SEF024
INTRODUCTION TO ENGINEERING
Chaper 1: Forces and moments in beams
A beam is any of various relatively long pieces of metal,
wood, stone, etc., manufactured or shaped especially for
use as rigid members or parts of structures or machines.
Pilar Garcia Souto
School of Engineering and Materials Science
m.garciasouto@qmul.ac.uk
Lectures: Thursdays 1-2pm, 3-5pm
Workshop: Thursdays 5-6 pm
Drop in sessions: Monday, Wednesday and Friday 12-2pm in FB 2.38
Drop in sessions with Pilar: Monday or Friday 12-2pm in FB 2.38 (to be
updated every week)
Examples of how beams can be found
A beam can be part of:
• a building,
• an aircraft wing,
• vehicle axle,
• artificial limb.
c) RIGIDLY SUPPORTED
a) SIMPLY SUPPORT
M
R
M
R
R
R
b) CANTILEVER
M
Where
R is the support reaction.
M is the support reaction.
R
QUESTION Find the support reactions and support moments in the
following beams
Steps:
10 N
3m
1)
SOLUTIONS
1) Define reference system for forces and moments
3m
+
R
10 N
2)
6m
5m
Note) Moment is defined as
MOMENT = FORCE x DISTANCE
2) There must be force and moment equilibrium, that means:
ΣFx = 0
12 N
3)
+
R
ΣFy = 0
1m
ΣM = 0
Balance of horizontal forces
Balance of vertical forces
Balance of moments at any point
Solving these equations we can solve all the beam problems in this
course, so it is worth to remember them!!!
R
R
1
SOLUTIONS (1)
SOLUTIONS (2)
10 N
3m
A
10 N
M
3m
6m
A
R1
R2
RESOLVING VERTICALLY (Balance of vertical forces)
ΣFy = 0
R1 + R2 - 10 = 0
ΣFy = 0
(eq. 1)
RESOLVING HORIZONTALLY (Balance of horizontal forces)
ΣFx = 0
R
RESOLVING VERTICALLY (Balance of vertical forces)
does not give us any extra information in this case
TAKING MOMENTS ABOUT A (Balance of moments)
ΣM = 0
- 10 . 3 + R2 . 6 = 0 (eq. 2)
SOLVING THE SYSTEM OF (eq. 1 & 2)
R - 10 = 0
R = 10 N
RESOLVING HORIZONTALLY (Balance of horizontal forces)
ΣFx = 0
does not give us any extra information in this case
TAKING MOMENTS ABOUT A (Balance of moments)
M - 10 x 6 = 0
M = 60 N.m
ΣM = 0
R2 = 5 N ; R1 = 5 N
SOLUTIONS (3)
12 N
ALTERNATIVELY, TAKING MOMENTS ABOUT B
5m
A
1m
R1
B
- 6 R1 + 1
x
12 = 0
R1 =
2N
R2
RESOLVING VERTICALLY (Balance of vertical forces)
ΣFy = 0
R1 + R2 - 12 = 0
(eq. 1)
RESOLVING HORIZONTALLY (Balance of horizontal forces)
ΣFx = 0
does not give us any extra information in this case
TAKING MOMENTS ABOUT A (Balance of moments)
ΣM = 0
- 12 . 5 + R2 . 6 = 0 (eq. 2)
SOLVING THE SYSTEM OF (eq. 1 & 2)
R2 = 10 N ; R1 = 2 N
UNIFORMLY DISTRIBUTED LOAD (UDL)
Finding the reactions (Force and moment)
The load may be spread evenly over all or part of the beam rather
than applied just at a point. For example the floor load in:
• a building
• an airplane's cargo hold
• a lorry trailer
100 N
W = 120 Kg
Convert the UDLs into point forces and place them in the centre of
pressure of each UDL. Then calculate reactions as normal
100 N
Equivalent
free body
diagram
F = 1200 N
M
3m
2m
1m
Case
A
6m
1m
B
100 N
Free
body
diagram
M
R
F = 200 N/m * 6 m = 1200 N
RESOLVING VERTICALLY (Balance of vertical forces)
UDL 200 N/m
R - 1200 - 100 = 0
5m
(FBD)
1m
R = 1300 N
TAKING MOMENTS ABOUT A (Balance of moments)
M - 1200 x 3 - 100 x 5 = 0
M = 4100 N.m
R
2
SHEAR FORCE DIAGRAM (SFD)
This diagram represents the shear forces within the beam which occur
as a result of applying external forces on it. Shear force is plot against
position along the beam.
