This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTION {12.4)
Known: A torsion bar has a known length and diameter. Find: (all Estimate the change in shear stress when one end of the rod rotates through TD“
relative to the other end. (b) Estimate the change in torque. Schematic and Given Data: d=[}.312 in. Assumptions: 1. The bar remains straight and the torque is applied about the longitudinal axis. 2. The material is homogeneous and perfectly elastic within the stress range
involved. Analysis:
1. From Table 5.1, for a torsional ease (Case 2]: u. L ‘ 1 : @123
{i — JG , therelote , "l L 2. El = [BUIIBUME =13? rad
.l= IEd4i32 = :rt(ﬂ.312}4f32 = D.ﬂﬂi}93 in.4
From [Appendix (31), G = 11.5 it ID‘E' psi
1.39 x anongs 311.5 x to‘”
50
4. From Eq. (4.4], for a solid round rod.
3 _ “WEE.33.} 1:15”de _  3:49.9151
n(0.312)' I 3. Therefore, T = a 29132 lb in. l SOLUTION (12.13)
Known: A helical coil spring with given D and d is wound with a known pitch value.
The material is ASTM A227 cold drawn carbon steel. Find: If the spring is compressed solid, would you expect it to return to its original
freelength when the force is removed? Schematic and Given Data: ASTM A227
Cold drawn carbon steel Assumptions: 1. There are no unfavorable residual stresses. 2. Both end plates are in contact with nearly a full turn of wire.
3. The end plate loads coincide with the spring axis. Analysis:
1. Force to compress spring solid can be calculated by using Eq. (12.7).
F = d4Go
soils where MN = p  d =10  5.5 = 4.5 mm
G = 79 X 109 PA (Appendix (31) = (5.5 x 103)4(79 x 109)(4.5 x 103) F = 325 N
am x 10333 I
2. The correSponding stress can be calculated by using Eq. (12.6).
1: = LP]? KS
rid for C = Dr’d = 5015.5 = 9.09
Ks = 1.05 (Fig. 12.4) _ 3(325)(50 x 103)
3.15.5 x 103)3
3. From Eq. (12.9),173 5 0.45 Su
From Fig. 12.7, Su 2: 1300 MPa Thus, 0.45 Su : 585 MPa 4. Since 261 MPa < 585 MP3, no set should occur; therefore, spring should return to
original length. I 1: (l.05)=261MPa Comment: Even considering the curvature (stress concentration) factor of the inner surface by using Kw = 1.17, the inner surface stress is only 290 MPa which is still less
than 585 MPa. SOLUTIO { .14) Known: A ceiled compression spring is leaded against a support. A pint of the
resulting forcendeﬂeetion curve is shown. Find: Brieﬂy state the reasons the curve changes at points A, E, and C. Schematic and Given Data: Deflection —II Assumptions:
1. There are no unfavorable residual stresses. 2. Beth end plates are in centact with nearly a full turn of wire.
3. The end plate leads coincide with the spring axis. Analysis: 1. .i—‘tt A, the furee overcomes the spring preload. causing deﬂectien to begin. 2. At B, significant yielding of the spring begins. [111 must instances, this would
indicate an improperly designed spring. 3. At C, the spring closes stilid. SOLUTION (12.13) Known: A. machine uses a pair of concentric helical compression springs to support a
known Stalin: load. Both springs are made of steel and have the same length when
loaded and when unloaded. Find: Calculate the deﬂection and the maximum stress in each spring. Schematic and Given Data: F: 3.1:! kN W
W D0245 mm (1.128 mm
NuS Assumptions:. There are no unfavorable residual stresses. 2. Both end plates are in contact with nearly a full turn of wile.
3. The end plate loads coincide with the spring axis. Analysis:
d4o
soils
where G = 79 X 109 Pa for steel. (Appendix Cl}
k0 = (a mrn]“(79.ﬂ{lﬂ Nimﬂll
3(45 mmﬁsi ates rnrn]3{l{l) I. From Eq. (12.8), k = = 38.7? Htmtn k; = = 39.50 Nintnt 2_ From Eq. (12.3), k e E 0r 5 :3
3.000
a ' (39.5 + Sgt1e)” ' 2139 mm
Using F = kt], we can calculate the force on each spring.
F0 = kn?) = (38.7??? N!mm)(23.39 mm) = EUTﬁ N
Fi = kiﬁ = (39.50 mem}{23.39 mm) = 924 N
4. Using Fig. 12.4, we can ﬁnd K; values for each spring.
Fer enter spring, C = 458 = 5.511% = 1.09
For inner spring, C : 255 = 5.0!]. Ks : 1.]{] LB 5. From Eq. (12.6), 1: = m K.
15:13 to = Siﬁpﬁﬁiiﬁ) {my} = 50:5 MPa n18)“
_ Beams)
‘tl _ —— J
n15) (1.10)=513 Mpa moment (12, ) Known: A helical compression Spring with squareti and ground ends is to he designed with given force and deﬂection requirement. Presetting is to be used. The loading is
static. Find: Determine appropriate values for D. N . and Lt. Check. for possible buckling. Schematic and Given Data:
son N lﬂmN 1 F.
i like. Music wire Clash allowance
with cl = 5 mm Assumptions: 1. Both end plates are in contact with nearly a full turn of wire.
2. The end plate loads coincide with the spring axis. 3. The clash allowance is 10% of the maximum deﬂection. Analysis:
I. From EL]. {12.3}, k = k _ 5F_ lﬂﬂﬂ  SUD = 3.33 Nimm _ as _ 6G
2. Since class allowance is 113% of maximum deﬂection.
F5: 1 113% oflﬂﬂﬂN= IIIJCIN
3. From Eq.{12.9}, 1:; = 0135 Su
4. From Fig. 12.7, Si. — lt‘SSﬂ MP3 for music wire with d = 5 mm.
5. Thus, 125 = ﬂﬁSUESﬂ) = 161'? MPa. _ 8F _ end2
6. From Eq. (12.5), 'c — m—,, CR. or CK.5 _ — nd 31:
_ {1ﬂ?3}(n}(5)2 _
cits a W e 9.53
Therefore. C = 9.0 (Fig. 12.4) and D = Cd = 9.D(5} = 451} mm I
7. From Eq. {12.3}, N = where o = 79 GPa [Appendix on
N = missus = 3.13 S(S.33)(9.U}3 I For ends squared and ground, L5 = Hid = (N + 2)d IFig. 12.3w” L; = {8.13 + 25 = 50.65 mm Lf= L3 + Fy'k = 50.!55 + H’U'DJ'SEE =182.7 mm I erD = 182.?l'45 = 4.1 aﬁlLr=(%}1132.T =u.?2 From Fig. 12.10, we see that thn: and plates must be constrained parallel (case A)
to avoid buckling. I ...
View
Full
Document
This note was uploaded on 05/05/2011 for the course MEM 431 taught by Professor Zhou during the Spring '08 term at Drexel.
 Spring '08
 ZHOU

Click to edit the document details