Mecanismos Solucionario
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Transcript of Mecanismos Solucionario
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Problem 11-02
The joists of a floor in a warehouse are to be selected using square timber beams made of oak. If each
beam is to be designed to carry 1.5 kN/m over a simply supported span of 7.5 m, determine thedimension a of its square cross section to the nearest multiples of 5mm. The allowable bending stress
is σ allow = 32 MPa and the allowable shear stress is τ allow = 0.875 MPa.
Given: σallow 32MPa:= L 7.5m:=
τallow 0.875MPa:= w 1.5kN
m:=
Solution:
Support Reactions : By symmetry, R L=R R =R
ΣF y=0; 2R w L⋅− 0= R 0.5w L⋅:=
Maximum Moment and Shear:
Vmax R := Vmax 5.63 kN=
Mmax R 0.5L( )⋅ w 0.5L( )⋅ 0.25L( )⋅−:= Mmax 10.55 kN m⋅=
Section Property : Ia4
12= Qmax 0.5a a⋅( ) 0.25⋅ a=
Bending Stress: cmax 0.5a=
a
36Mmax
σallow
:=σmax
M cmax⋅
I= σallow
12Mmax 0.5a( )⋅
a4
=
a 125.52 mm= (Use 130mm) Ans
Shear Stress : Ia4
12:= Qmax 0.5a a⋅( ) 0.25⋅ a:=
τmax
Vmax Qmax⋅
I a⋅:= τmax 0.536 MPa=
< τallow =0.875 MPa (O.K.!)
x 0 0.01 L⋅, L..:= V x( ) R w x⋅−( )1
kN⋅:= M x( ) R x⋅ w x⋅ 0.5x( )⋅−[ ]
1
kN m⋅:=
0 2 4 6
Distance (m)
S h e a r ( k N )
V x( )
x0 2 4 6
5
10
Distance m)
M o m e n t ( k N - m )
M x( )
x
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Problem 11-12
Determine the minimum width of the beam to the nearest multiples of 5mm that will safely support the
loading of P = 40 kN. The allowable bending stress is σ allow = 168 MPa, and the allowable shear stress
is τ allow = 105 MPa.
Given: σallow 168MPa:= L1 2m:=
τallow 105MPa:= L2 2m:=
P 40kN:= h 150mm:=
Solution: L L1 L2+:=
Support Reactions : Given
+ ΣF y=0; A B+ P− 0= (1)
ΣΜ B=0; A L1⋅ P L⋅− 0= (2)
Solving Eqs. (1) and (2): Guess A 1kN:= B 1kN:=
A
B
⎛ ⎝
⎞ ⎠
Find A B,( ):=A
B
⎛ ⎝
⎞ ⎠
80
40−
⎛ ⎝
⎞ ⎠
kN=
Maximum Moment and Shear:
Vmax P:= Vmax 40kN=
Mmax P L1⋅:= Mmax 80kN m⋅=
Section Property : I b h
3⋅
12=
SxI
0.5h= Sx
b h2
⋅
6=
Qmax 0.5h b⋅( ) 0.25⋅ h=
Bending Stress: Assume bending controls the design.
Sreq'd
Mmax
σallow=
b h2
⋅
6
Mmax
σallow= b
6Mmax
h2( )σallow
:=
b 126.98 mm= (Use 130mm) Ans
Check Shear : I b h
3⋅
12
:= Qmax
0.5h b⋅( ) 0.25⋅ h:=
τmax
Vmax Qmax⋅
I b⋅:= τmax 3.150 MPa=
< τallow =105 MPa (O.K.!)
