Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of [email protected]
ENGI 1313 Mechanics I
Lecture 22: Equivalent Force Systems and Distributed Loading
2 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Lecture 22 Objective
to demonstrate by example equivalent force systemsto determine an equivalent force for a distributed load
3 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Example 22-01
Handle forces F1 and F2are applied to the drill. Replace this system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form.
a 0.15 m=
b 0.25 m=
c 0.3 m=
F1
6
3−
10−
⎛⎜⎜⎝
⎞⎟⎟⎠
N=
F2
0
2
4−
⎛⎜⎜⎝
⎞⎟⎟⎠
N=
4 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Equivalent Distributed Loads
Reduce to Single Equivalent LoadMagnitude and position
ApplicationsEnvironment• Wind• Fluid
Dead load• Weight• Snow, sand• Objects
5 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Distributed Loads
Load IntensityPressure (Pa)• Force per unit area• N/m2 or lb/ft2
Infinite # parallel vertical loads
Reduce from Area to Line Load
( ) ( ) ⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡=
mNxwm
mNaxpw 2
6 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Distributed Loads (cont.)
Consider Element Length (dx)
Element force (dF) acts on element length (dx)The line load w(x) represents force per unit length
( )dxxwdF =
7 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Distributed Loads (cont.)
Magnitude Net Force (FR)Summation of all element forces
Area under line load curve
( ) dxxwdFFR ∑∑ ==
( ) AdAdxxwFAL
R === ∫∫
8 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Distributed Loads (cont.)
Resultant Moment (MRo)Element force moment
Total moment
Equivalent force
dFxMdFO =
( )∫∫∑ ===LL
ORo dxxwxdFxMMdF
( )∫==L
RRo dxxwxxFM
9 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Distributed Loads (cont.)
CentroidGeometric center of areaResultant force line of action
( )∫==L
RRo dxxwxxFM
( ) AdAdxxwFAL
R === ∫∫
( )
( ) ∫
∫
∫
∫==
A
A
L
L
dA
dAx
dxxw
dxxwxx
10 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Example 22-02
Determine equivalent resultant force and location for the following distributed load.
( )2Lxft5
lb4000
lbft2x400
lb4000
dxxwx
lb4000
dAx
dA
dAxx
10
0
210
0
10
0
A
A =⇒===== ∫∫∫
∫
( ) wLFlb4000ftlbx400dx400dAdxxwF R
10
0
10
0ALR =⇒===== ∫∫∫
L/2
11 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Example 22-03
Determine equivalent resultant force and location for the following distributed load.
( )3L,
3L2xm4
N1800
mN3x100
N1800
dxxwx
N1800
dAx
dA
dAxx
6
0
36
0
6
0
A
A =⇒===== ∫∫∫
∫
( )2
wLFN1800mNx50dxx100dAdxxwF R
6
0
26
0ALR =⇒===== ∫∫∫
L/3
12 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Comprehension Quiz 22-01
What is the location of FR or distance d?
A) 2 mB) 3 mC) 4 mD) 5 mE) 6 m
Answer: D
FR
3 m 3 m
d
13 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
Example 22-04
The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position aon the beam such that the resultant force and couple moment acting on the beam are zero.
w1 40lbft
=
w2 60lbft
=
c 10ft=
d 6 ft=
14 ENGI 1313 Statics I – Lecture 22© 2007 S. Kenny, Ph.D., P.Eng.
References
Hibbeler (2007)http://wps.prenhall.com/esm_hibbeler_engmech_1
ENGI 1313 Mechanics I Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng. Fall 2007
Page 1 of 2
EXAMPLE 22-01 (PROBLEM 4-131)
Handle forces F1 and F2 are applied to the electric drill. Replace this system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form.
F1
6
3−
10−
⎛⎜⎜⎝
⎞⎟⎟⎠
N=
F2
0
2
4−
⎛⎜⎜⎝
⎞⎟⎟⎠
N=
a 0.15 m=
b 0.25 m=
c 0.3 m=
ENGI 1313 Mechanics I Faculty of Engineering and Applied Science
Shawn Kenny, Ph.D., P.Eng. Fall 2007
Page 1 of 2
EXAMPLE 22-04 (PROBLEM 4-150)
The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are zero.
w1 40lbft
=
w2 60lbft
=
c 10ft=
d 6 ft=
