Problem 3:One hollow cylinder is shrink-fitted inside another. Both cylinders have length L and both the flat faces of eachcylinder are constrained in the axial direction. They are free to move in radial and tangential directions. Bothcylinders are made of the same material steel. New steel material will be created.CylinderInner CylinderFIGURE 1 SHRINK FIT ASSEMBLYOutside CylinderSteel: E = 30 x 106 psiInside Radiusr;=4"R₁=6"FIGURE 2 EXPLODED VIEW OF SHRINKFIT ASSEMBLYOutside Radiusr=6.005"R₂=8"LengthL = 5"L = 5"An internal pressure P = 30,000 psi is applied on the inner surface of the inner cylinder.What to Submit:1. Find the tangential stress distribution at the outside surface of the outer cylinder (r = 8").2. Find the tangential stress distribution at the inner surface of the inner cylinder (r = 4")3. Find the contact pressure developed at the interference location.

One hollow cylinder is shrink-fitted inside another. Both cylinders have length L and both the flat faces of each cylinder are constrained in the axial direction. They are free to move in radial and tangential directions. Both cylinders are made of the same material steel. New steel material will be created. Cylinder Inner Cylinder FIGURE 1 SHRINK FIT ASSEMBLY Outside Cylinder Steel: E = 30 x 106 psi Inside Radius r₁=4" R₁=6" FIGURE 2 EXPLODED VIEW OF SHRINK FIT ASSEMBLY Outside Radius r=6.005" R₂=8" Length L=5" L=5" An internal pressure P = 30,000 psi is applied on the inner surface of the inner cylinder. What to Submit: 1. Find the tangential stress distribution at the outside surface of the outer cylinder (r = 8"). 2. Find the tangential stress distribution at the inner surface of the inner cylinder (r = 4") 3. Find the contact pressure developed at the interference location.

One hollow cylinder is shrink-fitted inside another. Both cylinders have length L and both the flat faces of each cylinder are constrained in the axial direction. They are free to move in radial and tangential directions. Both cylinders are made of the same material steel. New steel material will be created. Cylinder Inner Cylinder FIGURE 1 SHRINK FIT ASSEMBLY Outside Cylinder Steel: E = 30 x 106 psi Inside Radius r₁=4" R₁=6" FIGURE 2 EXPLODED VIEW OF SHRINK FIT ASSEMBLY Outside Radius r=6.005" R₂=8" Length L=5" L=5" An internal pressure P = 30,000 psi is applied on the inner surface of the inner cylinder. What to Submit: 1. Find the tangential stress distribution at the outside surface of the outer cylinder (r = 8"). 2. Find the tangential stress distribution at the inner surface of the inner cylinder (r = 4") 3. Find the contact pressure developed at the interference location.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Q: Attention !!!! Please clarify this . When they go to calculate both the normal stress of the bolt…

A: Step 1: Step 2: Step 3: Step 4:

Q: Please help. Written on paper not typed on computer or keyboard please. Need help on all questions.…

A: Pin-Connected FrameNote:-we analyze it as a frame (subjected to forces on joints and external forces…

Q: Required Information NOTE: This is a multi-part question. Once an answer is submitted, you will be…

Q: Figure 8-4 Full Alternative Text 8-5. The gas storage tank is fabricated by bolting together two…

A: As per Expert Guidelines we are supposed to answer only the first question of the given multiple…

Q: What is Ix

A: According to the given data, we need to find a moment of inertia about the X-axis.

Q: Please write it out. I don’t like the computer typed answers. Need help on all questions. Please…

Q: 4. A 25 mm diameter steel rotating shaft in a machine is loaded as follows: $ = 25mm Rotating M T…

Q: L2 L3 A 03 C L1 L4

Q: Please solve in detail. Don't use chatgpt,The Free body diagram must be drawn.

A: Free-body Diagram: Note: I hope this helps. Thank you.

Q: calculate the theoretical stress using the flexure or bending stress formula. The experimental se is…

Q: Suppose you need to design a tension test machine capable of testing specimens that have nominal…

Q: Help!!! Answer it correctly!!! Make sure you solve work on hand!!! Answer correctly Please!!!!!!!

A: Thank you,Please rate my response.

