kuliah 3 internal force - wordpress.com · 2011. 9. 27. · ∑0 m 5 . 1 n 0 0 2 m1 =+ =mk k 1200 n...

Internal ForcesInternal ForcesInternal Forces Internal Forces

ENGR 221March 19, 2003March 19, 2003

Lecture GoalsLecture Goals

• Internal Force in StructuresInternal Force in Structures– Shear Forces

Bending Moment– Bending Moment • Shear and Bending moment Diagrams

Internal Forces and BendingInternal Forces and BendingInternal Forces and Bending Internal Forces and Bending MomentMoment

The bending moment, M.

Internal Forces and BendingInternal Forces and BendingInternal Forces and Bending Internal Forces and Bending MomentMoment

The shear force, V.

ExampleExample –– Internal Forces in aInternal Forces in aExample Example Internal Forces in a Internal Forces in a Frame ProblemFrame Problem

Determine the internal forces in member ACF at point J and in member BCD at point K.


Determine the forces atx Ex0F R= =∑

Ex

y Ey F

0 N0 2400 NR

F R R⇒ =

= = − + +∑

( ) ( )( )Ey F

E F

2400 N

0 4.8 m 2400 N 3.6 m

R R

M R

⇒ + =

= = −∑REx

R

F

Ey

1800 N600 N

RR

⇒ =⇒ =

REy RF


Look at the member BCD

y By Cy0 2400 N

2400 N

F F F

F F

= = − + +

⇒ + =∑

( ) ( )( )By Cy

B Cy

2400 N

0 2.4 m 2400 N 3.6 m

3600 N

F F

M R

R

⇒ + =

= = −

⇒ =∑

Cy

By

3600 N

1200 N

R

R

⇒ =

⇒ = −


Look at the member ACF

y Ay

Ay

0 3600 N 1800 N

1800 N

F F

F

= = − + +

⇒ =∑

Ay


Look at the member ABE

y 0 1800 N 1200 N 600 N

0 N

F

F

= = − + +

⇒∑

∑ y 0 NF⇒ =∑


Take a look at section ACF at point J

1 o5.4 mtan 48.374.8 m

θ − ⎛ ⎞= =⎜ ⎟⎝ ⎠4.8 m⎝ ⎠

Which section would you

θ

ylike to have to compute the moments?

θ

Example Example –– Internal Forces in a Internal Forces in a F P blF P blFrame ProblemFrame Problem

Take a look at section ACF at point J

( )( )

JM 1800 N 1.2 m 0

1800 N 1 2 m

M

M

= − =

=∑

( )J 1800 N 1.2 m2160 N-m

M ==

( )′∑ ( )oJxJ

1800 N cos 41.63

1345.41 N

F F

F

= −

=∑

( )J

oy J 1800 N sin 41.63

1195 77 N

F V

V

′ = +∑J 1195.77 NV = −


D t i th i t l f i b BCD t i t KDetermine the internal forces in ember BCD at point K.( )

( )KM 1200 N 1.5 m 0M= + =∑

( )K 1200 N 1.5 m1800 N-m

M = −=

kx0

0 N

F F

F

′ = =∑K

y K

0 N1200 N

FF V=

′ = − −∑K 1200 NV = −

Beams Beams –– Definition Definition A beam is defined as a structural member designed primarily to support forces acting perpendicular to th i f th b Th i i l diffthe axis of the member. The principal difference between beams and the axially loaded bars and torsionally loaded shafts is in the direction of thetorsionally loaded shafts is in the direction of the applied load.

Beams Beams –– SupportsSupportsA beam have a variety of supports.

- roller ( 1-DOF)

- pinned ( 2-DOF)- pinned ( 2-DOF)

- fixed ( 3-DOF)

Beams Beams –– LoadingsLoadingsA beam have a variety of loads.

- point loads

- distributed loads- distributed loads

- applied moments

Beams Beams –– Types Types A beam can be classified as statically determinate beam, which

th t it b l d imeans that it can be solved using equilibrium equations, or it is ...

Beams Beams –– Types Types A beam can be classified as statically indeterminate beam,

hi h t b l d ithwhich can not be solved with equilibrium equations. It requires a compatibilityrequires a compatibility condition.

Beams Beams –– Types Types A combination beam can be either statically determinate

i d t i t Th tor indeterminate. These two beams are statically determinate because thedeterminate, because the hinge provides another location, where the moment is ,equal to zero.

Shear and Bending momentShear and Bending momentShear and Bending moment Shear and Bending moment DiagramsDiagrams

In order to generate a shear and bending moment diagram one needs to

• Draw the free-body diagram

moment diagram one needs to

y g• Solve for reactions• Solve for the internal forces (shear V and• Solve for the internal forces (shear, V, and

bending moment, M)

Beam SectionsBeam Sections

A beam with a simple load in the center of the beam. Draw the free-body diagram.

y 0F =∑

0M =∑ z 0M∑

Beam Sections

Starting from the left side a take a series of section of the beam a compute the shear and bending moment

f h bof the beam.

