1. Beams: Pure Bending (4.1-4.5)
MAE 314 – Solid Mechanics
Yun Jing
Beams: Pure Bending

2. Beams in Pure Bending
Prismatic beams subject to equal and opposite couples acting in the same plane are in pure bending.
Beams: Pure Bending

3. Pure vs. Non-Uniform Bending
Pure bending: Shear force (V) = 0 over the section
Non-uniform bending: V ≠ 0 over the section
Non-uniform bending • Moment → normal stresses • Shear force → shear stresses (Ch. 6)
Pure bending • Moment → normal stresses
Non-uniform: there is a force, leading to a shear stress
Beams: Pure Bending

4. Pure Bending: Assumptions
Beam is symmetric about the x – y plane
All loads act in the x – y plane
Beams: Pure Bending

5. Pure Bending: Curvature
Sections originally perpendicular to the axis of the member remain plane and perpendicular: “Plane sections remain plane.”
Sign convention
Positive bending moment: beam bends towards +y direction
Negative bending moment: beam bends towards -y direction
Right angle
Beams: Pure Bending

6. Pure Bending: Deformation
Since angles do not change (remain plane), there is no shear stress.
The top part of the beam contracts in the axial direction.
The bottom part of the beam expands in the axial direction.
There exists a line in the beam that remains the same length called the neutral line.
Set y = 0 at the neutral line.
ρ = radius of curvature
εx < 0 for y > 0 and εx > 0 for y < 0
Beams: Pure Bending

7. Pure Bending: Axial Strain
The length of DE is LDE = ρθ
The length of JK is LJK = (ρ-y)θ
Axial strain at a distance y from the neutral axis (εx):
Maximum compressive strain occurs on the upper surface.
Maximum tensile strain occurs on the lower surface.
LJK
LDE
Beams: Pure Bending

8. Pure Bending: Axial Strain
c = maximum distance between the neutral axis and the upper or lower surface
When c is the distance to the surface in compression
When c is the distance to the surface in tension
Beams: Pure Bending

9. Pure Bending: Transverse Strain
Recall there are no transverse stresses since the beam is free to move in the y and z directions.
However, transverse strains (in the y and z directions) exist due to the Poisson’s ratio of the material.
ρ' = radius of anticlastic curvature = ρ/υ
Beams: Pure Bending

10. Pure Bending: Normal Stress
Let us now assume that the beam is made of a linear-elastic material.
The normal stress varies linearly with the distance from the neutral surface.
Beams: Pure Bending

11. Pure Bending: Normal Stress
Recollecting that the applied loading is a pure moment, we calculate resultant loads on the cross-section.
The resultant axial force must be equal to zero.
…which is the definition of the centroid, so the neutral axis is just the centroid of the section.
Beams: Pure Bending

12. Pure Bending: Normal Stress
The resultant moment about the z-axis must be equal to the applied moment M.
Definition of the second moment of inertia, I
Beams: Pure Bending

13. Centroids and Moments of Inertia
Example Problem
A steel bar of 0.82.5-in. rectangular cross section is subjected to two equal and opposite couples acting in the vertical plane of symmetry of the bar. Determine the value of the bending moment M that causes the bar to yield. Assume m=36 ksi.
Centroids and Moments of Inertia

14. Centroids and Moments of Inertia (A.1-A.5)
MAE 314 – Solid Mechanics
Yun Jing
Centroids and Moments of Inertia

15. Centroids and Moments of Inertia
Review From Last Time
Beams in pure bending (no shear forces)
Plane sections remain plane
+M, upper surface in compression
+M, lower surface in tension
Transverse strains exist due to υ
Radius of Curvature
Neutral Line
Centroids and Moments of Inertia

16. Centroids and Moments of Inertia
To match with our axis convention use centroid in the y-direction.
Centroids for common shapes are located in the back cover of your textbook.
If a cross section shape has an axis of symmetry, then the centroid passes through that axis of symmetry.
Centroids and Moments of Inertia

17. Centroids for Composite Shapes
Choose a reference x-axis (normally the base of the shape).
Divide the composite shape into individual shapes.
Calculate the area for each shape.
Find the location of the centroid for each shape relative to the reference x-axis. y A1
h1b1
0.5h1
0.5h12b1 A2
h2b2
h1+0.5h2
h2b2(h1+0.5h2) A3
h3b3
h1+h2+0.5h3
h3b3(h1+h2+0.5h3) x
y=0
Centroids and Moments of Inertia

