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### Transcript of Cap­tulo 04 carga e anlise de tens£o

• Chapter 4

4-1

1

RC

RA RB

RD

C

A B

W

D

1

23

RB

RA

W

RB

RC

RA2

1

W

RA

RBx

RBx RBy

RBy

RB

2

1

1

Scale ofcorner magnified

W

A

B

(e)

(f)

(d)

W

A

RA RB

B

1

2

W

A

RA RB

B

11

2

(a) (b)

(c)

shi20396_ch04.qxd 8/18/03 10:35 AM Page 50

• Chapter 4 51

4-2(a) RA = 2 sin 60 = 1.732 kN Ans.

RB = 2 sin 30 = 1 kN Ans.

(b) S = 0.6 m = tan1 0.6

0.4 + 0.6 = 30.96

RAsin 135

= 800sin 30.96

RA = 1100 N Ans.RO

sin 14.04= 800

sin 30.96 RO = 377 N Ans.

(c)RO = 1.2tan 30 = 2.078 kN Ans.

RA = 1.2sin 30

= 2.4 kN Ans.

(d) Step 1: Find RA and REh = 4.5

tan 30= 7.794 m

+

MA = 09RE 7.794(400 cos 30) 4.5(400 sin 30) = 0RE = 400 N Ans.

Fx = 0 RAx + 400 cos 30 = 0 RAx = 346.4 NFy = 0 RAy + 400 400 sin 30 = 0 RAy = 200 N

RA =

346.42 + 2002 = 400 N Ans.

D

C

h

B

y

E xA

4.5 m

9 m

400 N

3

42

30

60

RAyRA

RAx

RE

1.2 kN

60

RARO

6090

30

1.2 kN

RARO

45 30.96 14.04

13530.96

30.96

800 N

RA

RO

O

0.4 m

45

800 N

0.6 m

A

s

RA

RO

B

6090

30

2 kNRA

RB

2

1

2 kN

6030

RA

RB

shi20396_ch04.qxd 8/18/03 10:35 AM Page 51

• 52 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

Step 2: Find components of RC on link 4 and RD+

MC = 0

400(4.5) (7.794 1.9)RD = 0 RD = 305.4 N Ans.Fx = 0 (RCx )4 = 305.4 NFy = 0 (RCy)4 = 400 N

Step 3: Find components of RC on link 2Fx = 0

(RCx )2 + 305.4 346.4 = 0 (RCx )2 = 41 NFy = 0

(RCy)2 = 200 N

4-3(a)

+

M0 = 018(60) + 14R2 + 8(30) 4(40) = 0R2 = 71.43 lbf

Fy = 0: R1 40 + 30 + 71.43 60 = 0R1 = 1.43 lbf

M1 = 1.43(4) = 5.72 lbf inM2 = 5.72 41.43(4) = 171.44 lbf inM3 = 171.44 11.43(6) = 240 lbf inM4 = 240 + 60(4) = 0 checks!

4" 4" 6" 4"

1.4341.43

11.43

60

40 lbf 60 lbf

30 lbf

x

x

x

OA B C D

y

R1 R2

M1M2 M3

M4

O

V (lbf)

M(lbf in)

O

CC

DB

A

B D

E

305.4 N

346.4 N

305.4 N

41 N

400 N

200 N400 N

200 N

400 N

Pin C30

305.4 N

400 N

400 N200 N

41 N

305.4 N

200 N

346.4 N

305.4 N

(RCx)2

(RCy)2

C

B

A

2

400 N

4

RD

(RCx)4

(RCy)4

D

C

E

Ans.

shi20396_ch04.qxd 8/18/03 10:35 AM Page 52

• Chapter 4 53

(b)

Fy = 0R0 = 2 + 4(0.150) = 2.6 kN

M0 = 0M0 = 2000(0.2) + 4000(0.150)(0.425)

= 655 N m

M1 = 655 + 2600(0.2) = 135 N mM2 = 135 + 600(0.150) = 45 N mM3 = 45 + 12600(0.150) = 0 checks!

(c) M0 = 0: 10R2 6(1000) = 0 R2 = 600 lbfFy = 0: R1 1000 + 600 = 0 R1 = 400 lbf

M1 = 400(6) = 2400 lbf ftM2 = 2400 600(4) = 0 checks!

