What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105 m/m from a load producing a unit tensile stress of 44,000 psi? A. 42,300 x 10⁶ psi B. 41,202 x 10⁶ psi C. 43,101 x 10⁶ psi D. 41,905 x 10⁶ psi
What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105 m/m from a load producing a unit tensile stress of 44,000 psi? A. 42,300 x 10⁶ psi B. 41,202 x 10⁶ psi C. 43,101 x 10⁶ psi D. 41,905 x 10⁶ psi
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
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What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105 m/m
from a load producing a unit tensile stress of 44,000 psi?
A. 42,300 x 10⁶ psi B. 41,202 x 10⁶ psi C. 43,101 x 10⁶ psi D. 41,905 x 10⁶ psi
Please solve the Problem elaborately and Refer Machine Elements and Stresses Equations from the figures or use other shortcut Methods you have or you may Solve both or in more different methods the better ?. Your solution will be use as reference for my Future Board Exam preparation/review/study. Thank you so much dear your work will be appreciated much and rated excellently.
Transcribed Image Text:where: J = polar moment of inertia = T Dª /32 (for solid shaft)
c = distance from neutral axis to the farthest fiber
T = torque
c = r (for circular cross section)
7. Bending Stress(S:) Si =
d = diameter
For rectangular beam S = 6 M
b h?
Mc
where: M = moment
F
c = distance of farthest fiber from neutral axis
| = moment of inertia about the neutral axis
| = bh/12 (for rectangular cross section)
I M
c Sf
L
Z = section modulus
Y
8. Strain and Elongation Strain =
Stress
FL
Stress =
A
E
Strain
FL
Y =
AE
AY
where: y = elongation due to applied load
L = original length
F = force
A = area
S = stress
9. Thermal Elongation; Stresses
y = kL (t2 - t1) S = E! = kE (t2 - t1)
where: k = coeficient of thermal expansion, m/m-°C
For steel k = 6.5 x 10-6 in/in-F = 11.7 x 106 m/m-C
E = 30 x 10° psi
Relation between shearing and tensile stress based on theory of failure:
Samax = Sty /2
Smax = Sty
10. Variable Stress
1 Sm Sa
FS Sy S,
Sy = yield point
where: FS = factor of safety
Sn = endurance limit
Smax + Smin
S. = variable component stress =
Smax - Smin
Sm = mean stress =
2
2
Smax = maximum stress
11. Poisson's Ratio(u) = is the ratio of lateral unit deformation to axial unit deformation.
Smin = minimum stress
E
F
E = strain =-
AE
where: G = shear modulus of elasticity
2G
W2 – W1
Ey
t2 - ty
Lateral Strain
Ey Ez
Ex
L2 -L,
Ex
L1
Ez
Longitudin al Strain
Ex
t,
Transcribed Image Text:V1
Machine Elements
1. Cylinders Rolling in opposite direction:
Á. Tangential speed
B. Relation of diameter and speed
V, = V2 = T D, N, = r D2 N2
DI N1 = D2 N2
Speed of Driver
C. Speed Ratio =
Speed of the Driven
D. Center Distance = R, + R2 =
D, + D2
2. Cylinders Rolling in the same direction
Vi = V2 = n D, N; = n D2 N2
D, N1 = D2 N2
A. Tangential speed
B. Relation of diameter and speed
Speed of Driver
Speed of the Driven
D2 - D1
C. Speed Ratio =
D. Center Distance = R2 -R, =
Stresses
1. Stress (S) = a total resistance that a material offers to an applied load, Ib/in? , kg/cm² , KN/m?
2. Ultimate stress (Su ) - is the stress that would cause failure
3. Yield stress(Sy) - maximum stress without causing deformation
4. Allowable stress(Sal) = Ultimate stress/Factor of Safety
Sy
5. Design stress(Sa) - stress used in determining the size of a member. Sa =
or Sa =
FS
FS
where: FS = factor of safety
1. Tensile Stress (S.)
S =
For solid circular cross-section: A =
A
D2
#0,² -D²)
For hollow circular cross-section: A =
For rectangular cross-section: A = base x height = b x h
Fa
2. Compressive Stress(Sc)S. =
A
3. Shearing Stress(S.)
F
A. For single bolt of rivet needed to join to plates together. S, =
For single rivet: A = t/4 D²
For double riveted joint: A = 2(TT/4 D²)
where:
D
B. Shearing due to punching of hole.
S =
where A = 1 Dt (for punching a hole) A = 4 St (for square hole)
Where: S = length of side of square
C. Pressure needed to punch a hole, F: F = d x t x 80, tons
t = plate thickness
t = thickness, in
where: A = DL
Where: d = hole diameter, in
-Shear area
4. Bearing Stress(Sb)
Sp = Fb IA
a. Based on yield strength FS = Sy / Sall
5. Factor of safety(FS)
b. Based on ultimate strength FS = S,/ Sall
D
6. Torsional Shear Stress(Ss)
Projected Area
D
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