Chapter 29, Problem 11A

To determine

(a)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $0.943^{″}\pm 0.005^{″}$.

Explanation of Solution

Given information:

The unilateral tolerance is $0.938^{″}{}_{-0.000^{″}}^{+0.010^{″}}$.

Write the expression for the total tolerance.

  $T_t=U+L$   ...........(I)

Here, the upper limit is $U$ and the lower limit is $L$.

Write the expression for the mean dimension.

  $M_d=B_d+\frac{T_t}{2}$   ...........(II)

Here, the basic dimension is $B_d$.

Write the expression for the bilateral tolerance.

  $T_b=M_d\pm \frac{T_t}{2}$   ...........(III)

Calculation:

Substitute $0.010^{″}$ for $U$ and $-0.000^{″}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.010^{″}+\left(-0.000^{″}\right)\\ =0.010^{″}\end{array}$

Substitute $0.010^{″}$ for $T_t$ and $0.938^{″}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=0.938^{″}+\frac{0.010^{″}}{2}\\ =0.938^{″}+0.005^{″}\\ =0.943^{″}\end{array}$

Substitute $0.010^{″}$ for $T_t$ and $0.943^{″}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=0.943^{″}\pm \frac{0.010^{″}}{2}\\ =0.943^{″}\pm 0.005^{″}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $0.943^{″}\pm 0.005^{″}$.

To determine

(b)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $1.687^{″}\pm 0.001^{″}$.

Explanation of Solution

Given information:

The unilateral tolerance is $1.686{}_{-0.000^{″}}^{+0.002^{″}}$.

Calculation:

Substitute $0.002^{″}$ for $U$ and $-0.000^{″}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.002^{″}+\left(-0.000^{″}\right)\\ =0.002^{″}\end{array}$

Substitute $0.002^{″}$ for $T_t$ and $1.686^{″}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=1.686^{″}+\frac{0.002^{″}}{2}\\ =1.686^{″}+0.001^{″}\\ =1.687^{″}\end{array}$

Substitute $0.002^{″}$ for $T_t$ and $1.687^{″}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=1.687^{″}\pm \frac{0.002^{″}}{2}\\ =1.687^{″}\pm 0.001^{″}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $1.687^{″}\pm 0.001^{″}$.

To determine

(c)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $2.998^{″}\pm 0.002^{″}$.

Explanation of Solution

Given information:

The unilateral tolerance is $3.000^{″}{}_{-0.004^{″}}^{+0.000^{″}}$.

Calculation:

Substitute $0.000^{″}$ for $U$ and $-0.004^{″}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.000^{″}+\left(-0.004^{″}\right)\\ =-0.004^{″}\end{array}$

Substitute $-0.004^{″}$ for $T_t$ and $3.000^{″}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=3.000^{″}+\frac{-0.004^{″}}{2}\\ =3.000^{″}-0.002^{″}\\ =2.998^{″}\end{array}$

Substitute $-0.004^{″}$ for $T_t$ and $2.998^{″}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=2.998^{″}\pm \left(\frac{-0.004^{″}}{2}\right)\\ =2.998^{″}\pm 0.002^{″}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $2.998^{″}\pm 0.002^{″}$.

To determine

(d)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $0.069^{″}\pm 0.004^{″}$.

Explanation of Solution

Given information:

The unilateral tolerance is $0.073^{″}{}_{-0.008^{″}}^{+0.000^{″}}$.

Calculation:

Substitute $0.000^{″}$ for $U$ and $-0.008^{″}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.000^{″}+\left(-0.008^{″}\right)\\ =-0.008^{″}\end{array}$

Substitute $-0.008^{″}$ for $T_t$ and $0.073^{″}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=0.073^{″}+\frac{-0.008^{″}}{2}\\ =0.073^{″}-0.004^{″}\\ =0.069^{″}\end{array}$

Substitute $-0.008^{″}$ for $T_t$ and $0.069^{″}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=0.069^{″}\pm \left(\frac{-0.008^{″}}{2}\right)\\ =0.069^{″}\pm 0.004^{″}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $0.069^{″}\pm 0.004^{″}$.

