1. Structure I Course Code: ARCH 208 Dr. Aeid A. Abdulrazeg

2. Beams Shear & Moment Diagrams

3. Contents Introduction Internal Forces in Members
Various Types of Beam Loading and Support
Shear and Bending Moment in a Beam
Relations Among Load, Shear, and Bending Moment

4. 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

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

6. Statically Indeterminate Beams
Continuous Beam
Propped Cantilever Beam

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. L P a b

8. Beam Shear
Magnitude (V) = sum of vertical forces on either side of the section
can be determined at any section along the length of the beam
Upward forces (reactions) = positive
Downward forces (loads) = negative
Vertical Shear = reactions – loads

9. Bending Moment
Bending moment: tendency of a beam to bend due to forces acting on it
Magnitude (M) = sum of moments of forces on either side of the section
can be determined at any section along the length of the beam
Bending Moment = moments of reactions – moments of loads

10. 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

11. 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

12. 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.

13. 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.

14. 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

15. 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

16. 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

17. 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.

18. The Relationship Between Load, Shear and Bending Moment

20. Equation (1) states that the slope of the shear diagram at a point (dV/dx) is equal to the intensity of the distributed load w(x) at the point.
Equation (2) states that the slope of the moment diagram (dM/dx) is equal to the intensity of the shear at the point.
Equ. (1)
Equ. (2)

21. By using integration
Equation (3) states that the change in the shear between any two points on a beam equals the area under the distributed loading diagram between the points.
Equation (4) states that the change in the moment between the two points equals the area under the shear diagram between the points
Equ. (3)
Equ. (4)

22. Common Relationships
Load
Constant
Linear
Shear
Parabolic
Moment
Cubic

23. Common Relationships
Load
Constant
Shear
Linear
Moment
Parabolic M

24. Example
Determine the shear and moment distributions produced in the simple beam by the 4-kN concentrated load.

29. 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 

30. 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)

31. Example
Draw the shear-force and bending-moment diagrams for the loaded beam and determine the maximum moment M and its location x from the left end.
2 kN/m
3 m
1.5 m
1.5 m

32. Example
Draw the shear-force and bending-moment diagrams for the loaded beam and determine the maximum moment M and its location x from the left end.