Shear stress is uniform across a section. It produces a uniform cutting
failure.
The calculation of the SFD leads to:
- a better understanding of what is happening within the beam
- find out which sort of bending would occur
- find out how likely is it of breaking under that pressure
When drawing the SFD make sure you fully define it, that is:
- indicate the position at which a significant change occurs
- the values of the forces at each point
Convention:
Force up on left is POSITIVE.
Force down on left is NEGATIVE.
For S.F.D. start left and follow the force arrows.
SOLUTIONS (1)
SOLUTIONS (2)
10 N
3m
10 N
M
3m
6m
A
RB = 5 N
RA = 5 N
R = 10 N
SFD
SFD
SF(N)
SF(N)
10 N
5 N
3
6
x (m)
3
6
x (m)
-5 N
BENDING MOMENT DIAGRAMS (BMD)
SOLUTIONS (3)
12 N
Bending is caused when a load is applied on the beam, in a location
remote from the support.
1m
5m
L
RB = 10 N
RA = 2 N
L
SFD
F(N)
2 N
5
- 10 N
6
x (m)
A bending moment diagram (BMD) is used to find the position where
bending is most critical.
• A BMD plots along the length of the beam the moment causing
bending.
3
• The position along the beam is plotted horizontally.
• The bending moment is plotted vertically.
A bending moment is negative when the moment on the left of the
section is anticlockwise and that on the right is clockwise.
• To plot a BMD you start at one end of the beam and calculate the
moment at key points as you move to the other end of the beam.
x
SIGN CONVENTION
A bending moment is positive when the moment on the left of the
section is clockwise and that on the right anticlockwise.
M
M
x
THIS IS ALSO CALLED HOGGING
x
M
M
Why bothering?
BENDING
FAILURE
x
THIS IS ALSO CALLED SAGGING
SOLUTIONS (1)
3m
A
BMD
M (N.m)
10 N
3m
C
15
B
3
x (m)
6
RB = 5 N
RA = 5 N
FROM EARLIER RA = RB
= 5N
SFD
Calculating the bending moment (working from left to right)
AT A
MOMENT = R1 x 0 = 0
AT C
MOMENT = R1 x 3 - 10 x 0
AT B
MOMENT = R1 x 6 - 10 x 3
Note that the
BMD is the
integral of the
SFD
SF (N)
5 N
= 15 N.m
3
6
x (m)
-5 N
= 30 - 30
= 0
SOLUTIONS (2)
10
M
A
C
3m
BMD
M (N.m)
B
A
3m
C
3
FROM EARLIER M = 60 Nm AND R = 10 N
x (m)
- 60 Nm
MOMENT = -M
= -60 N m
AT C
6
- 30 Nm
R
AT A
B
MOMENT = -60 + R x 3
SF (N)
Note that the
BMD is the
integral of the
SFD
SFD
10 N
= - 30 N m
AT B
MOMENT
= -60 + R x 6
3
6
x (m)
= 0Nm
4
SOLUTIONS (3)
BMD
M (N.m)
12 N
10
5m
A
1m
D
B
5
R1
x (m)
6
R2
FROM EARLIER R1 = 2N
R2
SFD
= 10N
Note that the
BMD is the
integral of the
SFD
SF (N)
AT A
MOMENT = R1 x 0
AT D
MOMENT = R1 x 5
=
2 N
0
5
= 10 N.m
AT B
MOMENT
x (m)
6
- 10 N
= R1 x 6 -12 x 1
= 12 – 12 = 0
10kN
IN SUMMARY
2m
M
2m
LOADING
5kN
R
5kN
S.F.D.
+ve
+5kN
S.F.D.
-5kN
B.M.D.
- ve
10kNm
B.M.D.
