2. Unit 2- Stresses in Beams
Topics Covered
 Lecture -1 – Review of shear force and bending
moment diagram
 Lecture -2 – Bending stresses in beams
 Lecture -3 – Shear stresses in beams
 Lecture -4- Deflection in beams
 Lecture -5 – Torsion in solid and hollow shafts.

3. Theory of simple
bending (assumptions)
 Material of beam is homogenous and isotropic => constant E in all
direction
 Young’s modulus is constant in compression and tension => to simplify
analysis
 Transverse section which are plane before bending before bending remain
plain after bending. => Eliminate effects of strains in other direction (next
slide)
 Beam is initially straight and all longitudinal filaments bend in circular arcs
=> simplify calculations
 Radius of curvature is large compared with dimension of cross sections =>
simplify calculations
 Each layer of the beam is free to expand or contract => Otherwise they will
generate additional internal stresses.

4. Bending in beams
Key Points:
1. Internal bending moment causes beam to deform.
2. For this case, top fibers in compression, bottom in
tension.

5. Bending in beams
Key Points:
1. Neutral surface – no change in length.
2. Neutral Axis – Line of intersection of neutral surface
with the transverse section.
3. All cross-sections remain plane and perpendicular to
longitudinal axis.

6. Bending in beams
Key Points:
1. Bending moment
causes beam to deform.
2. X = longitudinal axis
3. Y = axis of symmetry
4. Neutral surface – does
not undergo a change
in length

7. Bending Stress in beams
Consider the simply supported beam below:
Radius of Curvature, R
P
B Deflected
A
Shape
Neutral Surface
M M
RA RB
M M
What stresses are generated
within, due to bending?

8. Axial Stress Due to Bending: M=Bending Moment
σx (Compression)
M M
Neutral Surface σx=0
Beam
σx (Tension)
stress generated due to bending:
σx is NOT UNIFORM through
the section depth
σx DEPENDS ON:
(i) Bending Moment, M
(ii) Geometry of Cross-section

9. Bending Stress in beams

10. Bending Stress in beams

11. Stresses due to bending
y
R Strain in layer EF =
R
A’ C’
N’ N’ Stress _ in _ the _ layer _ EF
E F E=
B’ D’ Strain _ in _ the _ layer _ EF
σ
E= €
⎛ y ⎞
⎜ ⎟
⎝ R ⎠
σ E E
= σ= y
y R R
€

12. Neutral axis
dA
force on the layer=stress on layer*area of layer
dy = σ × dA
y E
= × y × dA
R
N A
Total force on the beam section
€ E
σx σx
= ∫ R
× y × dA
Stress diagram
E
=
R
∫ y × dA
x For equilibrium forces should be 0
M M ∫
€ y × dA = 0
Neutral axis coincides with the geometrical
axis
€

13. Moment of resistance
dA
force on the layer=stress on layer*area of layer
dy = σ × dA
y E
= × y × dA
R
N A
Moment of this force about NA
€ E
= × y × dA × y
σx σx R
Stress diagram E
= × y 2 × dA
R
E E
x Total moment M= ∫ R
× y 2 × dA = ∫ y 2 × dA
R
M M € ∫y 2
× dA = I
E M E
€ M= I⇒ =
R I R
€

14. Flexure Formula
M E σ
= =
I R y
€

15. Beam subjected to 2 BM
In this case beam is subjected to
moments in two directions y and z.
The total moment will be a resultant
of these 2 moments.
You can apply principle of superposition
to calculate stresses. (topic covered in
unit 1).
Resultant moments and stresses

16. Section Modulus
Section modulus is defined as ratio of moment of inertia about the neutral axis to
the distance of the outermost layer from the neutral axis
I
Z=
y max
M σ
=
I y
M σmax
=
I y max
I
M = σmax
y max
M = σmax Z
€

17. Section Modulus of
symmetrical sections
Source:-
http://en.wikipedia.org/wiki/Section_modulus

18. Section Modulus of
unsymmetrical sections
In case of symmetrical section neutral axis passes through geometrical center of
the section. But in case of unsymmetrical section such as L and T neutral axis
does not pass through geometrical center.
The value of y for the outermost layer of the section from neutral axis will not be
same.

19. Composite beams
Composite beams consisting of layers with fibers, or rods strategically placed to
increase stiffness and strength can be “designed” to resist bending.

20. Composite beams
t t σ1 σ2
=
E1 E 2
y E1
σ1 = σ
E2 2
d = mσ 2 m=modular ratio
σ
M= I
y
€
M = M1 + M 2
b σ σ
= 1 I1 + 2 I2
y y
σ
= 2 [ mI1 + I2 ]
y
Equivalent I (moment of inertia)= mI1 + I2
€