/*! 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 55 A material that has a fracture t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 55

A material that has a fracture toughness of \(33 \mathrm{MPa}-\mathrm{m}^{0.5}\) is to be made into a large panel that is \(2000 \mathrm{~mm}\) long by \(250 \mathrm{~mm}\) wide and \(4 \mathrm{~mm}\) thick. If the minimum allowable total crack length is \(4 \mathrm{~mm}\), what is the maximum tensile load in the long direction that can be applied without catastrophic failure with a safety factor of \(2.5\) ?

Short Answer

Expert verified
The maximum tensile load (F) that can be applied to the panel without catastrophic failure, with a safety factor of 2.5, can be found by following the above steps.

Step by step solution

01

Rearrange the Fracture Toughness Formula

We start by rearranging the formula for fracture toughness in order to solve for the stress. The rearranged formula becomes: \(\sigma = \frac{K_Ic}{Y\sqrt{\pi a}}\).
02

Substituting Given Values

Next, substitute all the given values into the rearranged formula: \(\sigma = \frac{33 MPa\sqrt{m}}{\sqrt{\pi 脳 4 mm}}\). Remember, for the dimensionless shape factor, we use the value \(Y = 1\).
03

Calculate the Stress

Evaluate the expression to solve for \(\sigma\). This will give us the maximum tolerable stress that the panel can withstand without catastrophic failure.
04

Adjust for Safety Factor

As a safety factor of 2.5 is given, we need to divide the stress obtained in the previous step by this safety factor to get the maximum stress that can be applied.
05

Calculate the Maximum Tensile Load

The tensile load (F) is given by the equation \(F=\sigma A\), where \(A\) is the cross-sectional area (i.e., width 脳 thickness of the panel). By substituting the values, we can calculate the maximum tensile load.

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 91影视!

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 differential element is subjected to the stresses (in \(\mathrm{kpsi}\) ): \(\sigma_{x}=10, \sigma_{y}=-5\), \(\tau_{x y}=15\). The material is uneven and has strengths (in kpsi) of \(S_{u t}=50, S_{y}=40\), and \(S_{u c}=90\). Calculate the safety factor and draw a \(\sigma_{a}-\sigma_{b}\) diagram showing the boundary for each theory with the stress state and load line using: (a) Coulomb-Mohr theory, and (b) Modified Mohr theory.

A differential element is subjected to the stresses (in MPa): \(\sigma_{1}=70, \sigma_{2}=0, \sigma_{3}=-140\). A ductile material has the strengths (in MPA): \(S_{u t}=350, S_{y}=280, S_{u c}=350\). Calculate the safety factor and draw \(\sigma_{1}-\sigma_{3}\) diagrams of each theory showing the stress state using: (a) Maximum shear stress theory (b) Distortion energy theory

A differential element is subjected to the stresses (in MPa): \(\sigma_{1}=70, \sigma_{2}=0, \sigma_{3}=-140\). A brittle material has the strengths (in MPA): \(S_{u t}=350, S_{u c}=630\). Calculate the safety factor and draw \(\sigma_{1}-\sigma_{3}\) diagrams of each theory showing the stress state using: (a) Coulomb-Mohr theory (b) Modified-Mohr theory

A 10-mm-ID steel tube carries liquid at \(7 \mathrm{MPa}\). The steel has \(S_{y}=400 \mathrm{MPa}\). Determine the safety factor for the wall if its thickness is: (a) \(1 \mathrm{~mm}\), (b) \(5 \mathrm{~mm}\).

A component in the shape of a large sheet is to be fabricated from 4340 steel, which has a fracture toughness \(\mathrm{K}_{\mathrm{c}}=98.9 \mathrm{MPa}-\mathrm{m}^{0.5}\) and a tensile yield strength of \(860 \mathrm{MPa}\). The sheets are inspected for crack flaws after fabrication, but the inspection device cannot detect flaws smaller than \(3 \mathrm{~mm}\). The part is too heavy as designed. An engineer has suggested that the thickness be reduced and the material be heat-treated to increase its tensile strength to \(1515 \mathrm{MPa}\), which would result in decreasing the fracture toughness to \(60.4 \mathrm{MPa}-\mathrm{m}^{0.5}\). Assuming that the stress level does not exceed one half the yield strength, is the suggestion feasible? If not, why not?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Turning Points in Physics

Read Explanation

Dynamics

Read Explanation

Radiation

Read Explanation

Torque and Rotational Motion

Read Explanation

Oscillations

Read Explanation
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.