Beam Sections

The plot of the resulting series of shear and bending moment are the shear and bending

di hmoment diagrams. The technique is a cutting methodmethod.

ExampleExample –– Shear and BendingShear and BendingExample Example Shear and Bending Shear and Bending Moment DiagramMoment Diagram

Obtain the shear and bending moment diagram f th bfor the beam.


Draw the free body diagram d l f ilib iand solve for equilibrium.

y By D0 20 kN 40 kNF R R= = − + − +∑ y By DBy D 60 kNR R⇒ + =

∑

( ) ( ) ( )B D0 20 kN 2.5 m 40 kN 3.0 m 5.0 m14 kN

M RR

= = − +

⇒ =∑

D

By

14 kN46 kN

RR

⇒ =⇒ =


Look at the sections 1-1.

20 kNF V∑ 1-1 1 1 20 kNF V −= = −∑

∑ ( )1-1 1-1 120 kN xM M= =∑


Look at the section 2-2.

2-2 2 2 20 kNF V −= = −∑

( )

2 2 2 2

2 2 2 20 20 kN 2.5 mM M= = +

∑

∑ ( )2-2 2-22-2

0 20 kN 2.5 m50 kN-m

M MM

+

⇒ = −∑


Look at the section at 3-3.

y 3 3

3 3

20 kN 46 kN

26 kN

F V

V−= = − +

⇒ =∑

( )

3 3

3 3

26 kN

0 20 kN 2 5 m

V

M M

−⇒

= = +∑ ( )B 3 33 3

0 20 kN 2.5 m50 kN-m

M MM

−

−

+

⇒ = −∑


Look at the section 4-4.

y 4 4 20 kN 46 kNF V −= = − +∑ y4 4 26 kNV −⇒ =

∑

( ) ( )4-4 4-44 4

20 kN 5.5 m 46 kN 3.0 m28 kN-m

M MM

== + −

⇒ =∑

4-4 28 kN mM⇒


Look at section 5-5.

y 5 5

5 5

20 kN 46 kN 40 kN

14 kN

F V

V−= = − + −

⇒ = −∑

( ) ( )

5 5 14 kN

0 20 kN 5 5 m 46 kN 3 0 m

V

M M

−⇒

= = + −∑ ( ) ( )5-5 5-55-5

0 20 kN 5.5 m 46 kN 3.0 m28 kN-m

M MM

= = +

⇒ =∑


Look at section 6-6

y 6 6

6 6

20 kN 46 kN 40 kN

14 kN

F V

V−

−

= = − + −

⇒ = −∑

( ) ( ) ( )6-6 6-60 20 kN 7.5 m 46 kN 5.0 m 40 kN 2.0 mM M= = + − +∑6-6 0 kN-mM⇒ =


Look at section end

y end 20 kN 46 kN 40 kN 14 kN

0 kN

F V

V

= = − + − +

⇒ =∑

( ) ( ) ( )

end

d d

0 kN

0 20 kN 7 5 m 46 kN 5 0 m 40 kN 2 0 m

V

M M

⇒ =

= = + − +∑ ( ) ( ) ( )end endend

0 20 kN 7.5 m 46 kN 5.0 m 40 kN 2.0 m0 kN-m

M MM

+ +

⇒ =∑


Draw the shear and bending moment diagrams.

Location (m) Shear (kN) Moment (kN-m)1 0 -20 02 2.5 -20 -503 2.5 26 -504 5.5 26 285 5.5 -14 286 7.5 -14 07 7.5 0 0

ClassClass –– Shear and BendingShear and BendingClass Class Shear and Bending Shear and Bending Moment DiagramMoment Diagram

D th h dDraw the shear and bending moment diagramdiagram.

Relations between Load, Shear,Relations between Load, Shear,Relations between Load, Shear, Relations between Load, Shear, and Bending Momentand Bending Moment

Look at a section

( )y 0F V w x V VV w x

= = − ∆ − + ∆

∆ = − ∆∑

{dd

V w xV Vw w

∆ ∆∆

= − → = −∆ {0 dxx x∆ →∆


Multiply by dx

Integrate over V to Vd

d dV w x= −

Integrate over Vc to Vdd dV

d dx

V w x= −∫ ∫c c

d

V

d

x

x

V V

∫ ∫

∫c

d c dx

V V w x− = −∫


Integrate over Vc to Vdd dV x

V

d dV w x= −∫ ∫c c

d

V x

x

d dV V w x− = −∫c

d cx

dV V w x∫The difference of the shear is the area under the load curve between c and d.


