Cap­tulo 04 carga e anlise de tens£o

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  • 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) ]