Lecture Notes of Truss (Method of Joints) Part I

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







Truss (Method of Joints)

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







Truss
A truss can be defined as a pin jointed frame that made of slender members (members with high ratio of length to cross sectional area) and capable of taking loads at its joints. Trusses are used in building roofs, steel structure frames, and bridges.

In making a perfect joint frame (truss), the number of members (m) shall have the following relationships with the number of joints (j):
M=2J-3
where:
M: number of members,
J: number of joints.
In analyzing a truss, the following assumption are necessary:
The external loads are applied only on the joints.
Members do not hold any bending.
The members’ self-weight is negligible
Members’ ends are pin joined.

Method of Joints
This method of analysis is used to determine the internal forces in each member of a truss. It can be done as follows:
Determine the value of the reactions using the equilibrium equations
??F_x =0 ? (+ve.)
??F_y =0 ? (+ve.)
??M_z =0 C.C.W (+ve.)
Divide the truss to number of joints and then determine the internal force of each member alone using only the equilibrium equations that are associated with forces:
??F_x =0 ? (+ve.)
??F_y =0 ? (+ve.)
If the internal force of a member is moving toward the joint it calls compression force (-).
If the internal force is moving far from the joint, it is tension force (+).




Example 1
Determine the external reactions as well as the internal force of each member of the following truss.







Solution
Draw free-body diagram for the truss.
?=tan^(-1)??4/4?=45°





For the whole truss:
??F_x =0 ? (+ve.)
?RA?_x=0
??M_z =0 at point A C.C.W (+ve.)
?RC?_y×8-500×4=0
?RC?_y=500×4/8=250N ?
??F_y =0 ? (+ve.)
?RA?_y+?RC?_y-500=0
?RA?_y+250-500=0
?RC?_y=250N ?






Joint A




??F_y =0 ? (+ve.)
?RA?_y-F_AD×sin?45=0
F_AD=250/sin?45 =353.5N (Compression)
??F_x =0 ? (+ve.)
F_AB-F_AD×cos?45=0
F_AB=353.5×cos?45=250N (Tension)
Joint C




??F_y =0 ? (+ve.)
?RC?_y-F_CD×sin?45=0
F_CD=250/sin?45 =353.5N (Compression)
??F_x =0 ? (+ve.)
F_CB-F_CD×cos?45=0
F_CB=353.5×cos?45=250N (Tension)




Joint D




??F_y =0 ? (+ve.)
F_DC×sin?45+F_DA×sin?45+F_DB-500=0
353.5×sin?45+353.5×sin?45+F_DB-500=0
250+250+F_DB-500=0
F_DB=0
??F_x =0 ? (+ve.)
F_DA×cos?45-F_DC×cos?45=0
250-250=0 (Equilibrium Check)
You may use Joint B for checking purposes!?
Example 2
Determine the external reactions as well as the internal force of member AD for the following truss.







Solution
Draw free-body diagram for the truss.
?=tan^(-1)??4/4?=45°











For the whole truss:
??F_x =0 ? (+ve.)
?RA?_x=0
??M_z =0 at point A C.C.W (+ve.)
?RC?_y×8-100×4=0
?RC?_y=100×4/8=50N ?
??F_y =0 ? (+ve.)
?RA?_y+?RC?_y-100-100=0
?RA?_y+50-200=0
?RC?_y=150N ?
Joint E




??F_x =0 ? (+ve.)
-F_ED×cos?45=0
F_ED=0
??F_y =0 ? (+ve.)
F_EG+F_ED×sin?45-100=0
F_EG+0×sin?45-100=0
F_EG=100N (Compression)
?
Joint G




??F_x =0 ? (+ve.)
-F_GD=0
F_GD=0
??F_y =0 ? (+ve.)
F_GA-F_GE=0
F_GA=F_GE=100N (Compression)
Joint A




??F_y =0 ? (+ve.)
?RA?_y-F_AG-F_AD×sin45=0
150-100-F_AD×sin45=0
F_AD=50/sin45=70.7N (Compression)