mecanismos solucionario

8/18/2019 Mecanismos Solucionario

1/16

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


2/16

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.!)


3/16

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,


4/16

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.!)


5/16

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,


6/16

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)


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

⋅=


7/16

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


8/16

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,


9/16

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)


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)


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.


10/16

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,


11/16

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)


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)


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


12/16

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,


13/16

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


14/16

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


15/16

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


16/16

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,

mecanismos solucionario

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