1. Shear Force & Bending Moment Diagrams
MOHAMMED ABDUL MOYEED
MECHANICS OF SOLIDS
II YEAR I SEM
DEPARTMENT OF MECHANICAL ENGINEERING

2. Beams
Members that are slender and support loads applied perpendicular to their longitudinal axis.
Concentrated Load, P
Distributed Load, w(x)
Longitudinal Axis
Span, L

3. Types of Beams Depends on the support configuration FH FV FH M Fv FH
Pin
Roller M Fv FH
Fixed
Pin
Roller FV FH

4. Statically Indeterminate Beams
Continuous Beam
Propped Cantilever Beam
Can you guess how we find the “extra” reactions?

5. Internal Reactions in Beams
At any cut in a beam, there are 3 possible internal reactions required for equilibrium:
normal force,
shear force,
bending moment. L P a b

6. Internal Reactions in Beams
At any cut in a beam, there are 3 possible internal reactions required for equilibrium:
normal force,
shear force,
bending moment.
Positive Directions Shown!!! V M N
Left Side of Cut
Pb/L x

7. Internal Reactions in Beams
At any cut in a beam, there are 3 possible internal reactions required for equilibrium:
normal force,
shear force,
bending moment.
Positive Directions Shown!!! V M N
Right Side of Cut
Pa/L
L - x

8. Finding Internal Reactions
Pick left side of the cut:
Find the sum of all the vertical forces to the left of the cut, including V. Solve for shear, V.
Find the sum of all the horizontal forces to the left of the cut, including N. Solve for axial force, N. It’s usually, but not always, 0.
Sum the moments of all the forces to the left of the cut about the point of the cut. Include M. Solve for bending moment, M
Pick the right side of the cut:
Same as above, except to the right of the cut.

9. Point 6 is just left of P and Point 7 is just right of P.
Example: Find the internal reactions at points indicated. All axial force reactions are zero. Points are 2-ft apart.
P = 20 kips 1 2 3 4 5 8 9 10 6 7
8 kips
12 kips
12 ft
20 ft
Point 6 is just left of P and Point 7 is just right of P.

10. V M 12 ft 20 ft P = 20 kips 8 kips 12 kips (kips) (ft-kips) 1 2 3 4 59 10 6 7
8 kips
12 kips
12 ft
20 ft
8 kips V
(kips) x
-12 kips 96 80 64 72 48 48 32 16 24 M
(ft-kips) x

11. V & M Diagrams 12 ft 20 ft P = 20 kips 8 kips 12 kips Vx
What is the slope of this line?
-12 kips
96 ft-kips
What is the slope of this line?
96 ft-kips/12’ = 8 kips b
-12 kips M
(ft-kips) a c x

12. V & M Diagrams 12 ft 20 ft P = 20 kips 8 kips 12 kips V M 8 kipsx
What is the area of the blue rectangle?
-12 kips
96 ft-kips
What is the area of the green rectangle?
96 ft-kips b
-96 ft-kips M
(ft-kips) a c x

13. Draw Some Conclusions
The magnitude of the shear at a point equals the slope of the moment diagram at that point.
The area under the shear diagram between two points equals the change in moments between those two points.
At points where the shear is zero, the moment is a local maximum or minimum.

14. The Relationship Between Load, Shear and Bending Moment

15. Common Relationships
Load
Constant
Linear
Shear
Parabolic
Moment
Cubic

16. Common Relationships
Load
Constant
Shear
Linear
Moment
Parabolic M

17. Example: Draw Shear & Moment diagrams for the following beam
12 kN
8 kN A C D B
1 m
3 m
1 m
RA = 7 kN  RC = 13 kN 

18. 12 kN 8 kN 1 m 3 m 1 m 2.4 m 8 7 8 7 V -15 -5 7 M -8 A C D B (kN)