Chapt 04a beams

Internal Forces7

Engineering Mechanics: Statics in SI Units, 12e

Copyright © 2010 Pearson Education South Asia Pte Ltd


Chapter Objectives

• Method of sections for determining the internal loadings in a member

• Develop procedure by formulating equations that describe the internal shear and moment throughout a member


Chapter Outline

1. Internal Forces Developed in Structural Members2. Shear and Moment Equations and Diagrams3. Relations between Distributed Load, Shear and

Moment


7.1 Internal Forces Developed in Structural Members

• The design of any structural or mechanical member requires the material to be used to be able to resist the loading acting on the member

• These internal loadings can be determined by the method of sections



• Force component N, acting normal to the beam at the cut session

• V, acting tangent to the session are normal or axial force and the shear force

• Couple moment M is referred as the bending moment



• For 3D, a general internal force and couple moment resultant will act at the section

• Ny is the normal force, and Vx and Vz are the shear components

• My is the torisonal or twisting moment, and Mx and Mzare the bending moment components



Procedure for AnalysisSupport Reactions• Before cut, determine the member’s support reactions• Equilibrium equations used to solve internal loadings

during sectioning

Free-Body Diagrams• Keep all distributed loadings, couple moments and

forces acting on the member in their exact locations• After section, draw FBD of the segment having the

least loads



Procedure for AnalysisFree-Body Diagrams (Continue)• Indicate the z, y, z components of the force, couple

moments and resultant couple moments on FBD• Only N, V and M act at the section• Determine the sense by inspection

Equations of Equilibrium• Moments should be summed at the section • If negative result, the sense is opposite


Example 7.3

Determine the internal force, shear force and the bending moment acting at point B of the two-member frame.


Solution

Support ReactionsFBD of each memberMember AC∑ MA = 0;-400kN(4m) + (3/5)FDC(8m)= 0

FDC = 333.3kN (C)+→∑ Fx = 0;-Ax + (4/5)(333.3kN) = 0

Ax = 266.7kN+↑∑ Fy = 0;Ay – 400kN + 3/5(333.3kN) = 0

Ay = 200kN


Solution

Support ReactionsMember AB+→∑ Fx = 0; NB – 266.7kN = 0

NB = 266.7kN+↑∑ Fy = 0; 200kN – 200kN - VB = 0

VB = 0∑ MB = 0; MB – 200kN(4m) + 200kN(2m) = 0

MB = 400kN.m


7.2 Shear and Moment Equations and Diagrams

• Beams – structural members designed to support loadings perpendicular to their axes

• A simply supported beam is pinned at one end and roller supported at the other

• A cantilevered beam is fixed at one end and free at the other



Procedure for AnalysisSupport Reactions• Find all reactive forces and couple moments acting on

the beam• Resolve them into components

Shear and Moment Reactions• Specify separate coordinates x• Section the beam perpendicular to its axis• V obtained by summing the forces perpendicular to

the beam• M obtained by summing moments about the sectioned

end



Procedure for AnalysisShear and Moment Reactions (Continue)• Plot (V versus x) and (M versus x)• Convenient to plot the shear and the bending moment

diagrams below the FBD of the beam


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.


Solution

Support ReactionsFBD of the shaft

mxkNMMkNVFy

.5.2;0

5.2;0

mkNxMxkNmxkNMM

kNVVkNkNFy

.)5.210(0)(5.2)2(5 ;0

5.2055.2 ;0


Solution

Shear diagramInternal 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.

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Chapt 04a beams