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. Why study stresses in
beams

4. What are beams
 A structural member which is long when compared
with its lateral dimensions, subjected to transverse
forces so applied as to induce bending of the
member in an axial plane, is called a beam.

5. Objective
 When a beam is loaded by forces or couples,
stresses and strains are created throughout the
interior of the beam.
 To determine these stresses and strains, the
internal forces and internal couples that act on the
cross sections of the beam must be found.

6. Beam Types
 Types of beams- depending on how they are
supported.


7. Load Types on Beams
 Types of loads on beam
 Concentrated or point load
 Uniformly distributed load
 Uniformly varying load
 Concentrated Moment

8. Sign Convention for
forces and moments
P M M
Q Q
“Happy” Beam is +VE
+VE (POSITIVE)

9. Sign Convention for
forces and moments
M M
P
Q Q
“Sad” Beam is -VE
-VE (POSITIVE)

10. Sign Convention for
forces and moments
 Positive directions are denoted by an internal shear
force that causes clockwise rotation of the member
on which it acts, and an internal moment that
causes compression, or pushing on the upper arm
of the member.
 Loads that are opposite to these are considered
negative.

11. SHEAR FORCES AND BENDING
MOMENTS
 The resultant of the stresses must be such as to
maintain the equilibrium of the free body.
 The resultant of the stresses acting on the cross
section can be reduced to a shear force and a
bending moment.
 The stress resultants in statically determinate
beams can be calculated from equations of
equilibrium.

12. Shear Force and Bending
Moment in a Beam

13. Shear Force and Bending
Moment
 Shear Force: is the algebraic sum of the vertical
forces acting to the left or right of the cut section
 Bending Moment: is the algebraic sum of the
moment of the forces to the left or to the right of the
section taken about the section

14. SF and BM formulas
Cantilever with point load
W
x Fx= Shear force at X
A Mx= Bending Moment at X
B
L
W SF Fx=+W
BM Mx=-Wx
WxL
at x=0=> Mx=0
at x=L=> Mx=-WL

15. SF and BM formulas
Cantilever with uniform distributed load
w Per unit
length x Fx= Shear force at X
Mx= Bending Moment at X
A B
L
Fx=+wx
BM at x=0 Fx=0
wL at x=L Fx=wL
Mx=-(total load on right portion)*
Distance of C.G of right portion
wL2/2 Mx=-(wx).x/2=-wx2/2
at x=0=> Mx=0
at x=L=> Mx=-wl2/2

16. SF and BM formulas
Cantilever with gradually varying load
Fx= Shear force at X
w wx/L
Mx= Bending Moment at X
A B
x
wx 2
L Fx =
2L
at x=0 Fx=0
at x=L Fx=wL/2
wL/2 €
Parabola
C
Mx=-(total load for length x)*
Distance of load from X
wx 3
Cubic Mx =
6L
at x=0=> Mx=0
at x=L=> Mx=-wl2/6
€

17. SF and BM formulas
Simply supported with point load
W
x Fx= Shear force at X
C
Mx= Bending Moment at X
A B
W W
RA =
2 L RB =
2
Fx=+W/2 (SF between A & C)
€ € Resultant force on the left portion
W/2 SF
Baseline B ⎛ W ⎞ W Constant force
A C ⎜ − W ⎟ = −
⎝ 2 ⎠ 2 between B to C
SF W/2
€
BM
WL/4
B C B

18. SF and BM formulas
Simply supported with point load
W
x Fx= Shear force at X
C
Mx= Bending Moment at X
A B
W W
RA = L RB = for section
2 2
between A & C
W
€ € M x = RA x = x
W/2 SF 2
Baseline B at A x=0=> MA=0 W L
A C at C x=L/2=> MC = ×
SF W/2 2 2
€ for section
between C & B
BM €W × ⎛ x − L ⎞ = W x − Wx + W L
M x = RA x − ⎜ ⎟
WL/4 ⎝ 2 ⎠ 2 2
W L
B C B = − x +W
2 2
WL W
MB = − L =0
2 2
€

19. SF and BM formulas
Simply supported with uniform distributed load
w Per unit length
x Fx= Shear force at X
Mx= Bending Moment at X
A B
RA C RB
L
wL
RA = RB =
BM 2
wL/2 wL
Fx = RA − w.x = − w.x
A C B 2
wL w.0 wL
x = 0 ⇒ FA = − =
wL/2 2 2 2
L wL wL
x = ⇒ FC = − =0
wL2 2 2 2
wL2/2
8 wL wL
x = L ⇒ FB = − wL =
2 2
€
€

20. SF and BM formulas
Simply supported with uniform distributed load
w Per unit length
x Fx= Shear force at X
Mx= Bending Moment at X
A B
C x
RA RB M x = RA x − w.x
L 2
wL w.x 2
BM = x−
wL/2 2 2
A C B wL w.0
x = 0 ⇒ MA = .0 − =0
2 2
wL/2 L
2
wL L w ⎛ L ⎞ wL2 wL2 wL2
x = ⇒ Mc = . − ⎜ ⎟ = − =
2 2 2 2 ⎝ 2 ⎠ 4 8 8
wL2/2 wL2
wL w
8 x = L ⇒ MB = L − L2 = 0
2 2
€
€

21. SF and BM diagram
P Constant Linear
Load
Constant Linear Parabolic
Shear
Linear Parabolic Cubic
Moment

22. SF and BM diagram
0 0 Constant
Load M
Constant Constant Linear
Shear
Linear Linear Parabolic
Moment

23. Relation between load, shear
force and bending moment
1 2
x w/m run
dF
A B = −w
dx
1 C 2
L The rate of change of shear force is equal
to the rate of loading
M M+dM €
F F+dF dM
dx =F
dx
The rate of change of bending moment
is equal to the shear force at the section
€