MMAN1300 Shear Force and Bending Moment Laboratory Experiment for Statics
Civil Engineering
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MECHANICS OF MATERIALS Bending Moment and Shear Force Diagrams
MECHANICS OF MATERIAL S
Introduction
Beams - structural members supporting loads at
various points along the member Transverse loadings of beams are classified as concentrated loads or distributed loads Applied loads result in internal forces consisting of a shear force (from the shear stress distribution) and a bending couple (from the normal stress distribution)
MECHANICS OF MATERIAL S
Types of Beam supports
MECHANICS OF MATERIAL S
Shear and Bending Moment Diagrams • We need maximum internal shear force and bending couple in order to find maximum normal and shear stresses in beams.
• Shear force and bending couple at a point are determined by passing a section through the beam and applying an equilibrium analysis on the beam portions on either side of the section. • Sign conventions for shear forces V and V’ and bending couples M and M’
MECHANICS OF MATERIAL S
Example 1 SOLUTION: • Treating the entire beam as a rigid body, determine the reaction forces
For the timber beam and loading shown, draw the shear and bendmoment diagrams
• Section the beam at points near supports and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples • Identify the maximum shear and bending-moment from plots of their distributions.
MECHANICS OF MATERIAL S
Example 1 SOLUTION: • Treating the entire beam as a rigid body, determine the reaction forces from F y 0 M B : R B 40 kN R D 14 kN • Section the beam and apply equilibrium analyses on resulting free-bodies V 1 20 kN F y 0 20 kN V 1 0 M 1 0 20 kN 0 m M 1 0 M 1 0 F y 0
20 kN V 2 0
V 2 20 kN
M 2 0 20 kN 2.5 m M 2 0 M 2 50 kN m V 3 26 kN M 3 50 kN m V 4 26 kN M 4 28 kN m V 5 14 kN M 5 28 kN m V 6 14 kN M 6 0
MECHANICS OF MATERIAL S
Example 1 • Identify the maximum shear and bendingmoment from plots of their distributions. V m 26 kN M m M B 50 kN m
MECHANICS OF MATERIAL S
Example 2 SOLUTION: Replace the 45 kN load with an equivalent force-couple system at D. Find the reactions at B by considering the beam as a rigid body.
The structure shown is constructed of a W 250x167 rolled-steel beam. Draw the shear and bending-moment diagrams for the beam and the given loading.
Section the beam at points near the support and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples.
MECHANICS OF MATERIAL S
Example 2 SOLUTION: Replace the 45 kN load with equivalent forcecouple system at D. Find reactions at B. Section the beam and apply equilibrium analyses on resulting free-bodies. From A to C :
F 0 M 0 y
1
V 45 x kN 45 x V 0 45 x 12 x M 0 M 22.5x 2 kNm
From C to D :
F 0 M 0 y
2
V 108 kN 108 V 0 108 x 1.2 M 0 M 129.6 108 x kNm
From D to B : V 153 kN
M 305.1 153 x kNm
MECHANICS OF MATERIAL S
Example 2
MECHANICS OF MATERIAL S
Relations Among Load, Shear, and Bending Moment • Relationship between load and shear: F y 0 : V V V w x 0 V w x dV dx
w x D
V D V C w dx xC
• Relationship between shear and bending moment: x 0 M C 0 : M M M V x w x M V x 12 w x 2 dM dx
V x D
M D M C
V dx
xC
2
MECHANICS OF MATERIAL S
Example 3 SOLUTION: • Taking the entire beam as a free body, determine the reactions at A and D. • Apply the relationship between shear and load to develop the shear diagram. Draw the shear and bending moment diagrams for the beam and loading shown.
• Apply the relationship between bending moment and shear to develop the bending moment diagram.
MECHANICS OF MATERIAL S
Example 3 SOLUTION: Taking the entire beam as a free body, determine the reactions at A and D.
0 A y 90 kN 54 kN 115.6 kN 52.8 kN A y 81.2 kN
Apply the relationship between shear and load to develop the shear diagram. dV dx
w
dV w dx
- zero slope between concentrated loads - linear variation over uniform load segment
MECHANICS OF MATERIAL S
Example 3 Apply the relationship between bending moment and shear to develop the bending moment diagram. dM dx
V
dM V dx
- bending moment at A and E is zero - bending moment variation between A, B, C and D is linear - bending moment variation between D and E is quadratic - net change in bending moment is equal to areas under shear distribution segments - total of all bending moment changes across the beam should be zero
MECHANICS OF MATERIAL S
Example 4 SOLUTION: • Taking the entire beam as a free body, determine the reactions at C . • Apply the relationship between shear and load to develop the shear diagram. Draw the shear and bending moment diagrams for the beam and loading shown.
• Apply the relationship between bending moment and shear to develop the bending moment diagram.
MECHANICS OF MATERIAL S
Example 4 SOLUTION: • Taking the entire beam as a free body, determine the reactions at C . RC 1 w0a F y 0 12 w0a RC 2 a a M C 0 12 w0 a L M C M C 12 w0 a L 3 3
Results from integration of the load and shear distributions should be equivalent. • Apply the relationship between shear and load to develop the shear diagram. x 2 x V B V A w0 1 dx w0 x a a 2 0 a
V B 1 w0 a area under load curve 2
- No change in shear between B and C. - Compatible with free body analysis
a
0
MECHANICS OF MATERIAL S
Example 4 • Apply the relationship between bending moment and shear to develop the bending moment diagram. a
a
x 2 x 2 x3 M B M A w0 x dx w0 2a 0 2 6a 0 M B 1 w0 a 2 3
L
M B M C 1 w0 a dx 1 w0a L a 2 2 a
M C 1 w0a3 L a 6
a w0 a L 2 3
Results at C are compatible with free-body analysis
