b. BEAM Gaya Internal, Dia

8/18/2019 b. BEAM Gaya Internal, Dia

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BEAMGAYA INTERNAL, DIAGRAM

GAYA GESER DAN MOMEN

1SUTRIMO


2/20

Definisi Beam

• -

loads applied at various points along its length.

• Beam can be subjected to concentrated loads or

distributed loads or combination of both.

• Beam design is two-step process:

1) determine shearing forces and bending

moments produced by applied loads

-

shearing forces and bending moments

2SUTRIMO


3/20

Apa itu Gaya Internal ?

Gaya Internal : gaya yang mengikat bersama berbagai

bagian struktur sehingga struktur tersebut menjadi kokoh

-

equilibrium under application of F and

-F .

• Internal forces equivalent to F and -F arerequired for equilibrium of free-bodies AC

and CB.

3SUTRIMO


4/20

Reaksi pada Beam

• Beams are classified according to way in which they are

supported.

• Reactions at beam supports are determinate if they

involve only three unknowns. Otherwise, they are

stat ca y n eterm nate.

4SUTRIMO


5/20

Gaya Geser dan Momen pada Beam

• Wish to determine bending moment

beam subjected to concentrated and

distributed loads.

• Determine reactions at supports by

treatin whole beam as free-bod .

• Cut beam at C and draw free-body

agrams or C an CB. y

definition, positive sense for internal

force-couple systems are as shown.

• From equilibrium considerations,

e erm ne an or an .

5SUTRIMO


6/20

Diagram Gaya Geser dan Momen pada Beam

• Variation of shear and bending

moment along beam may be

p otte .

• Determine reactions at.

• Cut beam at C and consider

member AC ,

22 Px M PV

• Cut beam at E and consider

mem er ,

22 x LP M PV

• For a beam subjected to

concentrated loads, shear is

constant between loading points

and moment varies linearly. 6SUTRIMO


7/20

Contoh Soal 1

Given: A beam is supported

y a nge a , a ro er a .

Force applied at C. Moment

a lied at D.

Find: Draw the shear and

Plan:

a) Draw a FBD of the beam.

b)Calculate support reactions.

c) Find equivalent internal force-couple systems for free-bodies formed

by cutting beam on either side of load application points.

7SUTRIMO


8/20

Contoh Soal 1 & jawaban

Ay By

ΣFy = 0

Ay + By – 600 = 0

ΣMB = 0

- Ay(20) + 0.6(10) – 4 = 0

0.1 + By – 0.6 = 0

B = 0.5 kip

20Ay = 6 - 4

A = 0.1 kip

8SUTRIMO


9/20

Potongan 1-1

1 3V

1 2 3x

0.1 kip 0.5 kip 0.1 kip0 ≤ x < 10 ft

ΣF1-1 = 0

0.1 – V = 0

ΣM1-1 = 0

M – 0.1 x = 0

V = 0.1 kip

x = 0 ft V = 0.1 ki

M = 0.1(x)

x = 0 ft M = 0.1 0 = 0 ki -ft

x = 10 ft VC = 0.1 kipx = 10 ft MC = 0.1(10) = 1 kip-ft

9SUTRIMO


10/20

Potongan 2-2

V

M

1 3

1 2 3

0.1 kip

10 ≤ x < 15 ft

x0.1 kip 0.5 kip

ΣF2-2 = 0

– – =

ΣM2-2 = 0

– – = . .

V = 0.1 – 0.6 kip

= -

. .

M – 0.1(x) + 0.6(x) – 6 = 0

= - .

x = 10 ft VC = - 0.5 kip

= =

.

x = 10 ft MC = - 0.5(10) + 6 = 1 kip-ft

= = = D - .D - . - . -

10SUTRIMO


11/20

Poton an 3-3

1 2 3

1 2 3

M

0.1 kip 0.5 kip 0.1 kip15 ≤ x < 20 ft

x

ΣF3-3 = 0

0.1 – 0.6 – V = 0

ΣM3-3 = 0

M – 0.1(x) + 0.6(x – 10) – 4 = 0

V = 0.1 – 0.6 kip

V = - 0.5 kip

M – 0.1(x) + 0.6(x) – 6 – 4 = 0

M = - 0.5(x) + 10x = 15 ft VD = - 0.5 kip

x = 20 ft VD = - 0.5 kip

x = 15 ft MD = - 0.5(15) + 10 = 2.5 kip-ft

x = 20 ft MB = - 0.5(20) + 10 = 0 kip-ft

11SUTRIMO


12/20

Dia ram Geser dan Momen

ΣM1-1 = 0

=

ΣF1-1 = 0

=.

MA= 0 kip-ft

MC= 1 kip-ft

.

VA = 0.1 kip

VC = 0.1 kip

ΣM2-2 = 0

M = - 0.5(x) + 6

ΣF2-2 = 0

V = - 0.5 kip

C= p-

MD= - 1.5 kip-ftC = - . p

VD = - 0.5 kip

ΣM3-3 = 0

M = - 0.5(x) + 10

MD= 2.5 kip-ft

ΣF3-3 = 0

V = - 0.5 kip

VD = - 0.5 kipMB= 0 kip-ft VB = - 0.5 kip

12SUTRIMO


13/20

Contoh Soal 2

SOLUTION:

• Taking entire beam as free-body,

calculate reactions at A and B.

• Determine equivalent internal force-

couple systems at sections cut within

Draw the shear and bending moment

diagrams for the beam AB. The

, , .

• Plot results.

distributed load of 7200 N/m. extends

over 0.3 m of the beam, from A to C ,

and the 1800 N load is a lied at E .

13SUTRIMO


14/20

SOLUTION:

• Taking entire beam as a free-body, calculate

reactions at and B.

:0 A M

0m55.0 N1800m15.0 N2160m8.0 y B

N1642 y B

:0 B M

0m8.0m25.0 N1800m65.0 N2160 A

N2318

:0F 0 B

• Note: The 1800 N load at E may be replaced by

a 1800 N force and 180 Nm. couple at D.

14SUTRIMO


15/20

• Evaluate equivalent internal force-couple systemsat sections cut within segments AC , CD, and DB.

From A to C :

:0 yF 072002318 V x

:01 M 072002318 21 M x x x

xV 72002318

236002318 x x M

From C to D:

:0 yF 021602318 V

N158V

:02 M 015.021602318 M x x

m N158324 x M

15SUTRIMO


16/20

• Evaluate equivalent internal force-couple

,

CD, and DB.

From D to B:

:0 yF 0180021602318 V

N1642V

:02 M

045.0180018015.021602318 M x x x

m N16421314 x M

16SUTRIMO


17/20

• Plot results.

From A to C :

xV 72002318

60002318 x x M

From C to D:

N158V

m N158324 x M

From D to B:

N1642V

m N16421314 x M

17SUTRIMO


18/20

Contoh Soal 3

for the beam and loading shown.

18SUTRIMO


19/20

Jawaban akhir

19SUTRIMO


20/20

TERIMA KASIH

SUTRIMO 20

Documents

b. BEAM Gaya Internal, Dia