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x1 0 0.01 L1⋅, L1..:= x2 L1 1.01 L1⋅, L..:=
V1 x1( ) P−( )1
kN⋅:= V2 x2( ) P− A+( )
1
kN⋅:=
M1 x1
( )P− x1⋅
( )
1
kN m⋅
⋅:=
M2 x2( ) P− x2⋅ A x2 L1−( )⋅+1
kN m⋅⋅:=
0 2 450
0
50
Distance (m)
S h e a r ( k N )
V1 x1( )
V2 x2( )
x1 x2,0 2 4
100
50
0
Distane (m)
M o m e
n t ( k N - m )
M1 x1( )
M2
x2( )
x1 x2,
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Problem 11-21
The steel beam has an allowable bending stress σ allow = 140 MPa and an allowable shear stress of
τ allow = 90 MPa. Determine the maximum load that can safely be supported.
Given: σallow
140MPa:= a 2m:=
τallow 90MPa:=
bf 120mm:= d f 20mm:=
tw 20mm:= d w 150mm:=
Solution: L 2a:=
Section Property : h d f d w+:=
yc
Σ yi Ai⋅( )⋅Σ Ai( )⋅
=
yc bf d f ⋅( ) 0.5d f ( )⋅ tw d w⋅( ) 0.5d w d f +( )⋅+
bf d f ⋅( ) tw d w⋅( )+:= yc 57.22 mm=
I1
12 bf ⋅ d f
3⋅ bf d f ⋅( ) 0.5d f yc−( )
2⋅+
1
12tw⋅ d w
3⋅ tw d w⋅( ) 0.5d w d f + yc−( )
2⋅+⎡
⎣⎤⎦
+:=
I 15338333.33mm4
=
Qmax h yc−( ) tw⋅ 0.5 h yc−( )⋅⋅:= Qmax 127188.27mm3
=
Support Reactions : By symmetry, R R = - P
+ ΣF y=0; R c P− P− 0=
R c 2P=
Maximum Load : Assume failure due to bending moment.
Mmax P a⋅= cmax h yc−:=
σallow
Mmax cmax⋅
I= σallow
P a⋅ h yc−( )⋅I
= PI σallow( )a h yc−( )⋅
:=
P 9.52 kN= Ans
Check Shear : Vmax P:= R c 2P:=
τmax
Vmax Qmax⋅
I tw⋅:= τmax 3.947 MPa=
< τallow =90 MPa (O.K.!)
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x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, L..:=
V1 x1( ) P−( )1
kN⋅:= V2 x2( ) P− R c+( )
1
kN⋅:=
M1 x1
( )P− x1⋅
( )
1
kN m⋅
⋅:=
M2 x2( ) P− x2⋅ R c x2 a−( )⋅+1
kN m⋅⋅:=
0 2 4
10
0
10
Distance (m)
S h e a r ( k N )
V1 x1( )
V2 x2( )
x1 x2,0 2 4
20
10
0
Distane (m)
M o m e
n t ( k N - m )
M1 x1( )
M2
x2( )
x1 x2,
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Problem 11-39
Solve Prob. 11-38 using the maximum-distortion-energy theory of failure with σ allow = 180 MPa.
