engineering mechanics: statics in si units, 12e · 2019. 2. 19. · 7.3 yayılı yük ,kesme ve...

Internal Forces 7

Engineering Mechanics:

Statics in SI Units, 12e

1

Bölüm Hedefleri

• Bir elemandaki iç kuvvetlerin kesim metoduyla bulunması

• Develop procedure by formulating equations that describe

the internal shear and moment throughout a member

• Analyze the forces and study the geometry of cables

supporting a load

2

Bölüm Özeti

1. Yapı elemanlarında oluşan iç kuvvetler

2. Kesme ve moment denklemleri ve diyagramları

3. Yayılı yük ,kesme kuvvet ve moment arasındaki

ilişkiler

4. Kablolar

3

7.1 Yapı elemanlarında oluşan iç kuvvetler

• Yapı veya makine elemanlarının tasarımı elemana etkiyen

yüklere dayanacak malzeme kullanımı gerektirir.

• Bu iç kuvvetler kesim metoduyla bulunabilir.

4


• N kuvvet bileşeni, kesitte kirişe normal (dik) etki

eder.Normal veya eksenel kuvvet denir

• V, kesite paralel etkir ve kesme kuvveti denir.

• Moment M eğilme momentidir.

5


• 3 boyutlu durumlarda,kesitte genel iç kuvvet ve moment

bileşkeleri etkir.

• Ny normal kuvvet, ve Vx ve Vz kesme kuvvet bileşenleri

• My burulma momenti,ve Mx ve Mz eğilme moment

bileşenleri

6


Analiz işlemi

Mesnet reaksiyonlar

• Kesimden önce , eleman mesnet reaksiyonlarını bulunuz.

• Denge denklemleri kullanılarak kesim yerinde iç kuvvetler

hesaplanır.

Serbest cisim digaramları

• Eleman üzerine etkiyen bütün yayılı yükler,kuvvet çifti

momentleri ve kuvvetleri tam yerinde tutulur.

• Kesitin bir tarafının SCD çizilir kesitteki kuvvet bileşenleri

belirtilir. 7


Analiz işlemi

SCD (Continue)

• Kuvvet ,moment ve bileşke momentlerin z, y, z bileşenleri

SCD gösterilmelidir.

• Kesitte sadece N, V ve M etkir.

• Determine the sense by inspection

Denge denklemleri

• Momentler kesitte toplanmalıdır.

• Sonuç negatifse yön pozitiftir.

8

Örnek 7.3

İki elemanlı çerçevede B noktasında iç kuvvet,kesme kuvveti

ve eğilme momentini bulunuz.

9

Çözüm

Mesnet tepkileri

SCD

AC elemanı

∑ MA = 0; -400kN(4m) + (3/5)FDC(8m)= 0

FDC = 333.3kN

+→∑ Fx = 0; -Ax + (4/5)(333.3kN) = 0

Ax = 266.7kN

+↑∑ Fy = 0; Ay – 400kN + 3/5(333.3kN) = 0

Ay = 200kN 10

Çözüm

Mesnet tepkileri

AB elemanı

+→∑ Fx = 0; NB – 266.7kN = 0

NB = 266.7kN

+↑∑ Fy = 0; 200kN – 200kN - VB = 0

VB = 0

∑ MB = 0; MB – 200kN(4m) – 200kN(2m) = 0

MB = 400kN.m

11

Örnek

12

Çözüm

13

örnek

14

7.2 Kesme kuvveti ve moment denklemleri ve

diyagramları

• Kirişler – eksenlerine dik uygulanan yüklemeleri taşımak

için dizayn edilen yapısal elemanlardır.

• Basit mesnetli kiriş ,bir ucunda pimli diğerinde küçük

tekerliklidir.

• Ankastre kiriş bir ucundan sabitlenmiştir diğer ucu ise

serbesttir.

15


diyagramları

Analiz işlemi Mesnet tepkileri

• Kirişe etkiyen bütün reaksiyon kuvvetleri ve momentleri bul.

• Bileşenlere ayır.

Kesme kuvvet ve moment reaksiyonları

• Specify separate coordinates x

• Section the beam perpendicular to its axis

• V obtained by summing the forces perpendicular to the beam

• M obtained by summing moments about the sectioned end

16


diyagramları

Analiz işlemi

Kesme kuvvet ve moment reaksiyonları

• Kesme kuvvet diyagramı(x e göre V) ve (x e göre M) çizilir.

• Kesme kuvvet ve eğilme momenti diyagramlar SCD altına

çizilir.

