8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 1/32
ME101: Engineering Mechanics
Lecture 12
28th January 2011
Forces in Beams and Cables
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 2/32
• Preceding chapters dealt with:
a) determining external forces acting on a structure and
b) determining forces which hold together the various members of astructure.
• The current chapter is concerned with determining the internal forces
Introduction
.e., tens on compress on, s ear, an en ng w c o toget er t evarious parts of a given member.
• Focus is on two important types of engineering structures:
a) Beams - usually long, straight, prismatic members designed to
support loads applied at various points along the member.
b) Cables - flexible members capable of withstanding only tension,
designed to support concentrated or distributed loads.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 3/32
• Straight two-force member AB is in
equilibrium under application of F and
-F .
• Internal forces equivalent to F and -F arerequired for equilibrium of free-bodies AC
and CB.
Internal Forces in Members
-
ibrium under application of cable and
member contact forces.
• Internal forces equivalent to a force-
couple system are necessary for equil-ibrium of free-bodies JD and ABCJ .
• An internal force-couple system is
required for equilibrium of two-force
members which are not straight.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 4/32
SOLUTION:
• Compute reactions and forces at
connections for each member.
• Cut member ACF at J . The internal
forces at J are represented by equivalent
force-couple system which is determined
Sample Problem 11.1
Determine the internal forces (a) in
member ACF at point J and (b) inmember BCD at K .
by considering equilibrium of either part.
• Cut member BCD at K . Determine
force-couple system equivalent to
internal forces at K by applyingequilibrium conditions to either part.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 5/32
SOLUTION:
• Compute reactions and connection forces.
:0=∑ E M
0m8.4m6.3N2400 =+− F N1800=F
Consider entire frame as a free-body:
Sample Problem 11.1
:0=∑ yF
0N1800N2400 =++− y E N E y 600=
:0=∑ xF 0= x E
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 6/32
Consider member BCD as free-body:
:0=∑ B M
( )( ) ( ) 0m4.2m6.3N2400 =+− yC N3600= yC
:0=∑ C M
( )( ) ( ) 0m4.2m2.1N2400 =+− y B N1200= y B
:0=∑ xF 0=+− x x C B
Sample Problem 11.1
Consider member ABE as free-body:
:0=∑ A M ( ) 0m4.2 = x B 0= x B
∑ = :0 xF 0=− x A B 0= x A
∑ = :0 yF 0N600 =++− y y B A N1800= y A
From member BCD,
:0=∑ xF 0=+− xC B 0=C
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 7/32
• Cut member ACF at J . The internal forces at J are
represented by equivalent force-couple system.
Consider free-body AJ :
:0=∑ J M
( )( ) 0m2.1N1800 =+− M mN2160 ⋅= M
Sample Problem 11.1
:0=∑ xF
( ) 07.41cosN1800 =°−F N1344=F
:0=∑ yF
( ) 07.41sinN1800 =°+−V N1197=V
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 8/32
• Cut member BCD at K . Determine a force-couple
system equivalent to internal forces at K .
Consider free-body BK :
:0=∑ K M
( )( ) 0m5.1N1200 =+ M mN1800 ⋅−= M
Sample Problem 11.1
:0=∑ xF 0=F
:0=∑ yF
0N1200=−−
V N1200−=
V
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 9/32
Various Types of Beam Loading and Support
• Beam - structural member designed to support
loads applied at various points along its length.
• Beam can be subjected to concentrated loads or
• Beam design is two-step process:
1) determine shearing forces and bending
moments produced by applied loads
2) select cross-section best suited to resist
shearing forces and bending moments
.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 10/32
Various Types of Beam Loading and Support
• 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 statically indeterminate.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 11/32
Shear and Bending Moment in a Beam
• Wish to determine bending moment
and shearing force at any point in a
beam subjected to concentrated and
distributed loads.
• Determine reactions at supports by
treating whole beam as free-body.
• Cut beam at C and draw free-body
diagrams for AC and CB. By
definition, positive sense for internal
force-couple systems are as shown.
• From equilibrium considerations,
determine M and V or M’ and V’.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 12/32
Shear and Bending Moment Diagrams
• Variation of shear and bending
moment along beam may be
plotted.
• Determine reactions atsupports.