18. Centroids and Moments of Inertia
Moment of inertia with respect to the x-axis
Moments of inertia for common shapes are located in the back cover of your textbook. x x’ x Ix x
Ix’ Ix Ix
Centroids and Moments of Inertia

19. Moments of Inertia for Composite Shapes
Divide the composite shape into individual shapes.
Calculate I of each shape about its own centroid.
Calculate I of each shape about the centroid of the entire composite shape using the parallel axis theorem: A1
b1h13/12
b1h1d12
b1h1(d12+h12/12) A2
b2h23/12
b2h2d22
b2h2(d22+h22/12)
x’’’ A3
b3h33/12
b3h3d32
b3h3(d32+h32/12) x
x’’ x’
Centroids and Moments of Inertia

20. Centroids and Moments of Inertia
Example Problem
Find the location of the centroid and the moment of inertia of the section
below (the moment is applied about the horizontal axis).
5 in
0.5 in
3 in
4 in
1.5 in
3 in
Centroids and Moments of Inertia

21. Stress in the Elastic Range
Revisiting the beam subject to pure bending
For positive bending moment
Maximum tensile stress = Mc2 / I
Maximum compressive stress = Mc1 / I
For negative bending moment
Maximum tensile stress = Mc1 / I
Maximum compressive stress = Mc2 / I
The ratio I/c1 (or I/c2) is known as S, elastic section modulus.
Larger values for S indicate a beam that is more resistant to bending. c1 c2
Centroids and Moments of Inertia

22. Deformations in the Elastic Range
Which shapes are most resistant to bending?
S-beam
rectangular
Why?
Large portion of the cross section is located far from the neutral axis, resulting in large values of S.
The deformation of a member due to bending can be calculated by
Centroids and Moments of Inertia

23. Centroids and Moments of Inertia
Example Problem
A beam with cross-section like the previous example is subjected to a
pure bending moment of M = in·lb. Find the maximum tensile and
compressive normal stresses.
5 in
0.5 in
3 in
4 in
1.5 in
3 in
Centroids and Moments of Inertia

24. Centroids and Moments of Inertia
Example Problem
Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B.
Centroids and Moments of Inertia

25. Centroids and Moments of Inertia
Example Problem
The beam shown is made of nylon with an allowable stress of 24 MPa in
tension and 30 Mpa in compression. Determine the largest couple M that
can be applied to the beam.
Centroids and Moments of Inertia

26. Centroids and Moments of Inertia
Example Problem
Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam.
Centroids and Moments of Inertia

27. Bending in Composite Materials (4.6)
MAE 314 – Solid Mechanics
Yun Jing
Centroids and Moments of Inertia

28. Normal Stress in Composite Materials
Strain is still the same as homogenous material because it is not dependent on material properties.
The stress distribution is no longer linear.
Use different material properties in each section.
Centroids and Moments of Inertia

29. Normal Stress in Composite Materials
Top section:
Bottom section:
Original Shape
New Shape
Centroids and Moments of Inertia

30. Bending in Composite Materials
The new “homogenous” beam produces the same response to bending as the composite beam.
We can now calculate the centroid and moment of inertia for this beam.
To calculate the axial stresses, use σx for material 1 and nσx for material 2.
Curvature of the composite member is
Centroids and Moments of Inertia

31. Reinforced concrete beam
Beams are reinforced by steel rods
Concrete is very weak in tension, the steel rods will carry the entire tensile load
Concrete in the beam acts effectively only in compression
Centroids and Moments of Inertia

32. Centroids and Moments of Inertia
Example Problem
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Determine the largest permissible
bending moment when the composite bar is bent about a horizontal
axis. Ea = 70 GPa, σall,a = 100 MPa, Eb = 105 GPa, σall,b = 160 MPa.
Centroids and Moments of Inertia

33. Example Problem
Knowing that the bending moment in the reinforced concrete beam is +100kip.ft and that the modulus of elasticity is 3.625106 psi for the concrete and 29 106 psi for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.
Beams: Pure Bending 33

34. Example Problem
A flitched beam consists of two 50mm×200mm wooden beams and a 12 mm×80mm steel plate. The plate is placed centrally between the wooden beams and recessed into each so that, when rigidly joined, the three units form a 100mm×200mm section as shown in the figure. Determine the bending moment the beam can withstand when the maximum bending stress in the timber is 12MN/m2. What will then be the maximum bending stress in the steel. (E for the timber is 10GPa, for steel is 120GPa)
Beams: Pure Bending 34