WEIGHT OF THE BEAM
QUESTION: solve the following case
The weight of the beam can not be neglected in many of the cases. It
can be considered like an extra UDL load such that it spreads evenly
over all the beam rather than being applied just at a point.
W = 1200 kg
A
B
6m
M
UDL 2 kN/m
FBD
6m
R
F = 12 kN
M
Equivalent
FBD
3m
3m
R
TOTAL FORCE = 1200 kg * 10 N/kg = 12000 N = 12 kN
UDL = LOAD / DISTANCE = 12 kN / 6m = 2 kN/m
LINE OF ACTION IS IT´S MIDPOINT, i.e. 3 m FROM A.
5
Finding reaction moment and reaction force
LOADING
RESOLVING VERTICALLY (Balance of vertical forces)
ΣFy = 0
R - 12 = 0
R = 12 kN
12 kN
RESOLVING HORIZONTALLY (Balance of horizontal forces)
ΣFx = 0
does not give us any extra information in this case
+ ve
TAKING MOMENTS ABOUT A (Balance of moments)
ΣM = 0
M - 12 . 3
=0
S.F.D.
M = 36 kN.m
Finding moments along the beam
AT A
MOMENT = -36 kN.m
AT 2m MOMENT = -36 + 12 x2 - (2x2) 1 (looking to the left)
AT 4m MOMENT = -36 + 12(4) – (4x2) 2 (looking to the left)
-4
- ve
B.M.D.
- 16
= -16 kN.m (looking to the right)
-36 kN.m
= -4 kN.m (looking to the right)
Similarly, for simply supported beam
TYPICAL S.F.D. AND B.M.D. with weight included
FBD
Chaper 1 (continuation):
Practical case when weight
of the beam is considered
S.F.D.
B.M.D
Finding reaction moment and reaction force
(1) CANTILEVER
RESOLVING VERTICALLY (Balance of vertical forces)
W = 1200 kg
A
B
6m
M
ΣFy = 0
ΣFx = 0
FBD
6m
ΣM = 0
Equivalent
FBD
R
TOTAL FORCE = 1200 kg * 10 N/kg = 12000 N = 12 kN
UDL = LOAD / DISTANCE = 12 kN / 6m = 2 kN/m
M - 12 . 3
=0
M = 36 kN.m
Finding moments along the beam
F = 12 kN
3m
does not give us any extra information in this case
TAKING MOMENTS ABOUT A (Balance of moments)
R
3m
R = 12 kN
RESOLVING HORIZONTALLY (Balance of horizontal forces)
UDL 2 kN/m
M
R - 12 = 0
AT A
MOMENT = -36 kN.m
AT 2m MOMENT = -36 + 12 x2 - (2x2) 1 (looking to the left)
= -16 kN.m (looking to the right)
AT 4m MOMENT = -36 + 12(4) – (4x2) 2 (looking to the left)
= -4 kN.m (looking to the right)
LINE OF ACTION IS IT´S MIDPOINT, i.e. 3 m FROM A.
6
(2) SIMPLY SUPPORTED
W = 1200 kg, lenght = 6 m
LOADING
Case
TOTAL FORCE = 1200 kg * 10 N/kg = 12000 N = 12 kN
UDL = LOAD / DISTANCE = 12 kN / 6m = 2 kN/m
12 kN
UDL 2 kN/m
FBD
+ ve
S.F.D.
R
R
F = 12 kN
-4
- ve
B.M.D.
3m
- 16
Equivalent
FBD
3m
R1
R2
-36 kN.m
LINE OF ACTION IS IT´S MIDPOINT, i.e. 3 m FROM A.
