1. CHAP6.
Structural Analysis

2. 6.1 Simple Trusses 1. Definition of Truss
A structure composed of slender members jointed together at their end points by bolting, welding or pinning.
Bolting
2. Planar Trusses
planar trusses lie in a single and one often need to support roofs and bridges.
Bridge truss
Roof truss

3. (1) all loading are applied at the joints
3. Assumption for Truss
(1) all loading are applied at the joints
a. members’ weights are neglected (W << external force)
b. members’ weights are included
(2) members are jointed together by smooth pin (No friction forces)
Because of the two assumptions, the truss member is a two-force member T C
Compression force
Tensile force T
Note: Compression member must be made thicken than tension member because of the buckling or column effect in compression member.

4. 4. Simple Truss
A simple truss is constructed by starting with a basic triangular element and connecting two members to form an additional element. E D B C A
unstable truss P C D A
Simple form of rigid or stable truss

5. 6.2 the Method of Joints 1. Objective
Evaluate the force acting on each member of a truss structure to analyze or design a truss
2. Concept
A truss is in equilibrium, so each of its joint also is in equilibrium
Force system acting at pin or joint is coplanar and concurrent
Therefore, we have
member equilibrium is automatically satisfied at each joint
Only force equilibrium ΣFx=0 and ΣFy=0 is necessary

6. For the given truss, only joint B is satisfied.
3. Analysis Procedure
Support A is pin constraint.
Support C is roller constraint.
(1) Draw the free-body diagram of a joint having at least one known force and at most two unknown forces.
For the given truss, only joint B is satisfied.
F.B.D of joint B

7. (2) Establish the sense of the unknown forces
(a) assume the unknown member forces to be in tension.
(b) Determine correct sense of unknown member forces by inspection

8. FBC=- 500/sin45°=-500√2N (in compression)
(3) Apply the force equilibrium equations
ΣFx= FBC sin45°=0
ΣFy=0 -FAB-FBCcos45°=0
(4) Solve the unknown forces and check their correct sense
FBC=- 500/sin45°=-500√2N (in compression)
FAB=-FBCcos45°=500N (in tension)
(5) Continue to analyze then joints by repeating steps (1) to (4)

9. choose joint C first, then joint A
F.B.D of joint C
equilibrium equation at joint C
ΣFx=0 500√2 cos45°-FAC=0
ΣFy=0 Cy-500√2 sin45°=0
FAC=500N (in tension); Cy=500N
F.B.D joint A
Ax=Ay=500N

10. 6-3 zero-force members 1. zero-force member
A member supports no loading, which is used to increase the stability of the truss and to provide support if the applied loading is changed.
2. Rules for determining zero-force members
(1) If only two members form a truss joint and no external load or support reaction is applied to the joint. F E C B A D
Ex.
Joint A and joint D are formed by two members AF &AB, ED&CD respectively and no loading is applied. Hence, members AF, A, ED and CD are zero-force members.

11. ΣFx=0 , so FAB=0 ΣFy=0 , so FAF=0 ΣFx=0, FED+FDC cosθ=0
F.B.D of joint A
Equations of equilibrium
FAF
ΣFx=0 , so FAB=0
ΣFy=0 , so FAF=0 A
FAB
Equations of equilibrium
F.BD of joint D
ΣFx=0, FED+FDC cosθ=0
ΣFy=0, FDC sinθ=0
So, FDC=0,FED=0 D
FED x
FDC y F E C P b B
The equivalent system
is given as shown.

12. (2) If three members form a truss joint for which two of the members are collinear , the third member is zero-force members provided no external force or support reaction is applied to the joint.
EX: P D C B E A
Joint D :3 members AD, ED and CD (ED &DC collinear)
Joint C :3 members AD, DC and CB (DC& CB collinear)
No external force or support reaction is applied to the joint D & C.
∴AD and AC zero-force members

13. 6-4 The method of section 1. Purpose
Determine the loadings acting within a body .
2. Principle
If a body is equilibrium , then any part of the body is also in equilibrium. T T F
In equilibrium
F=T
Internal tension force T
In equilibrium
Imaginary
section
Apply the equations of equilibrium to the sectioned part to determine the loading at the section.

14. Equations of equilibrium
3. Analysis procedure C A B 2m 45
500N
(1) Determined the external reactions at the constraints F.B.D of entire truss.
F.B.D. of entire truss
500N C Cy Ay x y Ax A 2m B
Equations of equilibrium +
500-Ax=0
Cy-Ax=0 +
500x2-Cyx2=0 +

15. (2) Cut or section the truss through not more than three members in which the forces are unknown. (Because three independent equilibrium equations for three unknowns.) C
500N A
(1)
(2)
(3)
three possible ways
of section for given truss
(3) Draw the free-body diagram of the part of sectioned truss which has the least number of forces acting on it.
500N B A C
Section(1)

16. + (4) Establish the sense of the unknown member forcey 45 x F AB
FBC
(5) Apply the equations of equilibrium and check the correct sense of solve forces. +
500N-FBCsin45=0
FBCcos45-FAB=0
FBC=500
(in compression)
FAB=500N
(in tension)
Moments should be summed about a point lying at the intersection of the lines of actions of two unknown forces. The third unknown is determined directly from the moment equation.
Forces may be summed perpendicular to the direction of the two unknown forces which are parallel.

17. (6) Continue to analyze other new sections by repeating steps(1)to(5).x y
FAC
F.B.D. of right part of section (2)
Equations of equilibrium +
So, FAC=500N

18. 6.6 Frames and Machines 1. Definition (1)Frames
A stationary structure composed of pin-connected
multiforce members is used to support loads.
Ex:
bicycle frame
hoist E
油壓缸
Engine jig.

19. (2) Machine
A structure composed of pin-connected multiforce
members with moving parts is designed to
transmit and alter the effect of forces.
Ex:crusher
CAN F
Moving part
2. Assumptions
(1) The structure is properly supported.
(2) The structure contains no more supports or members than necessary to prevent its collapse.
(3) The joint reactions of the structure can be determined from the equilibrium equations.

20. (1) Draw the free body Diagram of entire structure,
3. Analysis procedure 2m
200N C B A 3m 60
(1) Draw the free body Diagram of entire structure,
a part of structure, or each of its members.
(a) Isolate each part by drawing its outline shape.
Show all forces and/or couple moments act on
the part.
(b) Identify all the two-force member in the
structure.
(c) Forces common to any two contacting
members act with equal magnitudes but
opposite sense on respective members.

21. , Cx , Cy (3) (2) Apply the equations of equilibrium
Free body diagrams of members AB & BC. 60 A 2m
200N Cx Cy
By inspection, member AB is a two-force member.
(2) Apply the equations of equilibrium
(a) Count the total number of unknowns to make
sure that an equivalent number of equilibrium
equations can be written for solutions.
, Cx , Cy (3)
Unkoown:

22. (b) Moments should be summed about a point that
lies at the intersection of the lines of action of as
many unknown forces as possible.
Equations of equilibrium for member BC.
(2) Check the correct sense of the unknown forces.

23. 第六章 完