Presentation of Equilibrium II

University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)







Equilibrium

Undergraduate Level, 1st Stage



Mr. Waleed Ali Tameemi
College of Engineering/ Babylon University
M.Sc. Civil Engineering/ the University of Kansas/ USA



2016-2017

Course Outline
Introduction
Parallelogram Law
Forces Resolution
Resultant of a Cunccurent, Coplanar Force System
Moment
Couples
Resultant of a Non-concurrent, Coplanar Force System
Resultant of a Concurrent Non-coplanar Force System
Equilibrium
Fraction
Truss
Method of Joints
Method of Sections







Equilibrium
A body is called in equilibrium when the system of forces acting on it has a resultant equal to zero (the external forces equal to the internal forces).
??F_x =0
??F_y =0
??M_z =0 at any point
Free-Body Diagram
In mechanical engineering, free body diagram is important in investigating the equilibrium conditions of a body or system. In the free body diagram all external forces (gravity weight and lateral weight) as well as all internal forces (reactions) are replaced with forces and reactions and as follows:





Type Force Representation Case F.B.D
Body with weight (w) A vertical force equal to the body weight pass through the center of gravity


Roller – Wheel A reaction normal to the surface that the roller move on


Hinge Two perpendicular reactions, one is
normal to the surface and the other parallel to it



Pin
Two perpendicular reactions



Fixed Two perpendicular reactions and a moment


Cord – Cable - Rope Can be replaced by a force along the cable



Smooth Surface A reaction normal to the surface that the roller moves on



Rough Surface Two reactions (normal and parallel) to the surface that the roller move on



Action And Reaction




Action
Reaction



Application
Draw a free-body diagram for the following systems below:


















?
Example 1:
For the force system shown below, determine all unknown reaction.















Solution:
Draw a free-body diagram for each segment in the required system (the free-body diagram is shown above).

Start with segment AB,
Take sum of moment around point A,
??M_z =0 C.C.W (+ve.)
R_B×4-100×2=0
R_B=50 N ?

??F_y =0 ? (+ve.)
?R_A+R?_B-100=0
R_A=50 N ?

??F_x =0 ? (+ve.)
?RA?_x=0

Segment BC,
??F_y =0 ? (+ve.)
?R_C-R?_B-50=0
R_C=50+50=100 N ?

??F_x =0 ? (+ve.)
?RC?_x=0

Take sum of moment around point C,
??M_z =0 C.C.W (+ve.)
R_B×4+50×4+M_C=0
M_C=-400 N m=400 N m (C.W)

?









Example 2:
For the force system shown below, determine all unknown reaction. Body weight is equal to 250 N.







Solution:
Draw a free-body diagram for the force system.

?=tan^(-1)??8/4=63.4°?







Divided the forces into their component:
Force A (R_A)
R_Ax=R_A×cos??63.4°?
R_Ay=R_A×sin??63.4°?
Force B (R_B)
R_Bx=R_B×cos??63.4°?
R_By=R_B×sin??63.4°?
??F_y =0 ? (+ve.)
?R_Ay+R?_By-250 N=0
R_By=250 N-R_Ay (1)

??F_x =0 ? (+ve.)
R_Ax+R_Bx=0
R_Ax=R_Bx (2)

Take sum of moment around point B,
??M_z =0 C.C.W (+ve.)
250×4-R_Ay×8=0
R_Ay=250×4/8=125 N ?
Sub into Equ. (1)
R_By=250 N-125 N=125 N ?

Therefore:
R_A=R_Ay/sin?63.4 =125/0.89=139.8 N
R_B=R_By/sin?63.4 =125/0.89=139.8 N

Calculate the x Components:
R_Ax=R_A×cos??63.4°?=139.8×0.447=62.6 N ?
Sub. The value of R_Ax into Equ. (2):
R_Bx=R_Ax=62.6 N ?
Check the solution:
R_A=?(R_Ax^2+R_Ay^2 )=?(?62.6?^2+?125?^2 )=139.8 N
?=tan^(-1)??(R_Ay/R_Ax )?=tan^(-1)??(125/62.6)?=62.4°