• Cut beam at C and consider
member AC ,
22 Px M PV +=+=
• Cut beam at E and consider
member EB,
( ) 22 x LP M PV −+=−=
• For a beam subjected to
concentrated loads, shear is
constant between loading points
and moment varies linearly.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 13/32
SOLUTION:
• Taking entire beam as a free-body,
calculate reactions at B and D.• Find equivalent internal force-couple
systems for free-bodies formed by
cutting beam on either side of load
Sample Problem 11.2
Draw the shear and bending moment
diagrams for the beam and loading
shown.
application points.
• Plot results.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 14/32
SOLUTION:
• Taking entire beam as a free-body, calculate
reactions at B and D.
• Find equivalent internal force-couple systems atsections on either side of load application points.
∑ = :0 yF 0kN20 1 =−− V kN201 −=V
Sample Problem 11.2
∑ = :0 yF 220kN 0V − − =2 20kNV = −
1 0 : M =
∑ ( )( ) 0m0kN20 1=+
M 01=
M
2 0 : M =∑ ( ) ( ) 220kN 2.5m 0 M + = 2 50kN.m M = −
3 3
4 4
5 5
6 6
26kN 50kN m
26kN 28kN m
14kN 28kN m
14kN 0
V M
V M
V M
V M
= = − ⋅
= = + ⋅
= − = + ⋅
= − =
Similarly,
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 15/32
• Plot results.
Note that shear is of constant value
between concentrated loads and
bending moment varies linearly.
Sample Problem 11.2
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 16/32
ME101: Engineering Mechanics
Lecture 13
31th January 2011
Forces in Beams and Cables
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 17/32
SOLUTION:
• Taking entire beam as free-body,
calculate reactions at A and B.
• Determine equivalent internal force-
couple systems at sections cut within
Sample Problem 13.1
Draw the shear and bending moment
diagrams for the beam AB. The
distributed load of 7200 N/m extends
over 0.3 m. of the beam, from A
toC
,and the 1800-N load is applied at E .
segments AC , CD, and DB.
• Plot results.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 18/32
SOLUTION:
• Taking entire beam as a free-body, calculate
reactions at A and B.
:0=∑ A M
( ) ( )( ) ( )( ) 0m55.0N1800m5.0N2160m8.0 =−− y B
N1642= y B
Sample Problem 13.1
:0=∑ B M
( )( ) ( )( ) ( ) 0m8.0N25.0N1800m65.0N2160 =−+ A
N2318= A
:0=∑ F 0= B
• Note: The 1800 N load at E may be replaced by
a 1800 N force and 180 N . m couple at D.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 19/32
• Evaluate equivalent internal force-couple systems
at sections cut within segments AC , CD, and DB.
From A to C :
072002318 =−− V x
N)72002318( xV −=
Sample Problem 13.1
:01=
∑ M ( ) 072002318 2
1 =+−−M x x x
m-N)36002318( 2 x x M −=
:02 =∑ M ( ) 015.021602318 =+−+− M x x
( ) mN158324 ⋅+= x M
From C to D:
∑ = :0 yF 021602318 =−− V
N158=V
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 20/32
• Evaluate equivalent internal force-couple
systems at sections cut within segments AC ,
CD, and DB.
From D to B:
∑ = :0 yF 0180021602318 =−−− V
Sample Problem 13.1
:03 =∑ M
( ) ( ) 045.0180018015.021602318 =+−+−−+− M x x x
( ) mN16421314⋅−=
x M
N1642−=
V
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 21/32
• Plot results.
From A to C :N)72002318( xV −=
mN)36002318( 2−−= x x M
From C to D:
Sample Problem 13.1
=
( ) mN158324 ⋅+= x M
From D to B:
N1642−=V
( ) mN16421314 ⋅−= x M
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 22/32
• Relations between load and shear:
( )
wV
dx
dV
xwV V V
x−=
∆
∆=
=∆−∆+−
→∆ 0lim
0
( )curveloadunderarea−=−=− ∫ D x
x
C D dxwV V
Relations Among Load, Shear, and Bending Moment
• Relations between shear and bending moment:
( )
( ) V xwV x
M
dx
dM
x xw xV M M M
x=∆−=
∆
∆
=
=∆
∆+∆−−∆+
→∆→∆2100
limlim
02
( )curveshearunderarea==− ∫ D
C
x
x
C D dxV M M
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 23/32
• Reactions at supports,2
wL R R B A ==
• Shear curve,
−=−=−=
−=−=− ∫
x L
wwxwL
wxV V
wxdxwV V
A
x
A
0
Relations Among Load, Shear, and Bending Moment
• Moment curve,
( )
===
−=
−=
=−
∫
∫
0at8
22
2
max
2
0
0
V dx
dM M
wL M
x x Lwdx x Lw M
Vdx M M
x
x
A
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 24/32
SOLUTION:
• Taking entire beam as a free-body, determine
reactions at supports.
