1. We can visualize the factor of safety for an arbitrary stress using a surface in principal stress space. Fora ductile material that yields according to a von Mises criterion with a yield stress σy, sketch the vonMises surface in σ₁ - 02 space and sketch the stress surface that corresponds to a factor of safety FoS =2. For a brittle material that yields according to a max normal (Rankine) criterion with a tensile strengthGyt and a compressive strength σvc = 20yt, sketch the yield surface and the surface that correspondsto a factor of safety FoS = 2.

Step 1:Step 2: Solution is given above Step 3: Step 4:

Answered: 1. We can visualize the factor of…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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A cylindrical 1045 steel bar (see the figure below) is subjected to repeated compression- tension stress cycling along its axis. If the load amplitude is 23,000 N, calculate the minimum allowable bar diameter (in mm) to ensure that fatigue failure will not occur. Assume a factor of safety of 2.0. Maximum stress, S (MPa) 700 600 500 400 300 200 100 0 4340 steel 104 Ti-5Al-2.5Sn titanium alloy 70Cu-30Zn brass EQ21A-T6 Mg alloy 105 1045 steel Ductile cast iron 107 2014-T6 Al alloy 108 10⁹ 106 Cycles to failure, N Data taken from the following sources and reproduced with permission of ASM International, Materials Park, OH, 44073: ASM Handbook, Vol. I, Properties and Selection: Irons, Steels, and High-Performance Alloys, 1990; ASM Handbook, Vol. 2, Properties and Selection; Nonferrous Alloys and Special- Purpose Materials, 1990; G. M. Sinclair and W. J. Craig, "Influence of Grain Size on Work Hardening and Fatigue Characteristics of Alpha Brass," Transactions of ASM, Vol. 44, 1952.
We have a grounded 2 cm in diameter steel shaft which is 10 cm in length connected to a rigid surface. This shaft is subject to the stresses at the tip point as given below; • Tmean = 100 Nm, Talternating= 50 Nm, • Mmean = 30 Nm, Malternating= 40 Nm, Fmean = 4 kN, Falternating = 6 kN. This shaft has minimum properties of Sy = 800 Mpa and Sut = 1000 Mpa. Assume; Kf for bending=D1.5, Kf for tension = 1.4, and Kfs for torsion = 1.8. Using Modified Goodman criterion what can you say about the shaft? a) Does it have an infinite life?
The stresses on the surface of a hard bronze component are shown in the figure below. The yield strength of the bronze is σY = 345 MPa.a) What is the factor of safety predicted by the maximum-shear-stress theory of failure for the stress state shown? Does the component fail according to this theory?b) What is the value of the Mises equivalent stress for the given state of plane stress?c) What is the factor of safety predicted by the failure criterion of the maximum-distortion energy theory of failure? Does the component fail according to this theory?
Q1B
In the Haigh diagram depicted in figure, point A represents the fatigue stress in a standard screw. Suppose the same screw is cold I rolled and then subjected to thes same fatigue l We A moves: expect point C A B O in D O in B O in C 246595 03mchd 3246598 03mehade dreamstime 10 010704650s 03mchu 24 5036 5.03men 2-699643mchado21 O 070 6246585 03cnoco 246595 03oh
answer as soon as soon as possible
FN = 50
i li. الامتحان هو سؤالان فقط Q. 1: If a material is cast iron (ultimate tensile strength = 300 Mpa, ultimate compressive strength - 900 Mpa) and the stress state at a point is shown in Fig. 1. Using a graphical method, find safety factor against static failure using 100 Mpa (a) Rankine failure theory (b) Coloumb - Mohr theory (c) Modified Coloumb - Mohr theory 50 Мра إضافة ملف Q.2 For a rotating shaft of diameter 15 mm as shown in Fig. 2. If the material is AISI-1045 CD, Find the safety factor II >
Ip
A steel component is subjected to alternate cyclical loading. The steel follows Basquin's law for high cycle fatigue, o, x N = C, (where the stress amplitude is in MPa). Ignore the geometric detail and assume that Marin's modifying factors are all equal to 1. You are given the minimum stress ain = -213 MPa, the maximum stress omax = 213 MPa. The material data are Tensile strength oUTS = 539 MPa, Basquin's constant c, = 875 MPa, Basquin's exponent a = 0.085. a) Calculate the stress ratio R, the stress amplitude o, in MPa and the mean stress am in MPa. The answers are acceptable with a tolerance of 0.01 for R and of 1 MPa the stresses. R: MPa MPа b) Calculate the corresponding life, in 10° cycles, (tolerance of 0.1 106 cycles) N :
EXAMPLE 3.9 Finite-element²¹ analysis shows that a long structural component car- ries a uniform axial stress of 35 MPa (T) (Figure 3.29). A hole in the center needs to be drilled for passing cables through the structural component. The yield stress of the material is 200 MPa. If failure due to yielding is to be avoided, determine the maximum diameter of the hole that can be drilled using a factor of safety of 1.6. 10 mm H = 100 mm -d- Figure 3.29 Component geometry in Example 3.9.

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