To determine

(e)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $4.1880^{″}\pm 0.007^{″}$.

Explanation of Solution

Given information:

The unilateral tolerance is $4.1873^{″}{}_{-0.000^{″}}^{+0.0014^{″}}$.

Calculation:

Substitute $0.0014^{″}$ for $U$ and $-0.000^{″}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.0014^{″}+\left(-0.000^{″}\right)\\ =0.0014^{″}\end{array}$

Substitute $0.0014^{″}$ for $T_t$ and $4.1873^{″}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=4.1873^{″}+\frac{0.0014^{″}}{2}\\ =4.1873^{″}+0.007^{″}\\ =4.1880^{″}\end{array}$

Substitute $0.0014^{″}$ for $T_t$ and $4.1880^{″}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=4.1880^{″}\pm \frac{0.0014^{″}}{2}\\ =4.1880^{″}\pm 0.007^{″}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $4.1880^{″}\pm 0.007^{″}$.

To determine

(f)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $0.9984^{″}\pm 0.0037^{″}$.

Explanation of Solution

Given information:

The unilateral tolerance is $1.0021{}_{-0.0074^{″}}^{+0.000^{″}}$.

Calculation:

Substitute $0.000^{″}$ for $U$ and $-0.0074^{″}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.000^{″}+\left(-0.0074^{″}\right)\\ =-0.0074^{″}\end{array}$

Substitute $-0.0074^{″}$ for $T_t$ and $1.0021^{″}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=1.0021^{″}+\left(\frac{-0.0074^{″}}{2}\right)\\ =1.0021^{″}-0.0037^{″}\\ =0.9984^{″}\end{array}$

Substitute $-0.0074^{″}$ for $T_t$ and $0.9984^{″}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=0.9984^{″}\pm \frac{-0.0074^{″}}{2}\\ =0.9984^{″}\pm 0.0037^{″}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $0.9984^{″}\pm 0.0037^{″}$.

To determine

(g)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $1.0006^{″}\pm 0.0004^{″}$.

Explanation of Solution

Given information:

The unilateral tolerance is $1.0010^{″}{}_{-0.0008^{″}}^{+0.000^{″}}$.

Calculation:

Substitute $0.000^{″}$ for $U$ and $-0.0008^{″}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.000^{″}+\left(-0.0008^{″}\right)\\ =-0.0008^{″}\end{array}$

Substitute $-0.0008^{″}$ for $T_t$ and $1.0010^{″}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=1.0010^{″}+\left(\frac{-0.0008^{″}}{2}\right)\\ =1.0010^{″}-0.0004^{″}\\ =1.0006^{″}\end{array}$

Substitute $-0.0008^{″}$ for $T_t$ and $1.0006^{″}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=1.0006^{″}\pm \frac{-0.0008^{″}}{2}\\ =1.0006^{″}\pm 0.0004^{″}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $1.0006^{″}\pm 0.0004^{″}$.

To determine

(h)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $8.4660^{″}\pm 0.0011^{″}$.

Explanation of Solution

Given information:

The unilateral tolerance is $8.4649{}_{-0.000^{″}}^{+0.0022^{″}}$.

Calculation:

Substitute $0.0022^{″}$ for $U$ and $-0.000^{″}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.0022^{″}+\left(-0.000^{″}\right)\\ =0.0022^{″}\end{array}$

Substitute $0.0022^{″}$ for $T_t$ and $8.4649^{″}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=8.4649^{″}+\frac{0.0022^{″}}{2}\\ =8.4649^{″}+0.0011^{″}\\ =8.4660^{″}\end{array}$

Substitute $0.0022^{″}$ for $T_t$ and $8.4660^{″}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=8.4660^{″}\pm \frac{0.0022^{″}}{2}\\ =8.4660^{″}\pm 0.0011^{″}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $8.4660^{″}\pm 0.0011^{″}$.