Given: a 300mm:= r 50mm:= σallow 180MPa:=
PB 5kN:= PC 5kN:=
Solution: L 3a:=
Support Reactions :
In x-z plane : ΣF z=0; Az DZ+ PB− 0= (1)
ΣΜ D=0; Az 3a( )⋅ PB 2a( )⋅− 0= (2)
Solving Eqs. (1) and (2):
Az2
3PB:= Az 3.3333 kN=
Dz PB Az−:= Dz 1.6667 kN=
In x-y plane : ΣF y=0; Ay Dy+( ) PC− 0= (3)
ΣΜ D=0; PC− a⋅ Ay 3a( )⋅+ 0= (4)
Solving Eqs. (3) and (4):
Ay1
3PC:= Ay 1.6667 kN=
Dy PC Ay−:= Dy 3.3333 kN=
Torsion occurs in segment BC : TBC PB( ) r ⋅:= TBC 0.250 kN m⋅=
Critical Section : Located just to the left of gear C and just to the right of gear B, where
My Dz a⋅:= Mz Dy a⋅:= M My2
Mz2
+:= M 1.118 kN m⋅=
T TBC:= T 0.250 kN m⋅=
Maximum Distortion Energy Theory : Applying Eq. 9-5:
σ1 0.5 σx' σy'+( ) 0.5 σx' σy'+( ) 2
τx'y'2
++=
σ2 0.5 σx' σy'+( ) 0.5 σx' σy'+( )⋅ 2
τx'y'2
+−=
where σy'
0:=
σx'M c⋅
I=
M c⋅
π
4c4
⋅
=
4M c⋅
π c4
⋅=
τx'y'T c⋅
J=
T c⋅
π
2c4
⋅
=
2T c⋅
π c4
⋅=
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Let a' 0.5σx'= and b' 0.5σx'( )2
τx'y'2
+=
Then σ12
a' b'+( )2
= σ22
a 'b'−( )2
=
σ1
σ2
⋅ a' b'
+( ) a' b'
−( )
⋅= a'
2
b'
2−=
σ12
σ1 σ2⋅− σ22
+ a' b'+( )2
a'2
b'2
−( )− a' b'−( )2+= a'2 3b'2+=
Hence σ12
σ1 σ2⋅− σ22
+ σallow2
=
0.5σx'( )2
3 0.5σx'( )2
τx'y'2
+2
+ σallow2
=
σx'2
3τx'y'2
+ σallow2
=
4M c⋅
π c4
⋅
⎛
⎝
⎞
⎠
2
3
2T c⋅
π c4
⋅
⎛
⎝
⎞
⎠
2
+ σallow2
=
c
6
16M2
12T2
+
π2
σallow2
⋅:=
c 20.05 mm=
d o 2c:= d o 40.09 mm= Use d o 41mm= Ans
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x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 3a..:=
My1 x1( )Az x1( )⋅kN m⋅
:= My2 x2( ) Az x2( )⋅ PB x2 a−( )⋅− 1
kN m⋅⋅:=
My3 x3( ) Az x3( )⋅ PB x3 a−( )⋅− 1
kN m⋅
⋅:=
0 0.2 0.4 0.6 0.80
0.5
1
Distance (m)
M o m e n t ( k N - m )
My1 x1( )
My2 x2( )
My3 x3( )
x1 x2, x3,
Mz1 x1( )Ay− x1( )⋅kN m⋅
:= Mz2 x2( )Ay− x2( )⋅kN m⋅
:= Mz3 x3( ) Ay− x3( )⋅ PC x3 2a−( )⋅+ 1
kN m⋅⋅:=
0 0.2 0.4 0.6 0.8
1
0.5
0
Distance (m)
M
o m e n t ( k N - m )
Mz1 x1( )
Mz2 x2( )
Mz3 x3( )
x1 x2, x3,
Mx1 x1( ) 0:= Mx2 x2( )r − PB⋅
kN m⋅:= Mx3 x3( ) 0:=
0 0.2 0.4 0.6 0.80.5
0
Distance (m)
M o m e n t ( k
N - m )
Mx1 x1( )
Mx2 x2( )Mx3 x3( )
x1 x2, x3,
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L2 1.5m:=
Given:P1 1250N
:=P2 750N
:=r 150mm
:=
L1 0.3m:= L3 0.6m:=
τallow 84MPa:=
Solution: L L1 L2+ L3+:=
Support Reactions : Ps P1 P2+:=
Ans
In y-z plane :ΣF z=0; Az BZ+ P1− P2− 0= (1)
ΣΜ B=0; Az L⋅ P1 P2+( ) L2 L3+( )⋅− 0= (2)
Solving Eqs. (1) and (2):
Az1
LP1 P2+( ) L2 L3+( )⋅:= Az 1750 N=
Bz P1 P2+ Az−:= Bz 250 N=
In x-y plane : ΣF x=0; Ax Bx+ P1− P2− 0= (3)
ΣΜ B=0; Ax L⋅ P1 P2+( ) L3( )⋅− 0= (4)
Solving Eqs. (3) and (4):
Ax1
LP1 P2+( ) L3( )⋅:= Ax 500 N=
Bx P1 P2+ Ax−:= Bx 1500 N=
Torsion occurs in segment DC : T P1 P2−( ) r ⋅:= T 75 N m⋅=
Critical Section : Located just to the left of point C.
