EBK 3I-EBK: WELDING PRINCIPLES & APPLIC
8th Edition
ISBN: 9780176919764
Author: Jeffus
Publisher: VST
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Textbook Question
Chapter 12, Problem 15R
What are the main limitations of the rutile fluxes?
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An AISI 1018 steel ball with 1.100-in diameter is used as a roller between a flat plate
made from 2024 T3 aluminum and a flat table surface made from ASTM No. 30 gray
cast iron. Determine the maximum amount of weight that can be stacked on the
aluminum plate without exceeding a maximum shear stress of 19.00 kpsi in any of the
three pieces. Assume the figure given below, which is based on a typical Poisson's
ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress
occurs for these materials.
1.0
0.8
Ratio of stress to Pmax
0.4
90
0.6
στ
Tmax
0.2
0.5a
a
1.5a
2a
2.5a
За
Distance from contact surface
The maximum amount of weight that can be stacked on the aluminum plate is
lbf.
A carbon steel ball with 27.00-mm diameter is pressed together with an aluminum ball
with a 36.00-mm diameter by a force of 11.00 N. Determine the maximum shear
stress and the depth at which it will occur for the aluminum ball. Assume the figure
given below, which is based on a typical Poisson's ratio of 0.3, is applicable to estimate
the depth at which the maximum shear stress occurs for these materials.
1.0
0.8
Ratio of stress to Pma
9 0.6
στ
24
0.4
Tmax
0.2
0
0.5a
a
1.5a
Z
2a
2.5a
За
Distance from contact surface
The maximum shear stress is determined to be
MPa.
The depth in the aluminum ball at which the maximum shear stress will occur is
determined to be [
mm.
Show all work please
Chapter 12 Solutions
EBK 3I-EBK: WELDING PRINCIPLES & APPLIC
Ch. 12 - List some factors that have led to the increased...Ch. 12 - How is FCAW similar to GMAW?Ch. 12 - What does the FCA flux provide to the weld?Ch. 12 - What are the major atmospheric contaminations of...Ch. 12 - How does slag help an FCA weld?Ch. 12 - How can FCA welding guns be cooled?Ch. 12 - Excessive drive roller pressure causes what...Ch. 12 - List the advantages that FCA welding offers the...Ch. 12 - Describe the two methods of manufacturing FCA...Ch. 12 - Why are the large diameter electrodes not used for...
Ch. 12 - How do deoxidizers remove oxygen from the weld...Ch. 12 - What do fluxing agents do for a weld?Ch. 12 - Why are alloying elements added to the flux?Ch. 12 - How does the flux form a shielding gas to protect...Ch. 12 - What are the main limitations of the rutile...Ch. 12 - Why is it more difficult to use lime-based fluxed...Ch. 12 - What benefit does adding an externally supplied...Ch. 12 - How do excessive amounts of manganese affect a...Ch. 12 - Why are elements added that cause ferrite to form...Ch. 12 - Why must a flux form a less dense slag?Ch. 12 - Referring to Table 12-5, what is the AWS...Ch. 12 - Describe the meaning of each part of the following...Ch. 12 - What does the number 316 in E316T-1 mean?Ch. 12 - What is the advantage of using an argon- CO2 mixed...Ch. 12 - Why are some slags called refractory?Ch. 12 - What can happen to slag that solidifies on the...Ch. 12 - How is the electrode extension measured?Ch. 12 - What can cause porosity in an FCA weld?Ch. 12 - What happens to water in the welding arc?Ch. 12 - What is the thin dark gray or black layer on new...Ch. 12 - Why is uniformly scattered porosity hard to detect...Ch. 12 - What cautions must be taken when chemically...
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- Draw top, side, front view With pen(cil) and paper Multi view drawing and handwriting all of itarrow_forwardA wheel of diameter 150.0 mm and width 37.00 mm carrying a load 2.200 kN rolls on a flat rail. Take the wheel material as steel and the rail material as cast iron. Assume the figure given, which is based on a Poisson's ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. At this critical depth, calculate the Hertzian stresses σr, σy, σz, and Tmax for the wheel. 1.0 0.8 0, т Ratio of stress to Pmax 0.4 0.6 90 69 0.2 0.5b b 1.5b Tmax 2b Distance from contact surface The Hertizian stresses are as follows: 02 = or = -23.8 psi for the wheel =| necessary.) σy for the wheel =| MPa σz for the wheel = MPa V4 for the wheel = | MPa 2.5b ཡི 3b MPa (Include a minus sign ifarrow_forwardOnly question 3arrow_forward
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