• With uniform loadin between D and E the
• Between concentrated load applicationpoints, and shear is
constant.
0=−= wdxdV
Sample Problem 13.2
Draw the shear and bending-
moment diagrams for the beam
and loading shown.
shear variation is linear.• Between concentrated load application
points, The change
in moment between load application points is
equal to area under shear curve betweenpoints.
.constant== V dxdM
• With a linear shear variation between D
and E , the bending moment diagram is a
parabola.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 25/32
SOLUTION:
• Taking entire beam as a free-body,
determine reactions at supports.
∑=
:0 D M
( ) ( ) ( ) ( )
( ) 0m2.1
m15.0N720m4.0N500m8.0N500
=−
−+−
y AN410= A
Sample Problem 13.2
• Between concentrated load application points,
and shear is constant.0=−= wdxdV
• With uniform loading between D and E , the shear
variation is linear.
:0∑ = yF
0DN720N500N500N410 y =+−−−
N1310= y D
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 26/32
• Between concentrated load application
points, The change
in moment between load application points is
equal to area under the shear curve between
points.
.constant== V dxdM
mN12836
mN164164
⋅+=−=−
⋅+=+=− B A B M M M
Sample Problem 13.2
• With a linear shear variation between D
and E , the bending moment diagram is aparabola.
0108
mN108236
=+=−
⋅−=−=−
E D E
DC D
M M M
M M M
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 27/32
SOLUTION:
• The change in shear between A and B is equal
to the negative of area under load curve
between points. The linear load curve resultsin a parabolic shear curve.
• With zero load, change in shear between B
and C is zero.
Sample Problem 13.3
Sketch the shear and bending-
moment diagrams for the
cantilever beam and loading
shown.
• The change in moment between A and B isequal to area under shear curve between
points. The parabolic shear curve results in
a cubic moment curve.
• The change in moment between B and C isequal to area under shear curve between
points. The constant shear curve results in a
linear moment curve.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 28/32
SOLUTION:
• The change in shear between A and B is equal to
negative of area under load curve between points.
The linear load curve results in a parabolic shearcurve.
0,0,at wwdV
V A A −=−==
Sample Problem 13.3
• With zero load, change in shear between B and C is
zero.
awV V A B 021−=− awV B 02
1−=
0,at =−= wdx
dV B
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 29/32
• The change in moment between A and B is equal
to area under shear curve between the points.
The parabolic shear curve results in a cubic
moment curve.
22
0,0,at === V dx
dM M A A
Sample Problem 13.3
• The change in moment between B and C is equal
to area under shear curve between points. Theconstant shear curve results in a linear moment
curve.
( ) ( )a Law M a Law M M
ww
C BC
B A B
−−=−−=−
−=−=−
3061
021
0303
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 30/32
Determine the location x of the two supports so as to minimize themaximum bending moment in the beam. Specify the maximumbending moment.
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 31/32
8/7/2019 Lecture Slides Statics 2011 Lectures 12 13 [Compatibility Mode]
http://slidepdf.com/reader/full/lecture-slides-statics-2011-lectures-12-13-compatibility-mode 32/32
1. Vector Mechanics for Engineers – Statics & Dynamics, Beer &
Johnston; 7th edition
2. Engineering Mechanics Statics & Dynamics, Shames; 4th edition
3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics
Dynamics Vol. 2, Meriam & Kraige; 5th edition
Reference books
32
STATICS – MID SEMESTER – DYNAMICSTutorial: Thursday 8 am to 8.55 am
4. Schaum’s solved problems series Vol. 1: Statics; Vol. 2:Dynamics, Joseph F. Shelley