17

Örnek 7.7

Şaft için kesme kuvvet ve eğilme moment diyagramlarını

çiziniz. A mesnedi itme yatak C mesnedi kayma

18

Çözüm

Mesnet raksiyonları

SCD

mxkNMM

kNVFy

.5.2;0

5.2;0

mkNxM

xkNmxkNMM

kNV

VkNkNFy

.)5.210(

0)(5.2)2(5 ;0

5.2

055.2 ;0

19

Çözüm

Kesme kuvvet diyagramı

Internal shear force is always positive within the shaft AB.

Just to the right of B, the shear force

changes sign and remains at

constant value for segment BC.

Moment diagram

Starts at zero, increases linearly to

B and therefore decreases to zero.

20

Örnek

21

Çözüm

22

7.3 Yayılı Yük ,Kesme ve moment arasındaki ilişkiler

Yayılı yük

• Keyfi w = w(x) yayılı yüküne ve bir dizi tekil kuvvet ve

momente maruz AD kirişini dikkate alalım

• Yayılı yük yükleme aşağı doğru etkidiği zaman pozitiftir.

23


Yayılı yük

• Kirişin ∆x uzunluklu küçük bir parçası için SCD kiriş

boyunca tekil kuvvet veya momente maruz olmayan bir x

noktasında seçilmiş.

• Elde edilen sonuçlar tekil yükleme noktalarında

kullanılamaz.

• The internal shear force and

bending moments assumed

positive sense

24


Yayılı yük

• Distributed loading has been replaced by a resultant force

∆F = w(x) ∆x that acts at a fractional distance k (∆x) from

the right end, where 0 < k <1

2)()(

0)()(;0

)(

0)()(;0

xkxwxVM

MMxkxxwMxVM

xxwV

VVxxwVFy

25


Yayılı yük

Vdx

dM

)(xwdx

dV

Slope of the shear diagram

Negative of distributed load intensity

Slope of moment diagram

Shear moment diagram

VdxM BC

dxxwVBC )(

Change in shear Area under

shear diagram

Change in moment Area under shear diagram

26


Kuvvet ve Moment

• FBD of a small segment of the beam

• Change in shear is negative

• FBD of a small segment of the beam located at the couple

moment

• Change in moment is positive

27

Örnek 7.9

Draw the shear and moment diagrams for the overhang beam.

28

Çözüm

The support reactions are shown.

Shear Diagram

Shear of –2 kN at end A of the beam

is at x = 0.

Positive jump of 10 kN at x = 4 m

due to the force.

Moment Diagram

mkN 842004

MMMxx

29

örnek

30

7.4 Cables

• Cables and chains used to support and transmit loads

from one member to another

• In force analysis, weight of cables is neglected

• Assume cable is perfectly flexible and inextensible

• Due to its flexibility cables has no resistance to

bending

• Length remains constant before and after loading

31

7.4 Cables

Cable Subjected to Concentrated Loads

• For a cable of negligible weight, it will subject to

constant tensile force

• Known: h, L1, L2, L3 and loads P1 and P2

• Form 2 equations of equilibrium

• Use Pythagorean Theorem to relate the three

segmental lengths

32

Example 7.11

Determine the tension in each segment of the cable.

33

Example 7.11

FBD for the entire cable.

kNE

EkNkNkNkN

F

kNA

mknmkNmkNmA

M

EA

F

y

y

y

y

y

E

xx

x

10

0315412

;0

12

0)2(3)10(15)15(4)18(

;0

0

;0

34

Example 7.11

Consider leftmost section which cuts cable BC since sag

yC = 12m.

kNT

TkNkN

F

kNT

F

kNEA

mkNmkNmA

M

BCBC

BCBC

y

BCBC

x

xx

x

C

2.10,6.51

0sin412

;0

033.6cos

;0

33.6

0)5(4)8(12)12(

;0

35

7.4 Cables

Cable Subjected to a Distributed Load

• Consider weightless cable subjected to a loading

function w = w(x) measured in the x direction

36

7.4 Cables

Cable Subjected to a Distributed Load

• For FBD of the cable having length ∆

• Since the tensile force changes continuously, it is

denoted on the FBD by ∆T

• Distributed load is represented by second integration,

dxdxxwF

yH

)(1

37

Example 7.12

The cable of a suspension bridge supports half of the

uniform road surface between the two columns at A and

B. If this distributed loading wo, determine the maximum

force developed in the cable and the cable’s required

length. The span length L and, sag h are known.

