MECH 0003 Notes - Chapter 3 2020 PDF

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MECH 0003

Applied Mechanics - Statics

3. Analysis of Structures 3.1 Simple Trusses (Planar pin-jointed Frame) The concept of a plane pin-jointed frame is used widely in the design and analysis of engineering structures. The concept involves a planar configuration of bars that are joined at their ends by frictionless hinges (pins) – loads are applied only at the joints. Although the concept represents an obvious simplification of reality, it can provide a very good insight into the primary behaviour of triangulated structures such as trusses.

2 Force Member: • Force direction along the bar • Tension or compression

F

F

Each member of the simple truss acts as a “two -force member”. The forces at the ends of the members are directed along the axis of the member. 29

MECH 0003

Applied Mechanics - Statics

Analysing one hinge (pin of simple truss): FBD members

FBD Hinge

F5

F4 F5 F4

F5

F4

F3

F3

F3

F1 F2 F2 F1

F2

F1 Each member of the truss acts as a two-force-member with the direction of the force along its orientation. Typically the member is assumed to be in tension and does not require further analysis. From the action = reaction principle we know that the same force magnitude, but opposite direction acts on the hinge (FBD).

For the hinge we assume that all members of the truss are connected in the same point (frictionless pin), so their lines of action all intersect in this point and the moment equilibrium condition is always fulfilled. For each hinge we can then evaluate the force equilibrium conditions in two directions to find the force in the members attached to the hinge. ∑ M(C) = 0 (always fulfilled) ; ∑ Fx = 0; ∑ Fy = 0

A statically determinate pin-jointed frame is one which has just sufficient members to make it stable. A frame with more than this is often referred to as being redundant and one that has too few members is a mechanism (i.e. unable to resist any applied load). 30

MECH 0003

Applied Mechanics - Statics

Example of a pin-jointed truss The diagram below represents a statically determinate pin-jointed truss under load. The fixed support at A prevents movement in both the horizontal and vertical directions whilst the roller support at B prevents only vertical motion. The length of each truss member is 3m.

-

3 unknown support reactions FBD of system, 3 equilibrium conditions => find support reactions in A, B

-

Each member of truss, 1 unknown force [2 force member, direction of force along the member]

FBD of system => support reactions

60° F1 = 100 N Ay ∑ 𝐹𝑥 ∶ 𝐴𝑥 = 0

3m

3m

Ax

By

∑ 𝐹𝑦 ∶ 𝐴𝑦 + 𝐵𝑦 − 100𝑁 = 0 ∑ 𝑀(𝐴) : 𝐵𝑦 ∙ 3𝑚 − 100𝑁 ∙ 6𝑚 = 0 => 𝐵𝑦 = 100𝑁 ∙

6𝑚 = 200𝑁 3𝑚

=> 𝐴𝑦 = 100𝑁 − 𝐵𝑦 = −100𝑁

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MECH 0003

Applied Mechanics - Statics

3.2 Method of Joints Consider the FBD for each joint (node) and member in the structure:

Ax = 0 By = 200N

Ay = -100N •

Known direction of force in each member is along the direction of the member

•

Go systematically through each pin / joint

•

For each pin: FBD, equilibrium conditions (  Fx ,

•

Good point to start at pin where only 2 unknown forces, as can be solved

F

y

)

immediately (if possible) •

Pin C is a good starting point

S9

60° S9 S8

S8

60°

S9

S8 Equilibrium conditions for pin C:

100N

∑ Fy : −100𝑁 + 𝑆9 sin 60° = 0 => 𝑆9 =

100𝑁 100𝑁 = 115𝑁 = sin 60° √3 2

[tension]

∑ Fx : −𝑆8 − 𝑆9 cos 60° = 0 1 => 𝑆8 = −𝑆9 cos 60° = −115𝑁 ∙ = −57.5𝑁 2

[compression]

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MECH 0003

Applied Mechanics - Statics

•

Go systematically through structure

•

Equilibrium conditions for pin C: S6

S6 S9

S6 S7

S9

S7

60°

60°

S9 S7 ∑ 𝐹𝑦 : −𝑆7 sin 60° − 𝑆9 sin 60° = 0 => 𝑆7 = −𝑆9 = −115𝑁 [𝑐𝑜𝑚𝑝. ] ∑ 𝐹𝑥 : −𝑆6 − 𝑆7 cos 60° + 𝑆9 cos 60° = 0 => 𝑆6 = −𝑆7 ∙

