انت هنا الان : شبكة جامعة بابل > موقع الكلية > نظام التعليم الالكتروني > مشاهدة المحاضرة

Presentation of Truss (Section Method) II

الكلية كلية الهندسة     القسم  الهندسة البيئية     المرحلة 1
أستاذ المادة وليد علي حسن       21/01/2017 19:27:40
University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)







Truss (Section Method)

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







Method of Sections
This method of analysis is used to determine the internal forces 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.)
Use the cutting section to cut the truss to two parts through three unknown members. Then take one of the cut sections and use the equilibrium equations in determining the internal forces of the cut members.
??F_x =0 ? (+ve.)
??F_y =0 ? (+ve.)
??M_z =0 C.C.W (+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 CH and member CG of the following truss.





Solution
Draw free-body diagram for the truss.
?
For the whole truss:
Structure height:
tan60=h/2
h=tan60×2=3.46m
??F_x =0 ? (+ve.)
?RA?_x-100=0
?RA?_x=100N ?

??M_z =0 at point A C.C.W (+ve.)
?RE?_y×16+100×3.46-50×10-200×6=0
?RE?_y×16+100×3.46-50×10-200×6=0
?RE?_y=84.63N ?

??F_y =0 ? (+ve.)
?RA?_y-200-50+?RE?_y=0
?RC?_y=165.38N ?
Section 1-1: Pass through member CD, CH, and GH









whole structure with section 1-1









Section 1-1
??F_y =0 ? (+ve.)
F_CH×sin?60+165.38-200=0
F_CH×sin?60+165.38-200=0
F_CH=40N (Tension)
Section 2-2: Pass through member GC, GH, and BC









whole structure with section 2-2









Section 2-2
??F_y =0 ? (+ve.)
F_GH×sin?60+165.38-200=0
F_GH=40N (Compression)

Example 2
Determine the external reactions as well as the internal force of CF the following truss.













Solution
Draw free-body diagram for the truss.















?
Section 1-1: Pass through member CE, CF, and CD.
The angle of member CF with x-axis: ?=tan^(-1)??3/4=37?















?
??F_x =0 ? (+ve.)
F_CF×cos?37-10=0
F_CF=12.5N (Compression)


المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .