Chapter 1_Statics
Transcript of Chapter 1_Statics
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UTMUNIVERSITI TEKNOLOGI MALAYSIA
CHAPTER 1 – STATICS
Mohamed Ruslan Abdullah, PhDDepartment of Applied Mechanics
Faculty of Mechanical Engineering Universiti Teknologi Malaysia
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Introduction
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Introduction
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Unit
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Statics of Particle
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Statics of Particle
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Statics of Particle
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Statics of Particle
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Statics of Particle
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Statics of Particle
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Statics of Particle
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Statics of Particle
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Statics of Particle
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Statics of Particle : Example
Three forces, F1, F2 and F3 act on point A,determine the resultant, R.
Example 1
Three forces, F1, F2 and F3 act on point A. If the resultant is known to be on the a-a axis, determine the magnitude of F3 and the resultant R.
Example 2
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Statics of Rigid Bodies
Previously, only concurrent forces are considered in analyses when bodies are treated as particles. In reality, forces act at different points of application, so the size of the body will have to be taken into consideration. Thus the assumption that all forces are concurrent is no longer valid.
Most of the bodies considered in statics are assumed to be rigid. A rigid body is defined as a body that does not deform.
Each of the forces acting on a rigid body can impart a motion of translation or rotation, or both.
Forces exerted on rigid bodies can be categorised as
External forces: action of other bodies (causing it to move or ensure that it remains at rest) on the rigid body under consideration.
Internal Forces : forces that hold particles forming the rigid body together. For rigid body with several parts, the forces holding each part together.
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Statics of Rigid Bodies: The concept of moment
The moment of a force about a point is defined by
Observe that the moment of force F about point B does not depend on point A, but on the line of action of the force.
Direction of a moment is referred to as clockwise (CW) or counter clockwise (CCW). For the above example, the force F will generate a clockwise moment about point B. The usual sign convention is CCW +ve.
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The concept of moment: Varignon’s Theorem
The moment of a force about a point is equivalent to the summation of themoments of the force’s components about that point.
=
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The concept of moment: Example
Determine the moment of the 1000 N force about point B by resolving the force into its components and use the results to find the shortest distance between the line of force and point B.
Example 3
Example 4
Determine the moment of the 1000 N forceabout point O a. when the force is at A b. when the force is at B c. when the force is at C
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Statics of Rigid Bodies: Free-Body Diagram
4 - 19
First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram.
• Select the extent of the free-body and detach it from the ground and all other bodies.
• Include the dimensions necessary to compute the moments of the forces.
• Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body.
• Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.
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Reactions at Supports and Connections
• Reactions equivalent to a force with known line of action.
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Reactions at Supports and Connections
• Reactions equivalent to a force of unknown direction and magnitude.
• Reactions equivalent to a force of unknown direction and magnitude and a couple of unknown magnitude
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Free-Body Diagram: Example
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Cables/ wires/ ropes and pulleys
All cables are assumed to be inextensible. All forces acting from cables must direct outwards from point of analysis, i.e. in
tension. When a cable passes a pulley, the tension is the same as long as it is the
same cable. Pulleys are assumed to be smooth except stated otherwise. Dimensions of a pulley are usually neglected in calculations except stated
otherwise.
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Cables/ wires/ ropes and pulleys
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Example
Draw the Free Body Diagram for themechanism shown. Mass of bodies is negligible unless stated by m. Allcontacting surfaces are smooth unless otherwise stated.
Determine the tension in cables AB and BC for the system to be in equilibrium.
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Internal forces
• Straight two-force member AB is in equilibrium under application of F and -F.
• An internal force-couple system is required for equilibrium of two-force members which are not straight.
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Types of Beam Loading and Support
• Beams are classified according to way in which they are supported.
• Reactions at beam supports are determinate if they involve only three unknowns. Otherwise, they are statically indeterminate.
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Free-Body Diagram: Example
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POWERPOINT TEMPLATES
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EXAMPLE 5The hoist in the figure consists of the beam AB and attached pulleys, the cable, and the motor. Determine the resultant internal loadings acting on the cross section at C if the motor is lifting the 2000 N (approximate 200 kg) load W with constant velocity. Neglect the weight of the pulleys and beam.
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EXAMPLE 6
Determine the resultant internal loading on the cross section through point C of the pliers. There is a pin at A and the jaws at B are smooth.
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EXAMPLE 7
The sky hook is used to support the cable of a scaffold over the side of a building. If it consists of a smooth rod that contacts the parapet of a wall at points A, B, and C, determine the normal force, shear force, and moment on the cross section at points D and E.
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FRICTION
2 types of friction Dry friction*** Fluid friction