1 1 1 1 + 𝑆9 ∙ = −(−115𝑁) ∙ + 115𝑁 ∙ = 115𝑁 [tension] 2 2 2 2

Continue with Equilibrium conditions for Pin B: S4 , S5

Note that the forces defined by the arrows on each member are forces acting on the joints. The joints will be exerting an equal and opposite force on each member. A good procedure is to write the equations for joint equilibrium as though each member is in tension. Following this approach, the values of the forces resulting from the analysis will be positive if the members are in tension, and negative if in compression.

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MECH 0003

Applied Mechanics - Statics

3.3 Zero-Force Members The analysis of a simple truss can be significantly simplified if we manage to identify members which will carry no load. Such members are often included into a structure to make it more redundant:

S2

S2 S2 S1 𝐄𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 𝐜𝐨𝐧𝐝𝐢𝐭𝐢𝐨𝐧𝐬:

S1

60°

∑ 𝐅𝐲 : −𝐒𝟏 𝐬𝐢𝐧 𝟔𝟎° = 𝟎 => 𝑺𝟏 = 𝟎 ∑ 𝐅𝐱 : 𝐒𝟐 + 𝑺𝟏 𝐜𝐨𝐬 𝟔𝟎° = 𝟎 => 𝑺𝟐 = 𝟎

S1 Sufficient, if zero force member is identified from equilibrium condition at one end, as 2 force members, force needs to be balanced.

F2

zero force members

zero force member

F2

SA

𝛼

SB zero force member

∑ 𝐹𝑦 : 𝑆𝐵 sin 𝛼 = 0; => 𝑆𝐵 = 0

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MECH 0003

Applied Mechanics - Statics

3.4 Method of Sections The method of sections can sometimes be employed to simplify the analysis of a truss by „cutting“ or sectioning the members of the truss, in order to draw the free-body diagram of its two parts. Since all parts need to be at equilibrium for the complete system to be at equilibrium, three independent equilibrium conditions (in 2D) can be applied to each part of the truss. Deciding the position of the „cut“ and considering ways to write the equilibrium conditions, quite often equations can be written that yield a direct solution for the one of the unknown reactions.

Ax Ay

By S6

S6

S6

S6

S5 S5 S5 S4

Ax

S4

S4

F1

S5

S4 Ay

By

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MECH 0003

Applied Mechanics - Statics

•

Section / cut through structure, cutting not more than 3 members

•

Force in 3 members from FBD, equilibrium conditions, either left or right part

•

Often easy to use moment equilibrium conditions 1.5 m

∑ 𝐹𝑥 : 𝑆4 + 𝑆6 + 𝑆5 cos 60° = 0

3𝑚 ∙

∑ 𝐹𝑦 : 𝐴𝑦 − 𝑆5 sin 60° = 0;

𝐴𝑦 −100𝑁 = = −115𝑁 => 𝑆5 = sin 60° √3 2

3m [𝑐𝑜𝑚𝑝. ]

√3 − 𝐴𝑦 ∙ 1.5𝑚 = 0 2 𝐴𝑦 −100𝑁 = = −57.5𝑁 [𝑐𝑜𝑚𝑝. ] => 𝑆4 = √3 √3 1 1 => 𝑆6 = −𝑆4 − 𝑆5 ∙ = 57.5𝑁 + 115𝑁 = 115𝑁 2 2 ∑ 𝑀(𝐸) : 𝑆4 ∙ 3𝑚

E

1.5 m √3 2

S6 60° S5

3m S4

Ax = 0

Ay = -100 N

[𝑡𝑒𝑛𝑠𝑖𝑜𝑛]

Analysing the statics of structures, we are often faced with frames, members connected by pins but bearing distributed loads (e.g. weight) or loads not applied at the joints. In these cases, the analysis must follow the general procedure of drawing the free-body diagrams for each part, identifying two-force members, and applying the equations of equilibrium (often in scalar notation).

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