Chapter 20, Problem 45A

It is sometimes impractical to make engineering drawings full size. If the part to be drawn is very large or small a scale drawing is generally made. The scale, which is shown on drawing, compares the lengths of the lines on the drawing to the dimensions on the part. A scale on a drawing that states $\frac{1\text{'}\text{'}}{4}=1\text{'}\text{'}$means the drawing is one-quarter the size of the part. It is expressed as a ratio of 1:4 or $\frac{1}{4}$ . A scale drawing that states $2\text{'}\text{'}=1\text{'}\text{'}$means that drawing is double the size of the part. It is expressed as a ratio of 2:1 or $\frac{2}{1}$ . The actual dimensions of a steel support are given in Figure 20-11. All dimensions are in inches. Using the proportion given, compute the lengths for each unknown in the table. Round the answers to 3 decimal places where necessary.

   $\frac{\text{Numerator of Scale Ratio}}{\text{Denominator Of Scale Ratio}}=\frac{\text{Drawing Length}}{\text{Part Dimension}}$

To determine

(a)

Drawing length for (a).

Answer to Problem 45A

Drawing length for B is $8.031^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Concept used:

Relation of drawing length is given as follows:

  $\frac{\text{Drawing}\text{length}}{\text{Part}\text{}\text{dimension}}=\frac{\text{Number of scale ratio }}{\text{Denominator of scale ratio}}$ ........ (1)

Calculation:

For the length B:

Substitute $\frac{1}{2}$ for $\text{Number of scale ratio}$ and $16.062^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{16.062^{″}}=\frac{\frac{1}{2}}{1}\\ \text{Drawing}\text{length}=8.031^{″}\end{array}$

Thus, drawing length for B is $8.031^{″}$.

Conclusion:

Drawing length for B is $8.031^{″}$.

To determine

(b)

Drawing length for (b).

Answer to Problem 45A

Drawing length for G is $1.124^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length G:

Substitute $4$ for $\text{Number of scale ratio}$ and $0.281^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{0.281^{″}}=\frac{4}{1}\\ \text{Drawing}\text{length}=1.124^{″}\end{array}$

Thus, drawing length for G is $1.124^{″}$.

Conclusion:

Drawing length for G is $1.124^{″}$.

To determine

(c)

Drawing length for (c).

Answer to Problem 45A

Drawing length for B is $4.0155^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length B:

Substitute $\frac{1}{4}$ for $\text{Number of scale ratio}$ and $16.062^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{16.062^{″}}=\frac{\frac{1}{4}}{1}\\ \text{Drawing}\text{length}=4.0155^{″}\end{array}$

Thus, drawing length for B is $4.0155^{″}$.

Conclusion:

Drawing length for B is $4.0155^{″}$.

To determine

(d)

Drawing length for (d).

Answer to Problem 45A

Drawing length for C is $2.72^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length C:

Substitute $2$ for $\text{Number of scale ratio}$ and $1.36^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{1.36^{″}}=\frac{2}{1}\\ \text{Drawing}\text{length}=2.72^{″}\end{array}$

Thus, drawing length for C is $2.72^{″}$.

Conclusion:

Drawing length for C is $2.72^{″}$.

To determine

(e)

Drawing length for (e).

Answer to Problem 45A

Drawing length for A is $2.808^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length A:

Substitute $1\frac{1}{2}$ for $\text{Number of scale ratio}$ and $1.872^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{1.872^{″}}=\frac{1\frac{1}{2}}{1}\\ \text{Drawing}\text{length}=2.808^{″}\end{array}$

Thus, drawing length for A is $2.808^{″}$.

Conclusion:

Drawing length for A is $2.808^{″}$.

To determine

(f)

Drawing length for (f).

Answer to Problem 45A

Drawing length for E is $7.95^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length E:

Substitute $\frac{3}{4}$ for $\text{Number of scale ratio}$ and $10.6^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{10.6^{″}}=\frac{\frac{3}{4}}{1}\\ \text{Drawing}\text{length}=7.95^{″}\end{array}$

Thus, drawing length for E is $7.95^{″}$.

Conclusion:

Drawing length for E is $7.95^{″}$.

To determine

(g)

Drawing length for (g).

Answer to Problem 45A

Drawing length for H is $1.125^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length H:

Substitute $3$ for $\text{Number of scale ratio}$ and $0.375^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{0.375^{″}}=\frac{3}{1}\\ \text{Drawing}\text{length}=1.125^{″}\end{array}$

Thus, drawing length for H is $1.125^{″}$.