(d) +

MC = 010R1 + 2(2000) + 8(1000) = 0R1 = 1200 lbf

Fy = 0: 1200 1000 2000 + R2 = 0R2 = 1800 lbf

M1 = 1200(2) = 2400 lbf ftM2 = 2400 + 200(6) = 3600 lbf ftM3 = 3600 1800(2) = 0 checks!

2000 lbf1000 lbf

R1

O

O

M1M2

M3

R2

6 ft 2 ft2 ftA B C

y

M

1200

1800

200

x

x

x

6 ft 4 ftA

O

O

O

B

600

M1

M2

V (lbf)

1000 lbfy

R1 R2

400

M(lbf ft)

x

x

x

V (kN)

150 mm200 mm 150 mm

2.6

655

M(Nm)

0.6

M1M2

M3

2 kN 4 kN/my

AO

O

O O

B C

ROMO

x

x

x

shi20396_ch04.qxd 8/18/03 10:35 AM Page 53

• 54 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

(e) +

MB = 07R1 + 3(400) 3(800) = 0R1 = 171.4 lbf

Fy = 0: 171.4 400 + R2 800 = 0R2 = 1371.4 lbf

M1 = 171.4(4) = 685.7 lbf ftM2 = 685.7 571.4(3) = 2400 lbf ftM3 = 2400 + 800(3) = 0 checks!

(f) Break at A

R1 = VA = 1240(8) = 160 lbf

+

MD = 012(160) 10R2 + 320(5) = 0R2 = 352 lbf

Fy = 0160 + 352 320 + R3 = 0R3 = 128 lbf

M1 = 12160(4) = 320 lbf in

M2 = 320 12160(4) = 0 checks! (hinge)M3 = 0 160(2) = 320 lbf inM4 = 320 + 192(5) = 640 lbf inM5 = 640 128(5) = 0 checks!

40 lbf/in

V (lbf)

O

O

160

160128

192

M

320 lbf

160 lbf 352 lbf 128 lbf

M1

M2

M3

M4

M5

x

x

x

8"

5"2"

5"

40 lbf/in

160 lbf

O

A

y

B DC

A

320 lbf

R2 R3

R1 VA

AO

O

O

C

M

V (lbf)800

171.4571.4

3 ft 3 ft4 ft

800 lbf400 lbf

B

y

M1M2

M3

R1 R2

x

x

x

shi20396_ch04.qxd 8/18/03 10:35 AM Page 54

• Chapter 4 55

4-4(a) q = R1x1 40x 41 + 30x 81 + R2x 141 60x 181

V = R1 40x 40 + 30x 80 + R2x 140 60x 180 (1)M = R1x 40x 41 + 30x 81 + R2x 141 60x 181 (2)

for x = 18+ V = 0 and M = 0 Eqs. (1) and (2) give0 = R1 40 + 30 + R2 60 R1 + R2 = 70 (3)0 = R1(18) 40(14) + 30(10) + 4R2 9R1 + 2R2 = 130 (4)

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

x = 4+: V = 1.43 40 = 41.43, M = 1.43xx = 8+: V = 1.43 40 + 30 = 11.43

M = 1.43(8) 40(8 4)1 = 171.44x = 14+: V = 1.43 40 + 30 + 71.43 = 60

M = 1.43(14) 40(14 4) + 30(14 8) = 240 .x = 18+: V = 0, M = 0 See curves of V and M in Prob. 4-3 solution.

(b) q = R0x1 M0x2 2000x 0.21 4000x 0.350 + 4000x 0.50V = R0 M0x1 2000x 0.20 4000x 0.351 + 4000x 0.51 (1)M = R0x M0 2000x 0.21 2000x 0.352 + 2000x 0.52 (2)at x = 0.5+ m, V = M = 0, Eqs. (1) and (2) give

R0 2000 4000(0.5 0.35) = 0 R1 = 2600 N = 2.6 kN Ans.R0(0.5) M0 2000(0.5 0.2) 2000(0.5 0.35)2 = 0

with R0 = 2600 N, M0 = 655 N m Ans.With R0 and M0, Eqs. (1) and (2) give the same V and M curves as Prob. 4-3 (note for

