SolveSample Problem 4.1.All remain as presented, except for the following changes.The factor ofsafety is now 2, and not 3.As Presented, the thickness of the tube is uniform. In the new one that you are solving,let the thicknessesof the top and bottom be the same,ttb, and the thicknesses of theleft and right be the same,tlr.Theheightand the width of the tube are still respectively 5 in and 3.25 in.As you make thesethickness changes,you are to keep thetotal mass of the tube unchanged. Answer the same questions (a) and (b).Explain your findings REFLECT and THINK: Alternatively, we can calculate the radius ofcurvature using Eq. (4.9). Since we know that the maximum stress iso all 20 ksi, the maximum strain ɛm can be determined, and Eq. (4.9) gives=o all20 ksiE 10.6 × 106 psiEm =Em||CCEmp = 1325 in.==1.887 × 10-³ in./in.2.5 in.1.887 × 10-³ in./in.p = 110.4 ft 5 in.--M13.25 in.0-04-1 = 0.25 in.5 in.3.25 in.2.75 in.Fig. 1 Superposition for calculatingmoment of inertia.Fig. 2 Deformed shapeof beam.Sample Problem 4.1The rectangular tube shown is extruded from an aluminum alloy forwhich oy = 40 ksi, ou = 60 ksi, and E = 10.6 x 10° psi. Neglecting theeffect of fillets, determine (a) the bending moment M for which the fac-tor of safety will be 3.00 and (b) the corresponding radius of curvatureof the tube.STRATEGY: Use the factor of safety to determine the allowable stress.Then calculate the bending moment and radius of curvature using Eqs. (4.15)and (4.21).MODELING and ANALYSIS:Moment of Inertia. Considering the cross-sectional area of the tube as thedifference between the two rectangles shown in Fig. 1 and recalling the for-mula for the centroidal moment of inertia of a rectangle, write1 = (3.25) (5)³(2.75)(4.5)³ I = 12.97 in*4.5 in.MAllowable Stress. For a factor of safety of 3.00 and an ultimate stress of60 ksi, we havedall =McIGall1 M==PΕΙSince < oy, the tube remains in the elastic range and we can apply theresults of Sec. 4.2.a. Bending Moment. With c = (5 in.) = 2.5 in., we writeouF.S.M = -0all =60 ksi3.0020 ksip = 1325 in.12.97 in*(20 ksi)2.5 in.b. Radius of Curvature. Using Fig. 2 and recalling thatE = 10.6 x 10° psi, we substitute this value and the values obtainedfor I and M into Eq. (4.21) and find103.8 x 10³ lb-in.(10.6 x 10° psi) (12.97 in*)M = 103.8 kip-in.= 0.755 x 10 in-1P = 110.4 ft

Answered: SolveSample Problem 4.1.All remain as…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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SolveSample Problem 4.1.All remain as presented, except for the following changes.The factor ofsafety is now 2, and not 3.As Presented, the thickness of the tube is uniform. In the new one that you are solving,let the thicknessesof the top and bottom be the same,ttb, and the thicknesses of theleft and right be the same,tlr.Theheightand the width of the tube are still respectively 5 in and 3.25 in.As you make thesethickness changes,you are to keep thetotal mass of the tube unchanged. Answer the same questions (a) and (b).Explain your findings

REFLECT and THINK: Alternatively, we can calculate the radius of
curvature using Eq. (4.9). Since we know that the maximum stress is
o all 20 ksi, the maximum strain ɛm can be determined, and Eq. (4.9) gives
=
o all
20 ksi
E 10.6 × 106 psi
Em =
Em
||
C
C
Em
p = 1325 in.
=
=
1.887 × 10-³ in./in.
2.5 in.
1.887 × 10-³ in./in.
p = 110.4 ft

5 in.
-
-
M
1
3.25 in.
0-04-
1 = 0.25 in.
5 in.
3.25 in.
2.75 in.
Fig. 1 Superposition for calculating
moment of inertia.
Fig. 2 Deformed shape
of beam.
Sample Problem 4.1
The rectangular tube shown is extruded from an aluminum alloy for
which oy = 40 ksi, ou = 60 ksi, and E = 10.6 x 10° psi. Neglecting the
effect of fillets, determine (a) the bending moment M for which the fac-
tor of safety will be 3.00 and (b) the corresponding radius of curvature
of the tube.
STRATEGY: Use the factor of safety to determine the allowable stress.
Then calculate the bending moment and radius of curvature using Eqs. (4.15)
and (4.21).
MODELING and ANALYSIS:
Moment of Inertia. Considering the cross-sectional area of the tube as the
difference between the two rectangles shown in Fig. 1 and recalling the for-
mula for the centroidal moment of inertia of a rectangle, write
1 = (3.25) (5)³(2.75)(4.5)³ I = 12.97 in*
4.5 in.
M
Allowable Stress. For a factor of safety of 3.00 and an ultimate stress of
60 ksi, we have
dall =
Mc
I
Gall
1 M
==
P
ΕΙ
Since < oy, the tube remains in the elastic range and we can apply the
results of Sec. 4.2.
a. Bending Moment. With c = (5 in.) = 2.5 in., we write
ou
F.S.
M = -0all =
60 ksi
3.00
20 ksi
p = 1325 in.
12.97 in*
(20 ksi)
2.5 in.
b. Radius of Curvature. Using Fig. 2 and recalling that
E = 10.6 x 10° psi, we substitute this value and the values obtained
for I and M into Eq. (4.21) and find
103.8 x 10³ lb-in.
(10.6 x 10° psi) (12.97 in*)
M = 103.8 kip-in.
= 0.755 x 10 in-1
P = 110.4 ft

Expert Solution

Step 1 Given

The ultimate tensile strength is $σ_{u} = 60 k s i$ .

The yield strength is $σ_{y} = 40 k s i$ .

The elastic modulus is $E = 10.6 * 10^{6} p s i$ .

The height of the rectangular section is $h = 5 i n$ .

The width of the rectangular section is $w = 3.25 i n$ .

The factor of safety is $2$ .

The initial thickness of the rectangular side is $t = 0.25 i n$ .

The final thickness of the upper and lower portion is $t t b$ .

The final thickness of the left and right sides is $t l r$ .

The mass of the rectangular section is constant.

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