Conclusion:

Drawing length for H is $1.125^{″}$.

To determine

(h)

Drawing length for (h).

Answer to Problem 45A

Drawing length for F is $1.575^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length F:

Substitute $\frac{1}{8}$ for $\text{Number of scale ratio}$ and $12.60^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{12.60^{″}}=\frac{\frac{1}{8}}{1}\\ \text{Drawing}\text{length}=1.575^{″}\end{array}$

Thus, drawing length for F is $1.575^{″}$.

Conclusion:

Drawing length for F is $1.575^{″}$.

To determine

(i)

Drawing length for (i).

Answer to Problem 45A

Drawing length for E is $5.30^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length E:

Substitute $\frac{1}{2}$ for $\text{Number of scale ratio}$ and $10.60^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{10.60^{″}}=\frac{\frac{1}{2}}{1}\\ \text{Drawing}\text{length}=5.30^{″}\end{array}$

Thus, drawing length for E is $5.30^{″}$.

Conclusion:

Drawing length for E is $5.30^{″}$.

To determine

(j)

Drawing length for (j).

Answer to Problem 45A

Drawing length for G is $1.686^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length G:

Substitute $6$ for $\text{Number of scale ratio}$ and $0.281^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{0.281^{″}}=\frac{6}{1}\\ \text{Drawing}\text{length}=1.686^{″}\end{array}$

Thus, drawing length for G is $1.686^{″}$.

Conclusion:

Drawing length for G is $1.686^{″}$.

To determine

(k)

Drawing length for (k).

Answer to Problem 45A

Drawing length for F is $9.45^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length F:

Substitute $\frac{3}{4}$ for $\text{Number of scale ratio}$ and $12.6^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{12.6^{″}}=\frac{\frac{3}{4}}{1}\\ \text{Drawing}\text{length}=9.45^{″}\end{array}$

Thus, drawing length for F is $9.45^{″}$.

Conclusion:

Drawing length for F is $9.45^{″}$.

To determine

(l)

Drawing length for (l).

Answer to Problem 45A

Drawing length for C is $2.04^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length C:

Substitute $1\frac{1}{2}$ for $\text{Number of scale ratio}$ and $1.36^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{1.36^{″}}=\frac{1\frac{1}{2}}{1}\\ \text{Drawing}\text{length}=2.04^{″}\end{array}$

Thus, drawing length for C is $2.04^{″}$.

Conclusion:

Drawing length for C is $2.04^{″}$.

To determine

(m)

Drawing length for (m).

Answer to Problem 45A

Drawing length for F is $6.30^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length F:

Substitute $\frac{1}{2}$ for $\text{Number of scale ratio}$ and $12.6^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{12.6^{″}}=\frac{\frac{1}{2}}{1}\\ \text{Drawing}\text{length}=6.30^{″}\end{array}$

Thus, drawing length for F is $6.30^{″}$.

Conclusion:

Drawing length for F is $6.30^{″}$.

To determine

(n)

Drawing length for (n).

Answer to Problem 45A

Drawing length for G is $0.843^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length G:

Substitute $3$ for $\text{Number of scale ratio}$ and $0.281^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{0.281^{″}}=\frac{3}{1}\\ \text{Drawing}\text{length}=0.843^{″}\end{array}$

Thus, drawing length for G is $0.843^{″}$.

Conclusion:

Drawing length for G is $0.843^{″}$.

To determine

(o)

Drawing length for (o).

Answer to Problem 45A

Drawing length for B is $4.0155^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length B:

Substitute $\frac{1}{4}$ for $\text{Number of scale ratio}$ and $16.062^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{16.062^{″}}=\frac{\frac{1}{4}}{1}\\ \text{Drawing}\text{length}=4.0155^{″}\end{array}$

Thus, drawing length for B is $4.0155^{″}$.

Conclusion:

Drawing length for B is $4.0155^{″}$.

To determine

(p)

Drawing length for (p).

Answer to Problem 45A

Drawing length for A is $3.744^{″}$.

Explanation of Solution

Given:

Part with all dimensions is given as follows:

Calculation:

For the length A:

Substitute $2$ for $\text{Number of scale ratio}$ and $1.872^{″}$ for $\text{Part}\text{}\text{dimension}$ in equation (1).

  $\begin{array}{l}\frac{\text{Drawing}\text{length}}{1.872^{″}}=\frac{2}{1}\\ \text{Drawing}\text{length}=3.744^{″}\end{array}$

Thus, drawing length for A is $3.744^{″}$.

Conclusion:

Drawing length for A is $3.744^{″}$.