V, M0x1 has no physical meaning).(c) q = R1x1 1000x 61 + R2x 101

V = R1 1000x 60 + R2x 100 (1)M = R1x 1000x 61 + R2x 101 (2)

at x = 10+ ft, V = M = 0, Eqs. (1) and (2) give R1 1000 + R2 = 0 R1 + R2 = 1000

10R1 1000(10 6) = 0 R1 = 400 lbf , R2 = 1000 400 = 600 lbf0 x 6: V = 400 lbf, M = 400x6 x 10: V = 400 1000(x 6)0 = 600 lbf

M = 400x 1000(x 6) = 6000 600xSee curves of Prob. 4-3 solution.

shi20396_ch04.qxd 8/18/03 10:35 AM Page 55

• 56 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

(d) q = R1x1 1000x 21 2000x 81 + R2x 101V = R1 1000x 20 2000x 80 + R2x 100 (1)M = R1x 1000x 21 2000x 81 + R2x 101 (2)

At x = 10+ , V = M = 0 from Eqs. (1) and (2)R1 1000 2000 + R2 = 0 R1 + R2 = 3000

10R1 1000(10 2) 2000(10 8) = 0 R1 = 1200 lbf ,R2 = 3000 1200 = 1800 lbf

0 x 2: V = 1200 lbf, M = 1200x lbf ft2 x 8: V = 1200 1000 = 200 lbf

M = 1200x 1000(x 2) = 200x + 2000 lbf ft8 x 10: V = 1200 1000 2000 = 1800 lbf

M = 1200x 1000(x 2) 2000(x 8) = 1800x + 18 000 lbf ftPlots are the same as in Prob. 4-3.

(e) q = R1x1 400x 41 + R2x 71 800x 101V = R1 400x 40 + R2x 70 800x 100 (1)M = R1x 400x 41 + R2x 71 800x 101 (2)

at x = 10+ , V = M = 0R1 400 + R2 800 = 0 R1 + R2 = 1200 (3)

10R1 400(6) + R2(3) = 0 10R1 + 3R2 = 2400 (4)Solve Eqs. (3) and (4) simultaneously: R1 = 171.4 lbf, R2 = 1371.4 lbf0 x 4: V = 171.4 lbf, M = 171.4x lbf ft4 x 7: V = 171.4 400 = 571.4 lbf

M = 171.4x 400(x 4) lbf ft = 571.4x + 16007 x 10: V = 171.4 400 + 1371.4 = 800 lbf

M = 171.4x 400(x 4) + 1371.4(x 7) = 800x 8000 lbf ftPlots are the same as in Prob. 4-3.

(f) q = R1x1 40x0 + 40x 80 + R2x 101 320x 151 + R3x 20V = R1 40x + 40x 81 + R2x 100 320x 150 + R3x 200 (1)M = R1x 20x2 + 20x 82 + R2x 101 320x 151 + R3x 201 (2)M = 0 at x = 8 in 8R1 20(8)2 = 0 R1 = 160 lbfat x = 20+ , V and M = 0

160 40(20) + 40(12) + R2 320 + R3 = 0 R2 + R3 = 480160(20) 20(20)2 + 20(12)2 + 10R2 320(5) = 0 R2 = 352 lbf

R3 = 480 352 = 128 lbf0 x 8: V = 160 40x lbf, M = 160x 20x2 lbf in8 x 10: V = 160 40x + 40(x 8) = 160 lbf ,

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

shi20396_ch04.qxd 8/18/03 10:35 AM Page 56

• Chapter 4 57

10 x 15: V = 160 40x + 40(x 8) + 352 = 192 lbfM = 160x 20x2 + 20(x 8) + 352(x 10) = 192x 2240

15 x 20: V = 160 40x + 40(x 8) + 352 320 = 128 lbfM = 160x 20x2 20(x 8) + 352(x 10) 320(x 15)

= 128x + 2560Plots of V and M are the same as in Prob. 4-3.

4-5 Solution depends upon the beam selected.

4-6(a) Moment at center, xc = (l 2a)/2

Mc = w2

[l2

(l 2a) (

l2

)2]= wl

2

(l4

a)

At reaction, |Mr | = wa2/2a = 2.25, l = 10 in, w = 100 lbf/in

Mc = 100(10)2(

104

2.25)

= 125 lbf in

Mr = 100(2.252)

2= 253.1 lbf in Ans.

(b) Minimum occurs when Mc = |Mr |wl2

(l4

a)

= wa2

2 a2 + al 0.25l2 = 0

Taking the positive root

a = 12

[l + l2 + 4(0.25l2) ]