Look at a sectionx∆⎛ ⎞∑

( )C

02xM M w x

V M M

′

∆⎛ ⎞= = − + ∆ ⎜ ⎟⎝ ⎠

∆ ∆

∑( )

( )2 V x M M

xM V

− ∆ + + ∆

∆∆ ∆

( )2

d

M V x w

M x M

∆ = ∆ −

∆ ∆{

0

d2 dx

M x MV w Vx x∆ →

∆ ∆= − → =

∆


Multiply by dx

Integrate over M to Md

d dM V x=

Integrate over Mc to Mdd dM x

d dM V x=∫ ∫c c

d

M x

x

dM M V

∫ ∫

∫c

d cx

dM M V x− = ∫


Integrate over Mc to Mdd dM x

M

d dM V x=∫ ∫c c

d

M x

x

d dM M V x− = ∫c

d cx

dM M V x∫The difference of the moment is the area under the load curve between c and d.

ExampleExample -- Shear and BendingShear and BendingExample Example Shear and Bending Shear and Bending Moment DiagramMoment Diagram

Draw the shear and bending moment diagram.


Draw free-body diagram and use equilibrium equations.q q

y A B

A B

0F R wL R

R R wL

= = − +

⇒ + =∑

A BR R wL

L

⇒ +

⎛ ⎞∑ A B0 2&

LM wL R L

wL wLR R

⎛ ⎞= = − +⎜ ⎟⎝ ⎠

∑

B A&2 2R R⇒ = =


Shear diagram.

L( )y 2wLF V x wx= = −∑

Note that the area under the load diagram.g


Remember that

dMddM Vx=

Where will the maximum moment occur?


The maximum will occur where

dM

h i i h

d 0dMx=

The maximum moment is the positive area under the curve

212 2 2 8

wL L wLM ⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠2 2 2 8⎝ ⎠⎝ ⎠


The moment equation

( ) ( )∫( ) ( )2

M x V x dx

wL x

=

⎛ ⎞

∫

( )2 2

wL xx w⎛ ⎞

= − ⎜ ⎟⎝ ⎠

Note that the slope of the moment diagram is equal to the shear.


Draw the shear and bending moment diagram


Free-body diagram RAx

RAyRCx Ax0F R= =∑

( )y Ay CAy C

0 20 kN/m 6 m

120 kN

F R R

R R

= = − +

⇒ + =∑

( )( ) ( )A C0 20 kN/m 6 m 3 m 9 mM R= = − +∑ ( )( ) ( )A CC Ay40 kN & 80 kNR R⇒ = =

∑

Example Example -- Shear and Bending Shear and Bending M DiM DiMoment DiagramMoment Diagram

Look at the shear

( )A

B

80 kN80 kN 20 kN/m 6 m

VV

=

= −

40 kN= −

( )x 80 kN 20 kN/m x 0 kN4 m

Vx

= − =

⇒ =

-C 40 kNV = −C+

C 40 kN 40 kN 0 kNV = − + =


Look at the shear diagram g

( )A

B

80 kN80 kN 20 kN/m 6 m

VV

=

= −

40 kN= −

( )x 80 kN 20 kN/m x 0 kN4 m

Vx

= − =

⇒ =

-C 40 kNV = −C+

C 40 kN 40 kN 0 kNV = − + =


Find the moments

( )( )

0 m 0 kN-m1

M =

⎛ ⎞( )( )4 m1 80 kN 4 m2

160 kN-m

M ⎛ ⎞= ⎜ ⎟⎝ ⎠

=

( )( )6 m1160 kN-m 40 kN 2 m2

M ⎛ ⎞= + −⎜ ⎟⎝ ⎠

( )( )9 m120 kN-m120 kN-m 40 kN 3 mM

⎝ ⎠=

= + −( )( )9 m0 kN-m=


Draw the moment diagram g

( )( )

0 m 0 kN-m1

M =

⎛ ⎞( )( )4 m1 80 kN 4 m2

160 kN-m

M ⎛ ⎞= ⎜ ⎟⎝ ⎠

=

( )( )6 m1160 kN-m 40 kN 2 m2

M ⎛ ⎞= + −⎜ ⎟⎝ ⎠

( )( )9 m120 kN-m120 kN-m 40 kN 3 mM

⎝ ⎠=

= + −( )( )9 m0 kN-m=


D th h dDraw the shear and bending moment diagram for the knowdiagram for the know reactions

Homework (Due 3/26/03)Homework (Due 3/26/03)

Problems:

8 15 8 20 8 37 8 41 8 518-15, 8-20, 8-37, 8-41, 8-51

Bonus ProblemBonus Problem –– Shear andShear andBonus Problem Bonus Problem Shear and Shear and Bending Moment DiagramBending Moment Diagram


kuliah 3 internal force - wordpress.com · 2011. 9. 27. · ∑0 m 5 . 1 n 0 0 2 m1 =+ =mk k 1200 n...

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