Tài liệu shi20396 chương 4 pdf

b With straight rigid wires, the mobile is not stable.. Consider a wire of length l bent at its string support: for an upward bend can be found by changing the sign of W.. The moment wil

W A

B

(e)

(f) (d)

2

(c)

R O = 1.2tan 30 = 2.078 kN Ans.

R A = 1.2sin 30 = 2.4 kN Ans.

(d) Step 1: Find R A and R E

h = 4.5tan 30 = 7.794 m

R A

R B

Step 2: Find components of R C on link 4 and R D

+
M C = 0400(4.5) − (7.794 − 1.9)R D = 0 ⇒ R D = 305.4 N Ans.

D B

E

Ans.

4-4 (a) q = R1x−1− 40x − 4−1+ 30x − 8−1+ R2x − 14−1− 60x − 18−1

Solve (3) and (4) simultaneously to get R1 = −1.43 lbf, R2 = 71.43 lbf Ans.

From Eqs (1) and (2), at x = 0+, V = R1 = −1.43 lbf, M = 0

With R0 and M0, Eqs (1) and (2) give the same V and M curves as Prob 4-3 (note for

V, M0x−1 has no physical meaning)

R3 = 480 − 352 = 128 lbf

0≤ x ≤ 8: V = 160 − 40x lbf, M = 160x − 20x2 lbf· in

8≤ x ≤ 10: V = 160 − 40x + 40(x − 8) = −160 lbf,

M = 160x − 20x2+ 20(x − 8)2 = 1280 − 160x lbf · in

4-5 Solution depends upon the beam selected.

a = 12



−l +l2+ 4(0.25l2)

= l2

√

2− 1 = 0.2071l Ans.

for l = 10 in and w = 100 lbf, Mmin= (100/2)[(0.2071)(10)]2 = 214.5 lbf · in

4-7 For the ith wire from bottom, from summing forces vertically

(b) With straight rigid wires, the mobile is not stable Any perturbation can lead to all wires

becoming collinear Consider a wire of length l bent at its string support:

for an upward bend can be found by changing the sign of W The moment will no longer

be correcting A curved, convex-upward bend of wire will produce stable equilibriumtoo, but the equation would change somewhat

φ p = 1

2tan

− 1

43

x

6.4

18.6 27.5

φ p = 12

24.7

x

26.22 7.78

69.7

4-9 (a)

24.4

x

14.63 6.63

69.4

2s

(12, 7 cw )

C R D

30.96

x

12

22 14.04
x

2s (20, 8 cw )

C R

12.36 32.36

5 20.81

2 circles

C D

y

x

(8, 4 ccw )

Roots are: 9, 0, 0 kpsi

For equal stress, the model load varies by the square of the scale factor



= wl28

(a) Can solve by iteration or derive equations for the general case.

Find maximum moment under wheel W3

1 1

because the centroids are coincident.

(d) Use a as a negative area.

G a

B b

a C

(f) Let a= total area

(b) The moment is maximum and constant between A and B

x

τmax= 3400 psi Ans.

4-28 If support R B is between F1and F2at position x = l, maximum moments occur at x = 3 and l.

l2+ 0.7046l − 39.61 = 0 The positive solution for l is 5.95 in and the magnitude of the moment is

M = 9300 − 43 575/5.95 = 1976 lbf · in Placing the bearing to the right of F2, the bending moment would be minimized by placing

it as close as possible to F2 If the bearing is near point B as in the original figure, then we

need to equate the reaction forces From statics, R B = 14 525/l, and R A = 3100 − R B

For R A = R B , then R A = R B = 1550 lbf, and l = 14 575/1550 = 9.37 in.

The fractional increase in the magnitude is

F(a − 2b) − ( p1/2)(a − 2b)2− [( p1+ p2)/6a](a − 2b)3

Substituting into (5) yields

4-30 Computer program; no solution given here

a  2b

l F

θrd= C



L A

Ratio equals 1, twists are the same

t

b t

F

l

Note the weight ratio is

Torque carrying capacity reduces with r i However, this is based on an assumption of

uni-form stresses which is not the case for small r i Also note that weight also goes down with

11.5(106) = 1.0667(10−2) rad/in= 0.611◦/in Ans.

