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 Members
2. Shear and Moment Equations and Diagrams
3. 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 Mz
are the bending moment components

Procedure for Analysis
Support 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

Free-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 Reactions
FBD of each member
Member 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 Reactions
Member 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

Support 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

Shear 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 Reactions
FBD of the shaft
mxkNMM
kNVFy
.5.2;0
5.2;0


mkNxM
xkNmxkNMM
kNV
VkNkNFy
.)5.210(
0)(5.2)2(5;0
5.2
055.2;0





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.

Chapt 04a beams

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