Q: Please draw a FBD only for the circled yellow part known as a rubber base plate :) I don’t want the…

Q: The nail cutter consists of the handle and the two cutting blades (Eigure 1) Figure 6 mm 6 mm -36 mm…

A: According to the given data F=6 N We need to find Force on the fingernail by the cutter

Q: Pc = 40, Pif = -2, πc = 20, πif = 4. (P = hydrostatic pressure, π – oncotic pressure, c – capillary,…

A: capillary hydrostatic pressure, Pc = 40 interstitial fluid hydrostatic pressure, Pif = -2 capillary…

Q: dimensions, L1 = 6.8 in and L2 = 3.3 in, determine the polar moment of the weld about its centroid.…

Q: -2x Criven the shape above, Find the location of the Controid. Show all work.

A: Here, four shapes are visible so, No Shape Ai Xi Yi AiXi AiYi 1 Semi circle 5654.87 60 0…

Q: ▾ Part A Determine the normal stress at point A on the cross section of drill bit at section a-a.…

A: The force component is-The area and moment of inertia is-

Concept explainers

Die Stock

Die stock is expressed as the type of component which is utilized for fitting purposes while creating or shaping uniform threads on tubes, pipes, or screws. The die stock (die handle) has three screws around the adjustable part of a particular machine that helps to locate the die. Also, die stock can endure on a variety of hex and split adjustable dies of several dimensions. One of the primary reasons for using die stock in industries is because die stock can be operated by hands. Commonly, it is made up of high carbon steel.

Question

Problem 3:
One hollow cylinder is shrink-fitted inside another. Both cylinders have length L and both the flat faces of each
cylinder are constrained in the axial direction. They are free to move in radial and tangential directions. Both
cylinders are made of the same material steel. New steel material will be created.
Cylinder
Inner Cylinder
FIGURE 1 SHRINK FIT ASSEMBLY
Outside Cylinder
Steel: E = 30 x 106 psi
Inside Radius
r;=4"
R₁=6"
FIGURE 2 EXPLODED VIEW OF SHRINK
FIT ASSEMBLY
Outside Radius
r=6.005"
R₂=8"
Length
L = 5"
L = 5"
An internal pressure P = 30,000 psi is applied on the inner surface of the inner cylinder.
What to Submit:
1. Find the tangential stress distribution at the outside surface of the outer cylinder (r = 8").
2. Find the tangential stress distribution at the inner surface of the inner cylinder (r = 4")
3. Find the contact pressure developed at the interference location.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1: Write the given data and what is to find

VIEW

Step 2: Determine the contact pressure developed at the interference location:

VIEW

Step 3: Determine the tangential stress distribution at the outside surface of the outer cylinder:

VIEW

Solution

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 4 steps with 6 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

Hi, I have been given this pin-jointed frame analysis to solve (see pictures of the structure attached). Can you help me answer all the questions below? Especially the one about the bending moment and shear force. The Figure 2 attached is a simplified structure to undertake a structural analysis on a concept design for a low cost stroller/buggy for children. The buggy consists of a number of metal sections that are pin-jointed together. The buggy is designed to accommodate a child of mass no more than 20 kg, sat on the seat which is supported at four points (Fc) on the frame of the buggy. Assume that the weight of the child is equally distributed across all of the points indicated on Figure 2 as Fc. Assume that there are no additional forces applied to the handle of the buggy, that the buggy is static (stationary) and that the buggy, and the elements of the frame have no mass. At points A and G, there are roll supports. At Points B, F and D there are the pin supports. I am required…
Please help. Written on paper not typed on computer or keyboard please. Need help on all questions. Please include all units, steps to the problem and information such as its direction or if it is in compression or tension. Thx.
if the distance between the fulcrum and the center of the mass of the bottles is 0.75m and the distance between the fulcrum and the jar is 0.42 meters, what is the ideal mechanical advantage of this lever? Show work. (the lever is below, ignore the part about the hotspots.)
why is the cylinder longitudinal stress zero at the bottom?
Problem 1: The assembly below will be used for each problem. Lengths, diameters, and materials are provided. If the temperature when this system was assembled was 74F, what would be the stress in each cylinder at a temperature of 40F? Problem 2: If the fixed support at D were removed and the temperature change did not occur, how far would point D move if a 400 lb tension load was applied at the center of each cylinder? (Hint: what sections would you break up the assembly into to find the Normal forces?) Problem 3: Based on the results from problem 2 and the sections specified in your solution, how much would the diameter of each section shrink from the loads applied? (Hint: Look at Poisson's ratio for each of the sections listed. 304 Stainless steel 2014-T6 Aluminum C 86100 Bronze A 12 in. 8 in. D В C| 4 in. 4 ft 6 ft 3 ft