To determine

(i)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $44.31\text{mm}\pm 0.01\text{mm}$.

Explanation of Solution

Given information:

The unilateral tolerance is $44.30{\text{mm}}_{-0.00\text{mm}}^{+0.02\text{mm}}$.

Calculation:

Substitute $0.02\text{mm}$ for $U$ and $-0.00\text{mm}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.02\text{mm}+\left(-0.00\text{mm}\right)\\ =0.02\text{mm}\end{array}$

Substitute $0.02\text{mm}$ for $T_t$ and $44.30\text{mm}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=44.30\text{mm}+\frac{0.02\text{mm}}{2}\\ =44.30\text{mm}+0.01\text{mm}\\ =44.31\text{mm}\end{array}$

Substitute $0.010^{″}$ for $T_t$ and $44.31\text{mm}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=44.31\text{mm}\pm \frac{0.02\text{mm}}{2}\\ =44.31\text{mm}\pm 0.01\text{mm}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $44.31\text{mm}\pm 0.01\text{mm}$.

To determine

(j)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $10.02\text{mm}\pm 0.04\text{mm}$.

Explanation of Solution

Given information:

The unilateral tolerance is $10.06{\text{mm}}_{-0.08\text{mm}}^{+0.00\text{mm}}$.

Calculation:

Substitute $0.00\text{mm}$ for $U$ and $-0.08\text{mm}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.00\text{mm}+\left(-0.08\text{mm}\right)\\ =-0.08\text{mm}\end{array}$

Substitute $-0.08\text{mm}$ for $T_t$ and $10.06\text{mm}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=10.06\text{mm}+\left(\frac{-0.08\text{mm}}{2}\right)\\ =10.06\text{mm}-0.04\text{mm}\\ =10.02\text{mm}\end{array}$

Substitute $-0.08\text{mm}$ for $T_t$ and $10.02\text{mm}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=10.02\text{mm}\pm \left(\frac{-0.08\text{mm}}{2}\right)\\ =10.02\text{mm}\pm 0.04\text{mm}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $10.02\text{mm}\pm 0.04\text{mm}$.

To determine

(k)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $64.92\text{mm}\pm 0.03\text{mm}$.

Explanation of Solution

Given information:

The unilateral tolerance is $64.89{\text{mm}}_{-0.00\text{mm}}^{+0.06\text{mm}}$.

Calculation:

Substitute $0.06\text{mm}$ for $U$ and $-0.00\text{mm}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.06\text{mm}+\left(-0.00\text{mm}\right)\\ =0.06\text{mm}\end{array}$

Substitute $0.06\text{mm}$ for $T_t$ and $64.89\text{mm}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=64.89\text{mm}+\frac{0.06\text{mm}}{2}\\ =64.89\text{mm}+0.03\text{mm}\\ =64.92\text{mm}\end{array}$

Substitute $0.06\text{mm}$ for $T_t$ and $64.92\text{mm}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=64.92\text{mm}\pm \frac{0.06\text{mm}}{2}\\ =64.92\text{mm}\pm 0.03\text{mm}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $64.92\text{mm}\pm 0.03\text{mm}$.

To determine

(l)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $38.016\text{mm}\pm 0.028\text{mm}$.

Explanation of Solution

Given information:

The unilateral tolerance is $37.988{\text{mm}}_{-0.00\text{mm}}^{+0.056\text{mm}}$.

Calculation:

Substitute $0.056\text{mm}$ for $U$ and $-0.00\text{mm}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.056\text{mm}+\left(-0.00\text{mm}\right)\\ =0.056\text{mm}\end{array}$

Substitute $0.056\text{mm}$ for $T_t$ and $37.988\text{mm}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=37.988\text{mm}+\frac{0.056\text{mm}}{2}\\ =37.988\text{mm}+0.028\text{mm}\\ =38.016\text{mm}\end{array}$

Substitute $0.056\text{mm}$ for $T_t$ and $38.016\text{mm}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=38.016\text{mm}\pm \frac{0.056\text{mm}}{2}\\ =38.016\text{mm}\pm 0.028\text{mm}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $38.016\text{mm}\pm 0.028\text{mm}$.

