Chapter 20, Problem 27P

The drawing shown is of a mounting fixture to locate and orient a rod (not shown) through the large bore. The fixture will be bolted to a frame through the four bolt holes that are countersunk to recess the bolt heads. The bolt holes have too much clearance to properly align the rod, so the fixture will be aligned with two locating pins in the frame that will fit in the Ø6 hole and slot.

(a)    Determine the minimum diameter allowed for the countersink.

(b)    Determine the maximum depth allowed for the countersink.

(c)    Determine the diameter of the bolt holes at MMC.

(d)    Identify every feature that qualifies as a feature of size.

(e)    The width of the base is specified with a basic dimension of 60, with no tolerance. (Note that as a feature of size, it could have had a tolerance directly specified.) What are the minimum and maximum allowed dimensions for the base width? Explain how they are determined.

Problem 20–27

(f)    Describe the datum features A, B, and C. Describe their corresponding datums. Describe the datum reference frame that is defined by applying A, B, and C in that order. Describe how the part is stabilized by these datums. Explain why this is more appropriate for this application than using the edges of the base for datums B and C. (Notice that the basic dimensions are either measured from, or implied to be centered on, the datums of the datum reference frame.)

(g)    If datum feature B is produced with a diameter of Ø6.00, what is the diameter of the tolerance zone in which its axis must lie? What if it is produced at Ø6.05?

(h)    If the bolt holes are produced at Ø6.0, what is the diameter of the tolerance zones locating the bolt hole pattern with respect to the true position specified by the basic dimensions? What if the bolt holes are produced at Ø6.1?

(i)    If the bolt holes are produced at Ø6.0, what is the diameter of the tolerance zones locating the position of the bolt holes with respect to one another? What if the bolt holes are produced at Ø6.1?

(j)    Explain why the Ⓜ modifier is appropriate for the bolt hole position tolerance.

(k)    For the large bore, explain what provides control of each of the following: orientation, straightness of its center axis, and cylindricity of its surface.

(l)    Assume the part is cast, and the casting operation can provide a surface profile tolerance of less than 0.5. Which surfaces can likely be left in the as-cast condition without compromising any of the requirements of the drawing? How would this change if the drawing were modified to use the edges of the base as datum features B and C, while still maintaining the functional goals for the alignment of the rod?

To determine

The minimum diameter allowed for the countersink.

Answer to Problem 27P

The minimum diameter allowed for the countersink is $11.8\text{unit}$.

Explanation of Solution

Write the expression for the minimum diameter allowed for the countersink.

    $d_{\min }=d-T_n$                                                                  (I)

Here, the nominal diameter is $d$ and the nominal tolerance is $T_n$.

Conclusion:

The countersink diameter is specified as $12\pm 0.2$.

Substitute $12\text{unit}$ for $d$ and $0.2\text{unit}$ for $T_n$ in Equation (I).

    $\begin{array}{c}{d}_{\min }=12\text{unit}-0.2\text{unit}\\ =11.8\text{unit}\end{array}$

Thus, the minimum diameter allowed for the countersink is $11.8\text{unit}$.

To determine

The maximum countersink depth.

Answer to Problem 27P

The maximum countersink depth is $3.1\text{unit}$.

Explanation of Solution

Write the expression for the maximum countersink depth.

    $D_{\max }=D+T_n$                                                            (II)

Here, the nominal tolerance for counter sink depth is $T_n$ and the nominal depth is $D$.

Conclusion:

The countersink depth is specified as $3\pm 0.1$.

Substitute $3\text{unit}$ for $D$ and $0.1\text{unit}$ for $T_n$ in Equation (II).

    $\begin{array}{c}{D}_{\max }=3\text{unit}+0.1\text{unit}\\ =3.1\text{unit}\end{array}$

Thus, the maximum countersink depth is $3.1\text{unit}$.

To determine

The diameter of the bolt holes at MMC.

Answer to Problem 27P

The diameter of the bolt holes at MMC is $6\text{unit}$.

Explanation of Solution

Write the expression for the diameter of the bolt holes at MMC.

    $d_{bh}=d_b-T_l$                                                                         (III)

Here, the nominal diameter of the hole is $d_b$ and the lower tolerance is $T_l$.

Conclusion:

The bolt holes at MMC is specified as $6_{-0.0}^{+0.1}$.

Substitute $6\text{unit}$ for $d_b$ and $0.0\text{unit}$ for $T_l$ in Equation (III).

    $\begin{array}{c}{d}_{bh}=6\text{unit}-0.0\text{unit}\\ =6\text{unit}\end{array}$

Thus, the diameter of the bolt holes at MMC is $6\text{unit}$.

To determine

The feature that qualifies as a feature of size.

