7. Calculate the minimum wall thickness between the inside diameter and datum feature B shown in the figure below.Datum B @ LMC:Ø-Øminus I. D. @ LMC:minus Tolerance @ LMC: - Ø.=The wall thickness equals half of the differences in diameters =Ø2.500.0204.059 +.003Ø2.000 .020+0.0200 ABOBAUnless otherwise Specified:.XXX= .010ANGLES = ± 1°

The wall thickness equals half of the difference in diameters is 0.46

7. Calculate the minimum wall thickness between the inside diameter and datum feature B shown in the figure below. Datum B @ LMC: Ø -Ø minus I. D. @ LMC: minus Tolerance @ LMC: -0. = The wall thickness equals half of the differences in diameters =

7. Calculate the minimum wall thickness between the inside diameter and datum feature B shown in the figure below. Datum B @ LMC: Ø -Ø minus I. D. @ LMC: minus Tolerance @ LMC: -0. = The wall thickness equals half of the differences in diameters =

Refrigeration and Air Conditioning Technology (MindTap Course List)

8th Edition

ISBN:9781305578296

Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson

Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson

Chapter1: Heat, Temperature, And Pressure

Section: Chapter Questions

Problem 17RQ: Convert 22C to Fahrenheit.

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Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

7. Calculate the minimum wall thickness between the inside diameter and datum feature B shown in the figure below.
Datum B @ LMC:
Ø
-Ø
minus I. D. @ LMC:
minus Tolerance @ LMC: - Ø.
=
The wall thickness equals half of the differences in diameters =
Ø2.500
.020
4.059 +.003
Ø2.000 .020
+0.0200 ABO
B
A
Unless otherwise Specified:
.XXX= .010
ANGLES = ± 1°

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