EBK 3I-EBK: WELDING PRINCIPLES & APPLIC
8th Edition
ISBN: 9780176919764
Author: Jeffus
Publisher: VST
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Textbook Question
Chapter 12, Problem 7R
Excessive drive roller pressure causes what problems?
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A punch press with flywheel adequate to minimize speed fluctuation produces 120 punching strokes per minute, each providing an average force of 2000 N over a stroke of 50 mm. The press is driven through a gear reducer by a shaft rotating 200 rpm. Overall efficiency is 80%. a) What power (W) is transmitted through the shaft? b) What average torque is applied to the shaft?
1.58 The crankshaft of a single-cylinder air compressor rotates 1800 rpm. The piston area is 2000 mm2 and the piston stroke is 50 mm. Assume a simple “idealized” case where the average gas pressure acting on the piston during the compression stroke is 1 MPa, and pressure during the intake stroke is negligible. The compressor is 80% efficient. A flywheel provides adequate control of the speed fluctuation. a) What motor power (kW) is required to drive the crankshaft? b) What torque is transmitted through the crankshaft?
28. The shaft shown in Figure P5-28 is supported by bear-
ings at each end, which have bores of 20.0 mm. Design
the shaft to carry the given load if it is steady and the
shaft is stationary. Make the dimension a as large as pos-
sible while keeping the stress safe. Determine the required
d
20 mm
5.4 kN
d
D = ?
Length not
to scale
-α =
=
-125 mm
20 mm
a =
-250 mm-
FIGURE P5-28 (Problems 28, 29, and 30)
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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- The motor shown operates at constant speed and develops a torque of 100 lb-in during normal operation. Attached to the motor shaft is a gear reducer of ratio 5:1, that is, the reducer output shaft rotates in the same direction as the motor but at one-fifth motor speed. Rotation of the reducer housing is prevented by the "torque arm" pin-connected at each end as shown. The reducer output shaft drives the load through a flexible coupling. Neglecting gravity and friction, what loads are applied to (a) the torque arm, (b) the motor output shaft, and (c) the reducer output shaft? Motor Gear reducer Flexible coupling (To load) Torque arm- Torque arm Reducer output shaft Motor Reducer Shaft rotationarrow_forwardPlease can you help with ten attatched question?arrow_forwardAn 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.arrow_forward
- 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.arrow_forwardShow all work pleasearrow_forwardDraw top, side, front view With pen(cil) and paper Multi view drawing and handwriting all of itarrow_forward
- A 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_forwardOnly question 2arrow_forward
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