ANALYSIS OF STATICALLY DETERMINATE STRUCTURES ERT 256/2 THEORY OF STRUCTURES ANALYSIS OF STATICALLY DETERMINATE STRUCTURES MRS SITI KAMARIAH BINTI MD SA’AT SCHOOL OF BIOPROCESS ENGINEERING

Idealized Structure To develop the ability to model or idealize a structure so that the structural engineer can perform a practical force analysis of the members Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Support Connection Pin connection (allows some freedom for slight rotation) Roller support (allows some freedom for slight rotation) Fixed joint (allows no relative rotation)

Idealized Structure Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Idealized Structure In reality, all connections and supports are modeled with assumptions. Need to be aware how the assumptions will affect the actual performance. Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Type of connection

Idealized Structure Consider the jib crane & trolley, we neglect the thickness of the 2 main member & will assume that the joint at B is fabricated to be rigid The support at A can be modeled as a fixed support Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Idealized Structure Consider the framing used to support a typical floor slab in a building The slab is supported by floor joists located at even intervals These are in turn supported by 2 side girders AB & CD For analysis, it is reasonable to assume that the joints are pin and/or roller connected to girders & the girders are pin and/or roller connected to columns Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Idealized Structure Tributary Loadings 1-way system : (Ly/Lx ≥ 1.5)

Idealized Structure Tributary Loadings 2-way system : (Ly/Lx < 1.5)

Example 2.1 The floor of a classroom is supported by the bar joists. Each joist is 4.5m long and they are spaced 0.75m on centers. The floor is made from lightweight concrete that is 100mm thick. Neglect the weight of joists & the corrugated metal deck, determine the load that acts along each joist. Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Solution Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Principle of Superposition Total disp. (or internal loadings, stress) at a point in a structure subjected to several external loadings can be determined by adding together the displacements (or internal loadings, stress) caused by each of the external loads acting separately Linear relationship exists among loads, stresses & displacements 2 requirements for the principle to apply: Material must behave in a linear-elastic manner, Hooke’s Law is valid The geometry of the structure must not undergo significant change when the loads are applied, small displacement theory

Principle of Superposition =

STATIC EQUILIBRIUM

Static Equilibrium The state of an object when it is at rest or moving with a constant velocity. There may be several forces acting on the object. If they are canceling each other out and the object is not accelerating, then it is in a state of static equilibrium.

Equations of Equilibrium A structure or one of its members in equilibrium is called statics member when its balance of force and moment. In general this requires that force and moment in three independent axes, namely

Equations of Equilibrium In a single plane or 2D structures, we consider

Static Equilibrium * The assumed positive direction is as indicated. Since the externally applied force system is in equilibrium, the three equations of static equilibrium must be satisfied, i.e. +ve ↑ ΣFy = 0 The sum of the vertical forces must equal zero. +ve ΣM = 0 The sum of the moments of all forces about any point on the plane of the forces must equal zero. +ve → ΣFx = 0 The sum of the horizontal forces must equal zero. * The assumed positive direction is as indicated.

DETERMINACY AND STABILITY

Determinacy If the reaction forces can be determined solely from the equilibrium EQs  STATICALLY DETERMINATE STRUCTURE No. of unknown forces > equilibrium EQs  STATICALLY INDETERMINATE STRUCTURE Can be viewed globally or locally (via free body diagram)

Determinacy Determinacy and Indeterminacy For a 2D structure The additional EQs needed to solve for the unknown forces are referred to as compatibility EQs No. of components No. of reaction support

Reaction on a support connection For rolled support (r = 1) For pin support (r = 2) For fixed support (r=3) Fy Fy Fx Fy Fx M

Example 2.3 Classify each of the beams as statically determinate or statically indeterminate. If statically indeterminate, report the no. of degree of indeterminacy. The beams are subjected to external loadings that are assumed to be known & can act anywhere on the beams. Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Solution Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Example 2.4 Classify each of the pin-connected structures as statically determinate or statically indeterminate. If statically indeterminate, report the no. of degree of indeterminacy. The structures are subjected to arbitrary external loadings that are assumed to be known & can act anywhere on the structures. Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Solution Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Stability Stability - Structures must be properly held or constrained by their supports In general, a structure is geometrically unstable if there are fewer reactive forces then equations of equilibrium. An unstable structure must be avoided in practice regardless of determinacy.

Stability Partial constraints Fewer reactive forces than equilibrium EQs Some equilibrium EQs can not be satisfied Structure or Member is unstable

Stability Improper constraints In some cases, unknown forces may equal equilibrium EQs However, instability or movement of structure could still occur if support reactions are concurrent at a point

Stability Improper constraints Parallel Concurrent

Stability

Application of the Equations of Equilibrium Free–Body Diagram - disassemble the structure and draw a free–body diagram of each member. Equations of Equilibrium - The total number of unknowns should be equal to the number of equilibrium equations

Determine the reactions on the beam as shown. Example 2-7 Determine the reactions on the beam as shown. 135 kN 60.4 kN 173.4 kN 50.7 kN Ignore thickness 183.1 kN Ignore thickness Draw Free Body Diagram Use Eq of Equilibrium

Solution Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Example 2.10 The compound beam in Fig 2.30(a) is fixed at A. Determine the reactions at A, B & C. Assume the connections at B is a pin & C a roller. Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Solution Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Example 2.13 The side of the building subjected to a wind loading that creates a uniform normal pressure of 15kPa on the windward side & a suction pressure on the leeward side. Determine the horizontal & vertical components of reaction at the pin connections A, B & C of the supporting gable arch. Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Solution Since the loading is evenly distributed, the central gable arch supports a loading acting on the walls & roof of the dark-shaded tributary area. This represents a uniform distributed load of (15kN/m2)(4m)=60kN/m on the windward side and (5kN/m2)(4m)=20kN/m on the suction side as shown. Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Solution By applying equilibrium equations in the following sequence, Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

Solution Chapter 2: Analysis Statically Determinate Structures Structural Analysis 7th Edition © 2009 Pearson Education South Asia Pte Ltd

EXERCISES – discuss in tutorial 2-3 2-4 2-10 2-13 2-17 2-19 2-20 2-22 2-23 2-24 2-39 2-40 2-43 2-49 2-50

SUMMARY Difference between an actual structure and its idealized model Principle of superposition Equilibrium, determinacy and stability Analyzing statically determinate structures

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