King Fahd University of Petroleum & Minerals
Department of Civil and Environmental Engineering
CE 201- Statics
Key Solution of Assignment 2 (Chapter 3) - Term 151
(Reference: Engineering Mechanics - Statics by Hibbeler 13th Edition)
1. Question 3-15
2. Qu
CE 201 - Statics
Chapter 8
Sections 8.1 and 8.2
FRICTION
We
used to assume that surface of contact
between two bodies are smooth
Reaction is assumed normal to surface of
contact
F
In
reality, surfaces are rough
Ff
Fn
Characteristics of Dry Friction
What
CE 201- Statics
Chapter 9
Section 9.3
COMPOSITE BODIES
A
composite body consists of simpler bodies
connected together (triangle, rectangle, .)
A composite body can be sectioned into its
composite parts
Center of gravity of the composite body can be
determ
CE 201 - Statics
Chapter 6
Section 6.4
THE METHOD OF SECTIONS
The method of sections is
based on the fact that if a
body is in equilibrium, then
any part of the body is in
equilibrium. This method
can be applied by cutting the
body into two parts and
incl
CE 201 - Statics
Chapter 6
Sections 6.2 and 6.3
THE METHOD OF JOINTS
All joints are in equilibrium since the truss is
in equilibrium. The method of joints is
applied using equilibrium equations at each
joint of the truss.
To do this, the free-body diagram
CE 201 - Statics
Chapter 10
Sections 10.1, 10.2 and 10.5
MOMENTS OF INERTIA
Objective
Develop
a method for determining the moment of inertia
for an area
Moment of inertia is an important parameter to be
determined when designing a structure or a mechani
CE 201 - Statics
Chapter 7
Section 7.1
INTERNAL FORCES
Internal
Forces Developed in Structural
Members (2-D)
Internal
Forces Developed in Structural
Members (3-D)
Internal Forces Developed in Structural
Members (2-D)
To
design a member, the forces acting
CE 201 - Statics
Chapter 10
Lecture 1
MOMENTS OF INERTIA
Objective
Develop
a method for determining the
moment of inertia for an area
Moment of inertia is an important parameter
to be determined when designing a structure
or a mechanical part
Definition o
CE 201- Statics
Chapter 9
Sections 9.1 and 9.2
CENTER OF GRAVITY AND
CENTROID
The following will be studied
Location of center of gravity (C. G.) and
center of mass for discrete particles
Location of C. G. and center of mass for an
arbitrary-shaped body
L
CE 201 - Statics
Chapter 6
Section 6.6
FRAMES AND MACHINES
Frames
and machines are structures
composed of pin-connected members.
Those members are subjected to more than
two forces.
The forces acting at the joints and supports
can be determined by applyin
CE 201 - Statics
Chapter 7
Section 7.2
Shear and Moment Equations and
Diagrams
To
design a beam, you should know the
variation of the internal shear force ( V )
and bending moment ( M ) acting at each
point along the axis of the beam.
The
variation of ( V
mam/Wmmm @Mm
Department of Civil and Environmental Engineering
Semester:
Examination:
Date (Day):
CE 201 — Statics
142
Second Major
April 21, 2015 (Tuesday)
Student’s Name
Time: 07:00 — 09:30 p.m. _ _ _
Section 1 _ 2 3 4 5 6 7
_ instruct
[email protected] am
Department of Civil and Environmental Engineering
CE 201 — Statics
Semester: 132
Examination: Second
Date (Day): April 18, 2014 (Friday)
Time: 02:00 — 05:00 pm.
Section 1 2 i 3 6 7 8
Instructor AI
pun
wymwyﬂ‘m @QWM
Department of Civil and Environmental Engineering
CE 201 — Statics
Semester: 131
Examination: Second Major
Date (Day): November 26, 2013 (Tuesday)
Time: 07:00 — 09:00 p.m.
4 | 5 | 7 |
Instructor Hussain
I Student’s Name
£%y§%¢QZmeygﬂ@[email protected]%§ama
Department of Civil and Environmental Engineering
(TE 201 — Statics
Semester: I41
Examination: Second Major
Date (Day): December 02, 2014 (Tuesday)
Time: 07:00 — l0:00 p.m.