Answer to Problem 27P

The feature that qualifies as a feature of size are:

• The maximum diameter of the bolt is $6.01\text{units}$.
• The minimum diameter of the bolt is $6\text{units}$.
• The maximum thickness of the clamping part is $8.01\text{units}$.
• The minimum thickness of the clamping part is $7.9\text{units}$.

Explanation of Solution

The term “feature of size” refers to a feature that has a size that can be measured across two opposing points, such as a hole, diameter, or slot.

The features of size are hole diameters and counter bore diameter, the slot width, the base widths, the base thickness, and the centre protrusion width.

Conclusion:

Refer to Figure 20-27 to obtain the feature of size as:

• The maximum diameter of the bolt is $6.01\text{units}$.
• The minimum diameter of the bolt is $6\text{units}$.
• The maximum thickness of the clamping part is $8.01\text{units}$.
• The minimum thickness of the clamping part is $7.9\text{units}$.

To determine

The minimum allowed dimension for the base width.

The maximum allowed dimension for the base width.

Answer to Problem 27P

The minimum allowed dimension for the base width is $59.5\text{units}$.

The maximum allowed dimension for the base width is $60.5\text{units}$.

Explanation of Solution

Write the expression for the minimum allowed dimension for the base width.

    $w_{\min }=w_b-T_l$                                                                                      (IV)

Here, the base width is $w_b$ and the lower tolerance is $T_l$.

Write the expression for the maximum allowed dimension for the base width.

    $w_{\max }=w_b+T_u$                                                                                     (V)

Here, the upper tolerance is $T_u$.

Conclusion:

The default surface profile tolerance in the note provide a tolerance zone around all the surfaces of $0.5$. This provides the tolerance of $0.25$ outside the base dimension on each edge, for a total of $0.5$.

Substitute $60\text{units}$ for $w_b$ and $0.5\text{unit}$ for $T_l$ in Equation (IV).

    $\begin{array}{c}{w}_{\min }=60\text{units}-0.5\text{units}\\ =59.5\text{units}\end{array}$

Thus, the minimum allowed dimension for the base width is $59.5\text{units}$.

Substitute $60\text{units}$ for $w_b$ and $0.5\text{unit}$ for $T_u$ in Equation (V).

    $\begin{array}{c}{w}_{\max }=60\text{units}+0.5\text{units}\\ =60.5\text{units}\end{array}$

Thus, the maximum allowed dimension for the base width is $60.5\text{units}$.

To determine

The description for the datum feature A.

The description for the datum feature B.

The description for the datum feature C.

The description for the datum reference frame.

The description for the part which is stabilized by this datum’s.

Whether it is appropriate to locate a part with respect to locating pins or with respect to the edge of a part.

Explanation of Solution

• Datum feature A: - It is the bottom surface of the feature. Its datum is a theoretical plane corresponding to a datum feature simulator, such as machine table.
• Datum feature B: - It is the surface of the hole. Its datum is the centre axis corresponding to a datum feature simulator, such that the largest gauge pin just touching the high point of the hole while being held perpendicular to datum A.
• Datum feature C: - It is the surface of the slot. Its datum is the centre plane corresponding to a datum feature simulator, such that the largest gauge block just touching the high points of the slot while being held perpendicular to datum A.

The datum reference frame:-It is an important part of an object, such as hole, set of holes, pairs of surface and lines. It is used as a reference to define the geometry of the object. Here, the datum reference frame is made of the three mutually perpendicular planes, with the origin of the coordinate axes at the bottom centre of the datum B hole.

The description for the part, which stabilized by the datum’s for manufacturing or inspections are:

1. 1. Constrain the base to be in contact with a datum feature simulator corresponding to datum A.
2. 2. Constrain the translation of the part on datum A with a gauge pin in the datum feature B hole.
3. 3. Constrain the final rotation of the part with a gauge pin or block in the groove.

Fixture: The purpose of the fixture is to locate and orient the large bores. The bores are located with respect to the following datum’s.

1. 1. The bottom surface (datum A).
2. 2. The pin hole (datum B).
3. 3. The groove (datum C).

The edge of the base is not touching anything for the alignment, and can consequently have much looser tolerances. Hence, it is usually easier and cheaper to precisely locate a part with respect to locating pins than with respect to the edge of the part.

Thus, it is more appropriate to locate a part with respect to locating pins than with respect to the edge of the part.

To determine

The diameter of the tolerance zone at produced diameter of $\varphi 6.00$.

The diameter of the tolerance zone at produced diameter of $\varphi 6.05$.

Answer to Problem 27P

The diameter of the tolerance zone at produced diameter of $\varphi 6.00$ is $0.05\text{unit}$.

The diameter of the tolerance zone at produced diameter of $\varphi 6.05$ is $0.05\text{unit}$.