38

Solution

Note w(x) = wo

Perform two integrations

Boundary Conditions at x = 0

Therefore,

Curve becomes

This is the equation of a parabola

Boundary Condition at x = L/2

dxdxwF

y o

H

1

21

2

2

1CxC

xw

Fy o

H

0/,0,0 dxdyxy

021 CC

2

2x

F

wy

H

o

hy

39

Solution

For constant,

Tension, T = FH/cosθ

Slope at point B

Therefore

Using triangular relationship

2

2

2 4 and

8x

L

hy

h

LwF o

H

2

4 222

max

LwFT

oH

)cos( max

maxHF

T

H

o

LxH

o

Lx F

Lw

F

w

dx

dy

2tantan 1

max

2/

max

2/

40

Solution

For a differential segment of cable length ds,

Determine total length by integrating,

Integrating yields,

dxxL

hds

L

2/

0

2

2

812

L

h

h

L

L

hL 4sinh

4

41

2

1

2

dxdx

dydydxds

2

221

41

7.4 Cables

Cable Subjected to its Own Weight

• When weight of the cable is considered, the loading

function becomes a function of the arc length s rather

than length x

• FBD of a segment of the cable is shown

42

7.4 Cables


• Apply equilibrium equations to the force system

• Replace dy/dx by ds/dx for direct integration

dsswFdx

dy

dsswT

FT

H

H

)(1

)(sin

cos

1

2

22

dx

ds

dx

dy

dydxdsT

43

7.4 Cables


• Therefore

• Separating variables and integrating

2/1

2

2)(

11 dssw

F

dsx

H

2/1

2

2)(

11

dssw

Fdx

ds

H

44

Example 7.13

Determine the deflection curve, the length, and the

maximum tension in the uniform cable. The cable weights

wo = 5N/m.

45

Solution

For symmetry, origin located at the center of the cable.

Deflection curve expressed as y = f(x)

Integrating term in the denominator

2/12

1

2/11 CswF

dsx

oH

2/1

22/11 dswF

dsx

oH

46

Solution

Substitute

So that

Perform second integration

or

1/1 CswFu oH

dsFwdu Ho )/(

2

1sinh Cuw

Fx

o

H

21

1 1sinh CCsw

Fw

Fx o

Ho

H

47

Solution

Evaluate constants

or

dy/dx = 0 at s = 0, then C1 = 0

To obtain deflection curve, solve for s

dswFdx

dyo

H

1

1

1Csw

Fdx

dyo

H

x

F

w

w

Fs

H

o

o

H sinh

48

Solution

Hence

Boundary Condition y = 0 at x = 0

For deflection curve,

This equations defines a catenary curve.

x

F

w

dx

dy

H

osinh

3cosh CxF

w

w

Fy

H

o

o

H

o

H

w

FC 3

1cosh x

F

w

w

Fy

H

o

o

H

49

Solution

Boundary Condition y = h at x = L/2

Since wo = 5N/m, h = 6m and L = 20m,

By trial and error,

1cosh x

F

w

w

Fh

H

o

o

H

1

50cosh

/56

H

H

F

N

mN

Fm

NFH 9.45

50

Solution

For deflection curve,

x = 10m, for half length of the cable

Hence

Maximum tension occurs when θ is maximum at

s = 12.1m

m

mmN

mN

mN

mxy

2.24

1.12109.45

/5sinh

/5

9.45

2

1109.0cosh19.9

51

Solution

NNF

T

N

mmN

dx

dy

H

ms

9.758.52cos

9.45

cos

8.52

32.19.45

1.12/5tan

max

max

max

max

1.12

52

QUIZ

1. In a multi-force member, the member is generally

subjected to an internal _________.

A) Normal force B) Shear force

C) Bending moment D) All of the above.

2. In mechanics, the force component V acting tangent to,

or along the face of, the section is called the _________ .

A) Axial force B) Shear force

C) Normal force D) Bending moment

53

QUIZ

3. A column is loaded with a vertical 100 N force. At

which sections are the internal loads the same?

A) P, Q, and R B) P and Q

C) Q and R D) None of the above.

4. A column is loaded with a horizontal 100 N force. At

which section are the internal loads largest?

A) P B) Q

C) R D) S

54

QUIZ

5. Determine the magnitude of the internal loads

(normal, shear, and bending moment) at point C.

A) (100 N, 80 N, 80 N m)

B) (100 N, 80 N, 40 N m)

C) (80 N, 100 N, 40 N m)

D) (80 N, 100 N, 0 N m )

2. A column is loaded with a horizontal 100 N force. At

which section are the internal loads the lowest?

A) P B) Q

C) R D) S

55

engineering mechanics: statics in si units, 12e · 2019. 2. 19. · 7.3 yayılı yük ,kesme ve...

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