;[email protected]_13; s _4_T%'! a m 3 gr 6' maul
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CE 201 - Statics
Chapter 4
Sections 4.3 and 4.4
MOMENT OF A FORCE-VECTOR
FORMULATION
In scalar formulation, we have seen that the moment of a force F
about point O or a bout a moment axis passing through point O is
equal to:
+ MO = F d
where d is the perp
CE 201 - Statics
Chapter 3
Section 3.4
Three-Dimensional Force Systems
Particle will be in equilibrium if F = 0
In 3-D problems, this can be expressed as:
Fx i + Fy j + Fz k = 0
Therefore, for a particle to be in equilibrium:
Fx = 0
Fy = 0
Fz = 0
Procedur
CE 201 - Statics
Chapter 3
Section 3.3
Coplanar Force System
All forces lie in the x-y plane.
Each force can be resolved into
its two rectangular components Fx
and Fy.
If the particle is in equilibrium,
then,
F=0
So,
(Fx i + Fy j) = 0
Fx i + Fy j = 0
Fo
1026. Determine the moment of inertia of the
composite area about the y axis.
y
150 mm 150 mm
100 mm
100 mm
x
300 mm
75 mm
1031.
Determine the moment of inertia for the beams crosssectional area about the x axis.
y
y'
x
40 mm
40 mm
x'
C
40 mm
40 mm
x
120
.
.
.
.
.
.
.
958.
Determine the location y of the centroidal axis x - x of the
beams cross-sectional area. Neglect the size of the corner
welds at A and B for the calculation.
150 mm
15 mm
B
y
15 mm
150 mm
C
x
SOLUTION
A
'
yA = 7.5(15) (150) + 90(150) (1
King Fahd University of Petroleum & Minerals
Department of Civil and Environmental Engineering
CE 201- Statics
Assignment 2 (Chapter 3)
(Reference: Engineering Mechanics - Statics by Hibbeler 13th Edition)
Key Solution of Assignment 2 (Chapter 3) - Term 1
711. Determine the internal normal force, shear force,
and moment at points D and E in the compound beam.
Point E is located just to the left of the 10-kN concentrated
load. Assume the support at A is fixed and the connection at
B is a pin.
10 kN
2 kN/m
B
819.
The 5-kg cylinder is suspended from two equal-length
cords. The end of each cord is attached to a ring of
negligible mass, which passes along a horizontal shaft. If the
coefficient of static friction between each ring and the shaft
is ms = 0.5, deter
King Fahd University of Petroleum & Minerals
Civil Engineering Department
. (3 _
_, OLU \ICNQ
CE 201 Statics
Second Semester 2007-2008 (072)
Final Examination gSection # 11
June 11, 2008 7:00 PM 9: 30 PM
Instructor
Dr. Shamshad Ahmad
Stude
Find the magnitude and direction of the resultant force.
Using Parallelogram Law
1 + 3 = 180 (30+45) = 180 75 = 105
Then 2 = 180 105 = 75
Apply cosine law
FR = (150)2 + (250)2 - 2150250cos (75)
= 22500+62500-19411.43=65588.57
FR = 256.1 N
Apply sine law
2
CE 201 - Statics
Chapter 3
Sections 3.1 and 3.2
EQUILIBRIUM OF A PARTICLE
CONDITION FOR THE EQUILIBRIUM OF A
PARTICLE
A particle is in EQUILIBRIUM if:
1. it is at rest, OR
2. it is moving with constant velocity
The term "EQUILIBRIUM" is often used to
desc
CE 201 - Statics
Chapter 2
Section 2.9
Contents
Dot
Product
DOT PRODUCT
Dot product is the method of multiplying two
vectors and is used to solve three-dimensional
problems.
If A and B are two vectors
then,
A . B = AB cos ()
where 0 180
Laws of Operation
CE 201 - Statics
Lecture 3
Section 2.4
ADDITION OF A SYSTEM OF COPLANNAR
FORCES
If we have more than two forces, the
resultant can be determined by successive
applications of the parallelogram law.
How about finding the components of each force
along cert
CE 201 - Statics
Chapter 2
Sections 2.7 and 2.8
Contents
Position Vectors
Force Vector
Directed along a Line
POSITION VECTORS
If a force is acting between two points, then
the use of position vector will help in
representing the force in the form of
Carte
CE 201 - Statics
Chapter 2
Sections 2.5 and 2.6
CARTESIAN VECTORS
We knew how to represent vectors in the
form of Cartesian vectors in two
dimensions. In this section, we will look
into how vectors can be represented in
Cartesian vectors form and in three