INSTRUCTIONS: Calculate the required wall thicknesses from the information provided below. 26. .990 .988 1. MAX WALL 2. MIN WALL \text{0}.750^{+010}_{-002}
Added by Christopher F.
Close
Step 1
The maximum outer diameter is 0.990. The minimum inner diameter is 0.750 - 0.002 = 0.748. The maximum wall thickness is (0.990 - 0.748)/2 = 0.121. Show more…
Show all steps
Your feedback will help us improve your experience
Penny Riley and 57 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A steel pipe of 100 -mm diameter is to support the loading shown. Knowing that the stock of pipes available has thicknesses varying from $6 \mathrm{mm}$ to $24 \mathrm{mm}$ in 3 -mm increments, and that the allowable normal stress for the steel used is $150 \mathrm{MPa}$, determine the minimum wall thickness $t$ that can be used.
A cast-iron tube is used to support a compressive load. It is given that E = 69 GPa and the maximum allowable change in length is 0.040%. Determine the minimum wall thickness for a load of 7.2 kN if the outside diameter of the tube is 50 mm. The minimum wall thickness of the tube is mm.
Penny R.
A steel pipe of 100 -mm diameter is to support the loading shown. Knowing that the stock of pipes available has thicknesses varying from 6 to $24 \mathrm{mm}$ in 3 -mm increments, and that the allowable normal stress for the steel used is $150 \mathrm{MPa}$, determine the minimum wall thickness $t$ that can be used.
Josee P.
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD