Chapter 4 - Structures

DJJ3053 ENGINEERING MECHANICS CHAPTER 4 : STRUCTURES Luqman Nul Hakim Bin Juwara Jabatan Kejuruteraan Mekanikal Politeknik Muadzam Shah [email protected]

Structures  This topic introduces the concept to analyze truss structures by using methods of joints and methods of sections. Learning Outcomes (LO)  Upon completion of this topic, students should be able to : 4.1 Explain plane trusses 4.1.1 Define a truss 4.1.2 Explain plane trusses 4.2 Analyze force in trusses by using related method 4.2.1 Analyze a truss by using the method of joints 4.2.2 Analyze the force in the members of a truss by using method of sections

Structures Consist of three categories of Engineering Structures. 1. TRUSSES  Framework composed of members

joined at their ends to form a rigid structures. 2. FRAMES  Stationary and can support external

loads, which contain at least one multi-force member. 3. MACHINES  Structures containing moving parts

designed to transmit and modify forces.

Trusses TRUSS  Framework composed of members joined at their ends to form a rigid structures. PLANE TRUSS Member of truss lie in same plane. - A planar truss lies in a single plane. In plane truss, both the truss structure and the applied loads lie in the same plane. The analysis of forces will be in 2 dimension.

Trusses

June 2015 Session : 2(a) Describe a plane truss. (3 marks)

Simple Trusses  3 bars joined with pins at end

Bar

Pin

Non-Rigid Rigid Bars Bars

form a triangular truss.  Rigid Bars (non-collapsible).  Non-Rigid Bars can be made rigid.  SIMPLE TRUSS  Attaching 2 or more members & connecting these members by expanding the basic triangular truss will form a larger truss

Assumption For Design 1. All loadings are applied at the joints. 2. Weight can be included. 3. The members are joined together by smooth pins

(bolting or welding). 4. Two force members – equilibrium only in two forces either TENSION or COMPRESSION. 5. Two force are applied at the end; they are equal, opposite and collinear for equilibrium.

Method to determine the correct SENSE of an unknown member force SENSE – unknown member force can be assumed.  POSITIVE – indicates the sense is correct.  NEGATIVE – indicates the sense is reversed. Always assume the unknown member force acting on the joint FBD as tension (pull) on the pin.  POSITIVE – in Tension (T)  NEGATIVE – in Compression (C)

Two Methods To Analyze Force In Simple Truss

Method of JOINTS  Satisfying the conditions of equilibrium for the forces acting

on the connecting pin of each joint.  Each joint is subjected to a force system that is coplanar and concurrent.  Only 2 independent equilibrium equations are solved, ∑Fx = 0 and ∑Fy = 0. Simple Beam Truss

Example 1 Determine the force in each member of the truss shown and indicate whether the members are in tension or compression.

Ans:

FBC  707 N (C) FBA  500 N (T) FCA  500 N (T)

C y  500 N Ax  500 N (T) Ay  500 N (T)

Solution: Joint B :

+ve  ∑Fx = 0; 500 N + FBC sin 45o = 0  FBC = -707 N = 707 N (C) +ve ↑ ∑Fy = 0; -FBC cos 45o – FBA = 0 -(-707 N) cos 45o = FBA  FBA = 500 N (T)

Solution: Joint C :

+ve  ∑Fx = 0; -FCA + 707 N cos 45o = 0  FCA = 500 N (T) +ve ↑ ∑Fy = 0; -707 sin 45o + Cy = 0  Cy = 500 N

Solution: Joint A :

+ve  ∑Fx = 0; 500 N – Ax = 0  Ax = 500 N (T)

+ve ↑ ∑Fy = 0; 500 N – Ay = 0  Ay = 500 N (T)

Example 2 The truss used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Set Set P1 = 800 N and P2 = 0 N. Ans: FAD  1131.4 N (C) FAB  800 N (T) FBD  0 FBC  800 N (T) FDC  1131.4 N (T) FDE  1600 N (C)

Example 3 Determine the force in each member of the truss and state if the members are in tension (T) or compression (C). Set P1 = 500 N and P2 = 100 N.