Mx Bx L3⋅:= Mz Bz L3⋅:= M Mx2
Mz2
+:= M 912.41 N m⋅=
Maximum Shear Stress Theory :
c
32
π τallow⋅M
2T
2+⋅:= c 19.07 mm=
d o 2c:= d o 38.15 mm= Use d o 39mm=
Problem 11-42
The pulleys attached to the shaft are loaded as shown. If the bearings at A and B exert only horizontal
and vertical forces on the shaft, determine the required diameter of the shaft to the nearest mm. using
the maximum-shear-stress theory of failure. τ allow = 84 MPa.
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y1 0 0.01 L1⋅, L1..:= y2 L1 1.01 L1⋅, L1 L2+..:= y3 L1 L2+( ) 1.01 L1 L2+( )⋅, L..:=
Mz1 y1( ) Az y1⋅( )1
N m⋅⋅:=
Mz2 y2( ) Az y2⋅ Ps y2 L1−( )⋅−1
N m⋅⋅:=
Mz3 y3( ) Az y3⋅ Ps y3 L1−( )⋅−1
N m⋅
⋅:=
0 1 20
500
Distance (m)
M z ( N - m ) Mz1 y1( )
Mz2 y2( )
Mz3 y3( )
y1 y2, y3,
Mx1 y1( )Ax y1⋅
N m⋅:= Mx2 y2( )
Ax y2⋅
N m⋅:= Mx3 y3( ) Ax y3⋅ Ps y3 L1− L2−( )⋅−
1
N m⋅⋅:=
0 1 20
500
1000
Distance (m)
M x ( N - m ) Mx1 y1( )
Mx2 y2( )
Mx3 y3( )
y1 y2, y3,
My1 y1( ) 0:= My2 y2( )T
N m⋅:= My3 y3( ) 0:=
0 1 2
0
50
100
Distance (m)
M y ( N
- m ) My1 y1( )
My2 y2( )My3 y3( )
y1 y2, y3,
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Problem 11-45
The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowable
shear stress for the shaft is τ allow = 35 MPa, determine to the nearest millimeter the smallest diameter of
the shaft that will support the loading. Use the maximum-shear-stress theory of failure.
Given: La 250mm:= L b 500mm:= τallow 35MPa:=
r D 150mm:= r C 100mm:= θ 30deg:=
PD 0.20kN:= PC 0.35kN:=
∆PD 0.10kN:= ∆PC 0.15kN:=
Solution: L La L b+ La+:=
Support Reactions :
In y-z plane : ΣF z=0; Az BZ+ PC sin θ( )⋅− PD sin θ( )⋅− 0= (1)
ΣΜ A=0; Bz L( )⋅ PC sin θ( )⋅ L La−( )⋅− PD sin θ( )⋅ La⋅− 0= (2)
Solving Eqs. (1) and (2):
Bz PCL La−
L⋅ PD
LaL
⋅+⎛ ⎝
⎞ ⎠
sin θ( ):= Bz 0.15625 kN=
Az PC PD+( ) sin θ( )⋅ Bz−:= Az 0.11875 kN=
In x-y plane : ΣF x=0; Ax Bx+ PC cos θ( )⋅− PD cos θ( )⋅+ 0= (3)
ΣΜ A=0; PC cos θ( )⋅ L La−( )⋅ PD cos θ( )⋅ La⋅− Bx L⋅− 0= (4)
Solving Eqs. (3) and (4):
Bx PC
L La−
L⋅ PD
La
L⋅−
⎛
⎝
⎞
⎠cos θ( ):= Bx 0.18403 kN=
Ax PC PD−( ) cos θ( )⋅ Bx−:= Ax 0.05413− kN=
Torsion occurs in segment CD : TCD ∆PC( ) r C⋅:= TCD 0.015 kN m⋅=
Critical Section : Located just to the left of gear C, where.
