Essentials of Materials Science and Engineering, SI Edition
Essentials of Materials Science and Engineering, SI Edition
4th Edition
ISBN: 9781337672078
Author: ASKELAND, Donald R., WRIGHT, Wendelin J.
Publisher: Cengage Learning
Question
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Chapter 7, Problem 7.24P
Interpretation Introduction

(a)

Interpretation:

The value of bending stress for given condition should be calculated.

Concept introduction:

Stress is defined as the ratio of load per unit area. It is denoted by σ.

  σ=PA

Where,

P is the force in newton.

A is area in square meter.

Bending Stress:

Bending Stress is the amount of stress applied at a point of the body subjected to load results in bending. The formula for bending stress is shown below:

  σf=1.5FSt2×w

Where,

σf is the bending stress.

F is the load in newton.

S is the support span.

t is the thickness in mm.

w is the width in mm.

Standard Deviation:

Standard deviation is the value which defines the variation with respect to mean.

Calculation of value of n using the formula of standard deviation,

  σ= ( x i x ¯ )n

Where,

  σ= Population standard deviation

  xi is the value of variable from initial point to final point.

  x¯ = mean.

  n = number of variables.

MeanMean is defined as the ratio of the sum of total components with respect to the number of components. The formula for mean is given by,

  Mean=SumofallcomponentsNumberofcomponents

In mathematical terms the above equation is defined as,

  x¯=i=1n( x i + x 2 .............. x n )n

The term x¯ is defined as Mean.

Expert Solution
Check Mark

Answer to Problem 7.24P

The values of bending stress corresponding to load, thickness and width is shown below,

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  1

The requiredvalue of Standard Deviation is 554.663.

Explanation of Solution

Given information:

The formula for calculation:

  σf=1.5FSt2×w

Where,

σf is the bending stress.

F is the load in newton.

S is the support span.

t is the thickness in mm.

w is the width in mm.

The values of load correspond to length and thickness is shown in table,

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  2

The value of supporting span (S) is 20mm.

Based on given information,

Calculation of bending stress using spreadsheet using the formula of column,

  σf=1.5FSt2×w

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  3

Thus, the bending stress for the given condition of load is shown above.

Calculation of standard deviation using spreadsheet,

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  4

Thus, the required value of standard deviation is 554.663.

Interpretation Introduction

(b)

Interpretation:

Weibull plot using the given condition should be determined. Also, the value of Weibull modulus should be calculated.

Concept introduction:

Weibull plot is used to define the relationship between the fracture of probability with characteristic length.

The relationship is given below:

  σ0=ln(11F)

Where,

  σ0 is the characteristics strength.

F is the failure of probability.

The relation gives Failure of Probability,

  F=n-0.5N

Where,

N is the number of variable ranging from 1 to 25.

n is the value of specified strength ranging from lowest strength (1) to highest strength (25).

Expert Solution
Check Mark

Answer to Problem 7.24P

Thus, the required Weibull plot is shown below:

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  5

Thus, the required value of Weibull modulus is m=0.007463.

Explanation of Solution

Given information:

The formula for calculation:

  σ0=ln(11F)

Where,

  σ0 is the characteristics strength.

F is the failure of probability.

The relation gives Failure of Probability,

  F=n-0.5N

Where,

N is the number of variable ranging from 1 to 25.

n is the value of specified strength ranging from lowest strength (1) to highest strength (25).

The values of load correspond to length and thickness is shown in table,

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  6

The value of lowest strength (n) is 1.

The value of highest strength (n) is 25.

Based on given information,

Calculation of fracture probability using spreadsheet and the relationship used is given below,

  F=n-0.5N

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  7

Calculation of characteristics length using the data of probability of failure.

Formula used is given below:

  σ0=ln(11F)

Where,

  σ0 is the characteristics strength.

F is the failure of probability.

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  8

Based on the calculated values of characteristics length and probability of failure, the graph using linear regression is shown below:

  Essentials of Materials Science and Engineering, SI Edition, Chapter 7, Problem 7.24P , additional homework tip  9

Calculation of Weibull modulus,

Using the equation of straight line considering two points,

  m=y2y1x2x1

Where,

  x2=70x1=3y2=1y1=0.5

Substituting the values in the above formula,

  m=10.5703.m=0.007463

Thus, the required value of Weibull modulus is m=0.007463.

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