was the velocity \(v\) of the ball when it left the paddle? (Assume linearly elastic behavior of the rubber cord, and disregard the potential energy due to any change in elevation of the ball.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.