To determine

(m)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $124.9915\text{mm}\pm 0.085\text{mm}$.

Explanation of Solution

Given information:

The unilateral tolerance is $125.00{\text{mm}}_{-0.017\text{mm}}^{+0.00\text{mm}}$.

Calculation:

Substitute $0.00\text{mm}$ for $U$ and $-0.017\text{mm}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.00\text{mm}+\left(-0.017\text{mm}\right)\\ =-0.017\text{mm}\end{array}$

Substitute $-0.017\text{mm}$ for $T_t$ and $125.00\text{mm}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=125.00\text{mm}+\left(\frac{-0.017\text{mm}}{2}\right)\\ =125.00\text{mm}-0.085\text{mm}\\ =124.9915\text{mm}\end{array}$

Substitute $-0.017\text{mm}$ for $T_t$ and $124.9915\text{mm}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=124.9915\text{mm}\pm \left(\frac{-0.017\text{mm}}{2}\right)\\ =124.9915\text{mm}\pm 0.085\text{mm}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $124.9915\text{mm}\pm 0.085\text{mm}$.

To determine

(n)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $43.078\text{mm}\pm 0.013\text{mm}$.

Explanation of Solution

Given information:

The unilateral tolerance is $43.091{\text{mm}}_{-0.026\text{mm}}^{+0.00\text{mm}}$.

Calculation:

Substitute $0.00\text{mm}$ for $U$ and $-0.026\text{mm}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.00\text{mm}+\left(-0.026\text{mm}\right)\\ =-0.026\text{mm}\end{array}$

Substitute $-0.026\text{mm}$ for $T_t$ and $43.091\text{mm}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=43.091\text{mm}+\left(\frac{-0.026\text{mm}}{2}\right)\\ =43.091\text{mm}-0.013\text{mm}\\ =43.078\text{mm}\end{array}$

Substitute $-0.026\text{mm}$ for $T_t$ and $43.078\text{mm}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=43.078\text{mm}\pm \left(\frac{-0.026\text{mm}}{2}\right)\\ =43.078\text{mm}\pm 0.013\text{mm}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $43.078\text{mm}\pm 0.013\text{mm}$.

To determine

(o)

The dimension in the bilateral tolerance.

Answer to Problem 11A

The dimension in the bilateral tolerance is $98.8835\text{mm}\pm 0.0045\text{mm}$.

Explanation of Solution

Given information:

The unilateral tolerance is $98.879{\text{mm}}_{-0.00\text{mm}}^{+0.009\text{mm}}$.

Calculation:

Substitute $0.009\text{mm}$ for $U$ and $-0.00\text{mm}$ for $L$ in Equation (I).

  $\begin{array}{c}{T}_t=0.009\text{mm}+\left(-0.00\text{mm}\right)\\ =0.009\text{mm}\end{array}$

Substitute $0.009\text{mm}$ for $T_t$ and $98.879\text{mm}$ for $B_d$ in Equation (II).

  $\begin{array}{c}{M}_d=98.879\text{mm}+\frac{0.009\text{mm}}{2}\\ =98.879\text{mm}+0.0045\text{mm}\\ =98.8835\text{mm}\end{array}$

Substitute $0.009\text{mm}$ for $T_t$ and $98.8835\text{mm}$ for $M_d$ in Equation (III).

  $\begin{array}{c}{T}_b=98.8835\text{mm}\pm \frac{0.009\text{mm}}{2}\\ =98.8835\text{mm}\pm 0.0045\text{mm}\end{array}$

Conclusion:

The dimension in the bilateral tolerance is $98.8835\text{mm}\pm 0.0045\text{mm}$.