Explanation of Solution

For the given feature, the tolerance zone is regardless of the feature size as there is no any material modifier that can specify the tolerance zone associated with the hole. Hence, the diameter of the tolerance zone is specified as $0.05\text{unit}$ whether the hole diameter is $\varphi 6.00$ or $\varphi 6.05$, or anything between $0.00$ and $0.05$.

Conclusion:

Thus, the diameter of the tolerance zone at produced diameter of $\varphi 6.00$ is $0.05\text{unit}$.

Thus, the diameter of the tolerance zone at produced diameter of $\varphi 6.05$ is $0.05\text{unit}$.

To determine

The diameter of the tolerance zones at produced diameter of $\varphi 6.00$.

The diameter of the tolerance zones at produced diameter of $\varphi 6.1$.

Answer to Problem 27P

The diameter of the tolerance zones at produced diameter of $\varphi 6.00$ is $0.3\text{units}$.

The diameter of the tolerance zones at produced diameter of $\varphi 6.1$ is $0.4\text{units}$.

Explanation of Solution

When manufactured at MMC.

The top line of the position control of the bolt holes specifies the diameter of the tolerance zones at produced diameter of $\varphi 6.00$ is $0.3\text{units}$ at the maximum material condition.

When manufactured at LMC.

The diameter of the tolerance zone is $0.4\text{units}$. This is the summation of the specified tolerance and bottom tolerance of $0.1$.

Thus, the diameter of the tolerance zones at produced diameter of $\varphi 6.1$ is $0.4\text{units}$ at the low material condition

Conclusion:

Thus, the diameter of the tolerance zones at produced diameter of $\varphi 6.00$ is $0.3\text{units}$.

Thus, the diameter of the tolerance zones at produced diameter of $\varphi 6.1$ is $0.4\text{units}$.

To determine

The diameter of the tolerance zones locating the position of the bolts holes with respect to one another at produced diameter of $\varphi 6.00$.

The diameter of the tolerance zones locating the position of the bolts holes with respect to one another at produced diameter of $\varphi 6.1$.

Answer to Problem 27P

The diameter of the tolerance zones locating the position of the bolts holes with respect to one another at produced diameter of $\varphi 6.00$ is $0.1\text{units}$.

The diameter of the tolerance zones locating the position of the bolts holes with respect to one another at produced diameter of $\varphi 6.1$ is $0.2\text{units}$.

Explanation of Solution

The second line of the position control of the bolt holes specifies the tolerance of $0.1$ at maximum material condition.

For the bolt holes when manufactured at the MMC, the diameter of the tolerance zones locating the position of the bolts holes with respect to one another at produced diameter of $\varphi 6.00$ is $0.1\text{unit}$.

For the bolt holes when manufactured at the LMC, the diameter of the tolerance zone is $0.2\text{units}$. This is the summation of the specified tolerance and bottom tolerance of $0.1$ due to the deviation of the hole from its MMC.

Conclusion:

Thus, the diameter of the tolerance zones locating the position of the bolts holes with respect to one another at produced diameter of $\varphi 6.00$ is $0.1\text{units}$.

Thus, the diameter of the tolerance zones locating the position of the bolts holes with respect to one another at produced diameter of $\varphi 6.1$ is $0.2\text{units}$.

To determine

The reason of being that the (M) modifier is appropriate for the bolt hole position tolerance.

Explanation of Solution

The (M) modifier is appropriate for the bolt hole position tolerance as the MMC ensures a minimum clearance for the bolts in their holes. Though they allows greater deviations from ideal if the produced holes are larger.

To determine

The parameter to control the orientation and straightness of its centre axis.

The parameter to control the cylindricity of its surface.

Explanation of Solution

The orientation and the straightness of the axis are controlled by the position tolerance zone which is specified in the feature control frame directly under the diameter dimension for the bore.

The position tolerance specifies a cylindrical tolerance zone that the axis of the hole must be within, thus controlling its orientation and straightness.

Cylindricity is a control of the surface of the bore. It could be controlled either by the default profile of the surface control, specified with the note at the bottom of the drawing, or by Rule # 1.

Here, the Rule #1 has the tighter tolerance, so it controls the cylindricity of the bore.

To determine

The surfaces that are controlled by the default surface profile tolerance.

The changes in the drawing with the use of the edges of the base as datum feature.

Explanation of Solution

All the exterior surfaces are controlled by the default surface profile tolerance, but the bottom surface is not controlled by the default surface profile tolerance.

Hence, the as-cast is sufficient for controlling the surfaces.

The drawing is modified because the edges of the base were used as datum features. This modification also effects the controlling of the surfaces. These surfaces would need to be controlled more tightly in order to locate the critical holes from them. Also, the edges of the base could not be left as-cast.