Mx Bz La⋅:= Mz Bx La⋅:= M Mx2
Mz2
+:= M 0.060354 kN m⋅=
T TCD:= T 0.015 kN m⋅=
Maximum Shear Stress Theory :
c
3
2π τallow⋅
M2 T2+⋅:= c 10.42 mm=
d o 2c:= d o 20.84 mm= Use d o 21mm= Ans
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y1 0 0.01 La⋅, La..:= y2 La 1.01 La⋅, La L b+( )..:= y3 La L b+( ) 1.01 La L b+( )⋅, L..:=
Mx1 y1( )Az y1( )⋅kN m⋅
:= Mx2 y2( ) Az y2⋅ PD sin θ( )⋅ y2 La−( )⋅−1
kN m⋅:=
Mx3 y3( ) Az y3⋅ PD sin θ( )⋅ y3 La−( )⋅− PC sin θ( )⋅ y3 La− L b−( )⋅−1
kN m⋅⋅:=
0 0.2 0.4 0.6 0.80
0.02
0.04
Distance (m)
M o m e n t ( k N - m
)Mx1 y1( )Mx2 y2( )
Mx3 y3( )
y1 y2, y3,
Mz1 y1( )Ax y1( )⋅kN m⋅
:= Mz2 y2( ) Ax y2⋅ PD cos θ( )⋅ y2 La−( )⋅+1
kN m⋅:=
Mz3 y3( ) Ax y3⋅ PD cos θ( )⋅ y3 La−( )⋅+ PC cos θ( )⋅ y3 La− L b−( )⋅−1
kN m⋅⋅:=
0 0.2 0.4 0.6 0.8
0
0.05
Distance (m)
M o m e n t (
k N - m )
Mz1 y1( )
Mz2 y2( )Mz3 y3( )
y1 y2, y3,
My1 y1( ) 0:= My2 y2( )
T
kN m⋅:= My3 y3( ) 0:=
0 0.2 0.4 0.6 0.8
0
0.02
Distance (m)
M o m e n t ( k N - m )
My1 y1( )
My2 y2( )
My3 y3( )
y1 y2, y3,
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Problem 12-03
Determine the equation of the elastic curve for the beam using the x coordinate that is valid for
Specify the slope at A and the beam's maximum deflection. EI is constant.2/0 L x
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Problem 12-16
A torque wrench is used to tighten the nut on a bolt. If the dial indicates that a torque of 90 N·m is
applied when the bolt is fully tightened, determine the force P acting at the handle and the distance s theneedle moves along the scale. Assume only the portion AB of the beam distorts. The cross section is
square having dimensions of 12 mm by 12 mm. E = 200 GPa.
Given: b 12mm:= h 12mm:= L 0.45m:=
E 200GPa:= δ 75mm:= R 0.3m:=
Tz 90N m⋅:=
Solution:
Equations of Equilibrium :
ΣF y=0; Ay P− 0= (1)
ΣΜ B=0; Tz P L⋅− 0= (2)
Solving Eqs. (1) and (2): PTz
L:= Ay P:= Ay 200N=
P 200 N= Ans
Moment Function : M x( ) Ay x⋅ Tz−=
Section Property : I b h
3⋅
12:=
Slope and Elastic Curve :
E I⋅d
2v⋅
dx2
⋅ M x( )=
E I⋅d
2v
dx2
⋅ Ay x⋅ Tz−=
E I⋅dv
dx⋅
Ay x2
⋅
2Tz x⋅− C1+= (1)
E I⋅ v⋅Ay x
3⋅
6
Tz x2
⋅
2− C1 x⋅+ C2+= (2)
Boundary Conditions : Due to symmetry, dv/dx=0 at x=0, and v=0 at x=0.