Finding moments along the beam (continuation)
Finding reaction forces
RESOLVING VERTICALLY (Balance of vertical forces)
ΣFy = 0
AT 4m MOMENT = 6 x 2 - (2x2) x 1 = 8 kN.m (looking to the right)
R1 + R2 - 12 = 0
TAKING MOMENTS ABOUT A (Balance of moments)
- 12
ΣM = 0
x
3 + R2 x 6 = 0
F = 8 kN
1m 1m
R1 = 6 kN
Finding moments along the beam
AT A
4m
MOMENT = 0 kN.m (looking towards the left)
AT 2m MOMENT = 6 x 4 - (4x2) x2 = 8 kN.m (looking to the right)
F = 4 kN
F = 8 kN
1m
2m
2m
Equivalent
FBD
2m
6 kN
6 kN
AT 6m
MOMENT = 0 kN.m (looking towards the right)
Point of interest x = 3m
2m
Equivalent
FBD
4m
6 kN
F = 4 kN
R2 = 6 kN
6 kN
AT 3m MOMENT = 6 x 3 + (2x3) x 1.5 = 9 kN.m (looking to the right)
This is the maximum bending moment because it is when shear force
is zero
(3) TIPICAL CASE EXAMPLE: beam with weight and external loads
6 kN
S.F.D.
3m
A beam of length 5 m and weight 10 N is simply supported at the
corners. A pointed load of 8 N is placed at 1 meter from the left
corner. Find out the SFD and BMD. Which is the maximum bending
moment?
- 6 kN
TOTAL WEIGHT = 10 N
UDL = LOAD / DISTANCE = 10 N / 5m = 2 N/m
UDL=2 N/m
8N
1m
9 kN
8 kN
4m
A
C
FBD
B
2m
3m
4m
B.M.D
R1
R2
LINE OF ACTION OF THE WEIGHT IS IT´S MIDPOINT, i.e. 2.5 m FROM A.
7
8N
Step 2: Sketch the SFD
10 N
Equivalent
FBD
2 N/m
1m
A
C
B
R1
2.5 m
2.5 m
SFA+ = 11.4 N
SFB − = 11.4 N − 2
R2
SFB + = 9.4 N − 8 N = 1.4 N
N
SFC − = 1.4 N − 2 × 4m = 6.6 N
m
SFC + = 6.6 N − 6.6 N = 0 N
SFA− = 0 N
N
× 1m = 9.4 N
m
SOLUTION
F (N)
Step 1: Find reaction forces
11.4
RESOLVING VERTICALLY (Balance of vertical forces)
ΣFy = 0
9.4
R1 + R2 - 8 - 10 = 0
S.F.D.
1.4
TAKING MOMENTS ABOUT A (Balance of moments)
-8
ΣM = 0
x
1 – 10 x 2.5 + R2 x 5 = 0
A
R2 = 6.6 N
C
B
R1 = 11.4 N
To fully define the SFD, we need to find the point at which it
crosses zero
F (N)
-6.6
Step 3: Sketch the BMD
AT B
(1, 1.4)
S.F.D.
5
1
B.M . = (11.4 ×1) − (2 ×1) × 1
(5, -6.6)
We can solve by getting the equation of the line:
AT MIDPOINT
B.M . = (11.4 × 2.5) − ( 2 × 2.5) × 2.5
2) Get the coordinate in the origen
1) Get the slope
b = 1.4 + 2 ×1 = 3.4
2
− (8 × 4)
AT C
B.M . = (11.4 × 5) − (2 × 5) × 5
y = −2 x + 3 . 4
= 57 − 32 − 25
= 0
The point x at which the line crosses the axis is y=0, that is
− 3.4
= 1.7
−2
10.4
− (8 ×1.5)
b = y + 2x
y = −2 x + b
x=
2
= 28.5 − 6.25 − 12
= 10.25 N .m
using data of 1 of the points
− 6 .6 − 1 .4
= −2
5 −1
y = a.x + b
m=
Hence:
2
= 10.4 N .m
1.8
B.M.D.
10.25
REMEMBER
B.M. = Ʃ FORCE x DISTANCE
AREA UNDER S.F.D. = Ʃ FORCE x DISTANCE
POSITIVE S.F.D. = INCREASED B.M.
1.0 m
NEGATIVE S.F.D. = DECREASED B.M.
2.5 m
Step 4: What is the max. B.M?
HENCE
Maximum bending moment occurs where the SF is equal to zero
MAX B.M . WHEN S .F . = 0
AT 1.7 m
B.M. = (11.4 x 1.7) – (2 x 1.7) 1.7/2 - (8 x 0.7)
= 19.38 - 2.89 - 5.6
= 10.89 N.m
Maximum bending moment
8