Equivalent force systems: The wrench
A wrench: A wrench is a force and couple system in which the force and couple are
parallel.
Every force and couple system can be reduced to a wrench: As above, the
perpendicular part of the resultant moment can be repl
Statics-Sample Exam I
Note to the students: The following is a typical "Exam I" for statics. It should give you an
idea of the difficulty of the exam. Try to do the exam in 50 minutes without the aid of
your book or your notes, only using your one page of
Area Moment of inertia
The area moment of inertia is the second moment of area around a given axis.
For example, given the axis O-O and the shaded area shown, one calculates the
second moment of the area by adding together
for all the elements of
area dA
Equilibrium of rigid bodies
Static equilibrium for a rigid body: A body (or any part of it) which is currently
stationary will remain stationary if the resultant force and resultant moment are zero for
all the forces and couples applied on it.
F R = F = 0
Centre of gravity and centroid:
The center of gravity (G) is a point that locates the
resultant weight of a system of particles.
Particles with weights W1, W2, , Wn can be replaced
by a resultant force of weight W located at the center of
gravity G.
To fi
Equivalent force systems: Distributed loads
Replacing distributed loads by a resultant load and resultant couple applied at a
given point O :
Replacing a distributed load by single resultant load:
Note: Since the equation for d is the same as that for det
Direction cosines
F
F
Fx
cos = x =
, cos = y =
F
F
Fx2 + Fy2 + Fz2
Fy
Fx2 + Fy2 + Fz2
, cos =
Fz
=
F
Fz
Fx2 + Fy2 + Fz2
Prove:
cos 2 ( ) + cos 2 ( ) + cos 2 ( ) = 1
Solution:
2
2
Fy
F 2 + Fy2 + Fz2
Fx
Fz
cos + cos + cos = 2
+2
+2
= x2
=1
Fx + Fy2 + Fz2 Fx
Statics-Sample Exam III
Note to the students: The following is a typical "Exam III" for statics. It should give you
an idea of the difficulty of the exam. Try to do the exam in 50 minutes without the aid of
your book or your notes, only using your one pag
AREA MOMENT OF INERTIA
Problem 2:
Determine the moment of inertia of the trianglular area about the base x-axis. Use the parallel
axis transfer theorem to determine Ixo from Ix.
y
y =(-h/b)x + h
h
x0
h/3
x
b
Solution:
Choosing to use double integration an
AREA MOMENT OF INERTIA
Problem 1:
Find the moment of inertia about the x axis and also y axis of the following figure:
Calculating Ix:
I x = y 2 dA
dA = dxdy
h
b
by 3
I x = y dxdy =
3
y =0 x =0
y =h
=
2
Calculating Iy:
y =0
bh3
3
I y = x 2 dA
dA = dxdy
b
Test 5:
1. What is the unit vector along the line A-B, if the Cartesian coordinates of point A are (5,0,3) and the
Cartesian coordinate of point B are (0,5,5)?
Solution:
r = ( xB x A ) i + ( y B y A ) j + ( z B z A ) k
r = ( 0 5 ) i + ( 5 0 ) j + ( 5 3) k
Statics-Sample Exam II
Note to the students: The following is a typical "Exam II" for statics. It should give you
an idea of the difficulty of the exam. Try to do the exam in 50 minutes without the aid of
your book or your notes, only using your one page