4-37 Separate strips: For each 1 /16 in thick strip,

4.65 4.60

4.70 4.75 4.80 4.85

For each strip,

d4G θ

(π/16) τd3

= π360

d G θ τ

= π360

Square has greater θ by a factor of 1.13 Ans.

4-46 Text Eq (4-43) gives

Plot is a gentle convex-upward curve Roark uses a polynomial, which in our notation is

y  1.867x  3.061

2+ 52152 = 11 792 psi Ans.

4-50 As shown in Fig 4-34, the maximum stresses occur at the inside fiber where r = r i

There-fore, from Eq (4-51)

(120)2− (110)2 + 1

= 15.0 MPa Ans.

ravp

t

F

4-55 From Table A-20, S y = 57 kpsi; also, r o = 0.875 in and r i = 0.625 in

Solving, gives p o = 11 200 psi Ans.

4-56 From Table A-20, S y = 57 kpsi; also r o = 1.1875 in, r i = 0.875 in.

solving gives p i = 13 510 psi Ans.

4-57 Since σ t andσ r are both positive and σ t > σ r

− 1+ 3(0.24)

3+ 0.24 (0.0125)2

(10)− 6

δ o= 2.25(104)30(106) 0.5

Eqs (d ) and (e) just above Eq (4-59)

pmax = 11 576 psi Ans.

Eq (4-51) for outer member at r = 1 in

(a) Axial resistance

Normal force at fit interface

N = pA = p(2π Rl) = 2πpRl

Fully-developed friction force

F ax = f N = 2π f pRl Ans.

(b) Torsional resistance at fully developed friction is

T = f RN = 2π f pR2l Ans.

4-72 d = 1 in, r i = 1.5 in, r o = 2.5 in

From Table 4-5, for R = 0.5 in,

Section BB: Abscissa angle θ of line of radius centers is

4-75 Find the resultant of F1and F2

507 142

2000 lbf •in

A = (6 − 2 − 1)(0.75) = 2.25 in2

r c = 6+ 2

2 = 4 inSimilar to Prob 4-75,

(b) For the centroid, Eq (4-70) gives,

For I = 1 To 1000 Step 1Sum = Sum + DS / R

R = R + DSNext Icen = 4 / SumEnd FunctionFor eccentricity, Eq (4-71) gives,

S = S + DS

R = R − DSNext Iecc = SUM1 / SUM2End Function

In the spreadsheet enter the following,

which results in,

S = S + DSNext I

ecc = SUM1/SUM2End Function

4-80 100 and 1000 elements give virtually the same results as shown below.

Visual basic program for 100 elements:

Function ecc(RC)

DS = 1.6 / 100

S = −0.8 + DS / 2SUM1 = 0

SUM2 = 0For I = 1 To 25 Step 1SUM1 = SUM1 + DS * S / (RC − S)SUM2 = SUM2 + DS / (RC − S)

S = S + DSNext I

0.4"R

0.4"

0.4"

For I = 1 To 50 Step 1SUM1 = SUM1 + DS * S * (1 − 2 * Sqr(0.4 ^ 2 − S ^ 2)) / (RC − S)SUM2 = SUM2 + DS * (1 − 2 * Sqr(0.4 ^ 2 − S ^ 2)) / (RC − S)

S = S + DSNext I

For I = 1 To 25 Step 1SUM1 = SUM1 + DS * S / (RC − S)SUM2 = SUM2 + DS / (RC − S)

S = S + DSNext

ecc = SUM1 / SUM2End Function

and so, σ z = −251F1/3 MPa Ans.

τmax = 0.3(251)F1/3 = 75.3F1/3MPa Ans

4-84 Good class problem

4-85 From Table A-5, ν = 0.211

τmax

pmax = 1

2(σ1− σ3) = 1

2(1− 0.711) = 0.1445

4-86 From Table A-5:ν1 = 0.211, ν2 = 0.292, E1 = 14.5(106) psi, E2 = 30(106) psi, d1 = 6 in,

d2= ∞, l = 2 in

2(800)

0 .4& # 34; R

0 .4& # 34;

0 .4& # 34;