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Internal Forces7
Engineering Mechanics: Statics in SI Units, 12e
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Chapter Objectives
• Method of sections to find internal forces in a structure • Write equations to describe shear forces and moments for an entire member
Chapter Outline1. Internal Forces Developed in Structural Members 2. Shear and Moment Equations and Diagrams 3. Relations between Distributed Load, Shear and Moment
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7.1 Internal Forces Developed in Structural Members
• To design a part or member of a structure or machinery, the material of the member must be able to withstand or resist internal forces in the member
• The internal forces can be found by method of section • The Axial force N is perpendicular to the cut • The Shear force V is parallel to the cut and perpendicular to N • The Moment M or Bending moment
Positive sign convention
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7.1 Internal Forces Developed in Structural Members
• For 3D internal forces and forces are shown below • Ny is a normal force, Vx and Vz are shear force
• My is torisonal or twisting moment, Mx and Mz are bending moment
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7.1 Internal Forces Developed in Structural Members
Procedure for Analysis Support Reactions • Before cutting a section, find all support reactions
Free-‐Body Diagrams • Write all forces acting on FBD • Consider FBD ,where there are fewer forces for the ease of analysis • Write all known forces in x, y, z and moment • Write in internal forces as N, V, M at the cut and assume the directions
as shown in previous slide
Equations of Equilibrium • Find magnitude and directions from calculation +-‐
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Example 7.3
Determine the internal force, shear force and the bending moment acting at point B of the two-‐member frame.
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Solution
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Solution
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7.2 Shear and Moment Equations and Diagrams
• Beam is a structural member designed to take the load that are perpendicular to the member
• A simply supported beam has a pin support on one side and the other side is roller
• A cantilever beam has one support as a fix support and the other end is free
simply supported beam cantilever beam
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7.2 Shear and Moment Equations and Diagrams
• To diesign a beam, require finding shear and bending moment for an entire beam
• For every point on a beam, the shear V(x) and bending moment M(x) can be found by the method of section, expressed in a function of x
• V(x) and M(x) are not continuous at the point when there is a change in a distributed load or at a point where the concentrated force is located, hence need to consider equations at interval at these critical point
• The graph ofV(x) and M(x) are called Shear Force Diagram (SFD) and Bending Moment Diagram (BMD), respectively
SFD BMD
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7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Support Reactions • Find forces and moments acting on a beam and then find
components parallel and perpendicular to the beam
Shear and Moment Functions • Define x and then start from the left end of a beam and
write and equation for each interval, consider especially locations where concentrated forces action or where distributed load changed
• When cutting a beam at x draw an FBD and use the notation of V and M as explained earlier (sign convention)
• Find V and M from equilibrium of the FBD of an easy part
sign convention
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7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Shear and Moment Diagrams • Draw SFD for V and x and BMD for M and x. For the calculation with
positive value, plot above x and negative below x • Draw SFD & BMD after FBD of a beam
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Example 7.7
Draw the shear and bending moments diagrams for the shaft. The support at A is a thrust bearing and the support at C is a journal bearing.
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Solution
Support Reactions From equilibrium equations
FBD of the shaft
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Solution
Shear diagram Internal shear force is always positive within the shaft AB. Just to the right of B, the shear force changes sign and remains at constant value for segment BC.
Moment diagram Starts at zero, increases linearly to B and therefore decreases to zero. Mmax = 5 kN.m
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7.3 Relations between Distributed Load, Shear and Moment
Distributed Load
• For the beam AD under the distributed load w = w(x) and concentrated force (F1, F2) and moment (M1, M2)
• For vertical forces, the forces is positive when acting upwards
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Distributed Load
• Cut FBD for an infinitestimal part with the length ∆x and located at x, which does not have forces or moments acting
• Write V and M if they are upwards • The resultant force is ∆F = w(x) ∆x and located at k (∆x) from the right 0<k<1 (k=0.5 for uniform load)
7.3 Relations between Distributed Load, Shear and Moment
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7.3 Relations between Distributed Load, Shear and Moment
Relations between Distributed Load and Shear
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7.3 Relations between Distributed Load, Shear and Moment
Relations between Shear and Moment
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7.3 Relations between Distributed Load, Shear and Moment
Concentrated Force and Couple Moment
• If the external force F is upward, the shear force increases
• For a clockwise moment, the bending moment increases
FV =Δ
0MM =Δ
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Example 7.8
Draw the shear and moment diagrams for the cantilever beam.
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Solution
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Example 7.9
Draw the shear and moment diagrams for the overhang beam.
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Solution
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