Ans: FBA  285.71 N (T) FBC  383.86 N (T) FCA  271 N (C) C y  271.43 N

Example 4 Determine the force in each member of the truss shown. Indicate whether the members are in tension or compression. Ans:

C x  600 N (C) Ay  600 N (C) C y  200 N (C) FAB  750 N (C) FAD  450 N (T) Ay  500 N (T) FDB  250 N (T) FDC  200 N (C) FCB  600 N (C) C y  200 N (C)

Solution: Support Reaction :

+ve  ∑Fx = 0; 600N – Cx = 0

+ve

 Cx = 600 N (C)

∑Mc = 0; Ay(6m) – 400N(3m) – 600N(4m) = 0  Ay = 600 N (C)

+ve ↑ ∑Fy = 0; Ay – 400N – Cy = 0  Cy = 200 N (C)

Solution: Joint A :

+ve ↑ ∑Fy = 0; 600N + (⅘)FAB = 0  FAB = -750 N  FAB = 750 N (C) +ve  ∑Fx = 0; FAD + (⅗)FAB = 0

FAB

5

4 3

FAD Ay = 600N

 FAD = 450 N (T)

Solution: Joint D :

+ve  ∑Fx = 0; -450N – (⅗)FDB + 600N = 0  FDB = 250 N (T) +ve ↑ ∑Fy = 0; (⅘)FDB + FDC = 0

FDB 5

4

FDC

3

FAD = 450N

600 N

 FDC = -200 N  FDC = 200 N (C)

Solution: Joint C :

+ve  ∑Fx = 0; -600N – FCB = 0

 FCB = -600 N  FCB = 600 N (C)

+ve ↑ ∑Fy = 0; FDC– Cy = 0  FDC = Cy = 200 N (C)

Cy = 200N FC B

Cx = 600N FDC = 200N

Solution:

June 2015 Session : 2(c) Figure below shows a truss is subjected to a horizontal force of 500N. i. Calculate the force in each member of the truss. (16 marks) ii. Identify whether the members are in tension or compression form. (4 marks) Ans:

C x  500 N Ay  250 N C y  250 N FAD  353.553 N (C) FAB  250 N (T) FBC  250 N (T) FBD  0 N FCD  353.553 N (T)

Two Methods To Analyze Force In Simple Truss

Method of SECTIONS A truss is divided into two parts by taking an imaginary “cut” through the truss. 2. Decide how you need to cut the truss: a. Where you need to determine forces b. Where the total number of unknown does not exceed than 3. 3. Decide which side of the cut truss will be easier to work with (minimize the number of force you have to find). 4. If required, determine the necessary support reactions by drawing the FBD of the entire truss. 1.

Example 5 Determine the force in members GE, GC & BC of the truss shown. Indicate whether the members are in tension or compression.

Ans: FBC  800 N (T) FGE  800 N (C) FGC  500 N (T)

Solution:

+ve  ∑Fx = 0; 400N – Ax = 0 +ve

 Ax = 400 N

∑MA = 0; 1200N (8m) + 400N (3m) - Dy (12m) = 0  Dy = 900 N +ve ↑ ∑Fy = 0; Ay – 1200N + 900= 0  Ay = 300 N

Solution:

+ve ∑MG = 0; 300N(4m) + 400N (3m) - FBC (3m) = 0  FBC = 800 N (T)

+ve ∑Mc = 0; 300N (8m) - FGE (3m) = 0 +ve ↑ ∑Fy = 0; 300N - (3/5)FGC = 0

 FGE = 800 N (C)  FGC = 500 N (T)

Example 6 Determine the force in members BC, CG & GF of the Warren truss shown. Indicate whether the members are in tension or compression.

Ans: FGF  8.08 kN (T) FBC  7.70 kN (C) FCG  0.770 kN (C)

Example 7 Determine the force in members CD, CF & FG of the Warren truss shown. Indicate whether the members are in tension or compression.

Ans: FFG  8.08 kN (T) FCD  8.47 kN (C) FCF  0.770 kN (T)

Example 8 Determine the force in members BC, HC & HG of the bridge truss and indicate whether the members are in tension or compression.

Ans: FHG  29.0 kN (C) FBC  20.5 kN (T) FHC  12.0 kN (T)

Example 9 Determine the force in members CD, CF & GF of the bridge truss and indicate whether the members are in tension or compression.

Ans: FGF  29.0 kN (C) FCD  23.5 kN (T) FCF  7.78 kN (T)

Thank You For Listening QUESTION & ANSWER SESSION