From Eq. (1): 0 0 0− C1+= C1 0:=
From Eq. (2): 0 0 0− 0+ C2+= C2 0:=
The Elastic Curve : Substitute the values of C1 and C2 into Eq. (2),
E I⋅ v⋅Ay x
3⋅
6
Tz x2
⋅
2−= v
x2
6E I⋅Ay x 3Tz−( )⋅= (3)
At x=R, v=-s. From Eq. (3),
sR
2−
6E I⋅Ay R 3Tz−( )⋅:= s 9.11 mm= Ans
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Problem 12-17
The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft and at B
by a thrust bearing that exerts horizontal and vertical reactions on the shaft. Draw the bending-momentdiagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft's
centerline. Determine the equations of the elastic curve using the coordinates x1 and x2 . EI is constant.
Given: h 60mm:= L1 150mm:=
P 5kN:= L2 400mm:=
Solution: L L1 L2+:=
Support Reactions:
+
ΣF y=0; A B+ 0= (1)
ΣΜ B=0; P h⋅ A L2⋅+ 0= (2)
Solving Eqs. (1) and (2): AP− h⋅
L2:= A 0.75− kN=
B A−:= B 0.75 kN=
Moment Function : M1 x1( ) P h⋅:=
M2 x2( ) B x2⋅:=
Section Property : EI kN m2
⋅:=
Slope and Elastic Curve :
EId
2v1⋅
dx12
⋅ M1 x1( )= EId
2v2⋅
dx22
⋅ M2 x2( )=
EI
d 2v1
dx12
⋅ P h⋅= EI
d 2
v2
dx22
⋅ B x2⋅=
EIdv1
dx1⋅ P h⋅( ) x1⋅ C1+= (1) EI
dv2
dx2⋅
B
2x2
2⋅ C3+= (3)
EI v1⋅P h⋅ x1
2⋅
2C1 x1⋅+ C2+= (2) EI v2⋅
B x23
⋅
6C3 x2⋅+ C4+= (4)
Boundary Conditions :
v1=0 at x1=0.15m, From Eq. (2): 0P h⋅ 0.15m( )
2⋅
2
C1 0.15m( )⋅+ C2+= (5)
v2=0 at x2=0, From Eq. (4): 0 0 0+ C4+= C4 0:= Ans
v2=0 at x2=0.4m, From Eq. (4): 0B 0.4m( )
3⋅
6C3 0.4m( )⋅+ C4+=
C3B
6− 0.4m( )
2⋅:= C3 0.02− kN m
2⋅= Ans
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Continuity Condition:
dv1/dx1= - dv2/dx2 at A (x1=0.15m and x2=0.4m)
From Eqs. (1) and (3), P h⋅ 0.15m( )⋅ C1+B
20.4m( )
2⋅ C3+=
C1
B
2 0.4m( )
2⋅
C3+
P h⋅
0.15m( )⋅−:=
C1 0.005−
kN m
2⋅= Ans
From Eq. (5):C2
P h⋅ 0.15m( )2
⋅
2− C1 0.15m( )⋅−:= C2 0.00263− kN m
3⋅= Ans
The Elastic Curve : Substitute the values of C1 and C2 into Eq. (2), and C3 and C4 into Eq. (4),
v11
EI
P h⋅
2x1
2⋅ C1 x1⋅+ C2+
⎛ ⎝
⎞ ⎠
= Ans
v21
EI
B
6x2
3⋅ C3 x2⋅+ C4+
⎛ ⎝
⎞ ⎠
= Ans
BMD :
x'1 0 0.01 L1⋅, L1..:= x'2 L1 1.01 L1⋅, L..:=
M'1 x'1( )P h⋅
kN m⋅:= M'2 x'2( ) P h⋅ A x'2 L1−( )⋅+
1
kN m⋅⋅:=
0 0.2 0.40
0.2
0.4
Distane (m)
M o m e
n t ( k N - m )
M'1 x'1( )
M'2
x'2( )
x'1 x'2,