Type of BeamsType of Beams
Statically Determinate
Simply Supported Beam
1
Overhanging Beam
Cantilever Beam
Type of BeamsType of Beams
Statically Indeterminate
Continuous Beam
2
Propped Cantilever Beam
Fixed Beam
3
Example 1Example 1
Equilibrium equation for 0 x 3m:
A B
M
VF
x
Equilibrium equation for 0 x 3m:
* internal V and M should be assumed +ve
kNV
VF
Fy
9
0
0
=
=
=
)(9
0
0
kNmxM
MVx
M
=
=+
=
Shear DiagramShear Diagram
M
VF
x
Lecture 1 5
Sign convention: V= -9kN
Shear DiagramShear Diagram
M = -9x kN.m
V = -9 kNF
x
6
Sign convention: M= -9x kNmX=0: M= 0X=3: M=-27kNm
M=-9x
x
V=9kN
M=X
At cross section A-A
At section A-A
7
X
Example 2Example 2
1) Find all the external forces
8
kN5
01050F
kN5
0)2()1(10;0
y
====
====++++====
====
========
y
y
y
yA
A
A
C
CM
)(5
0
0
10
downkNV
VF
F
mx
y
=
=
=
)(5
10
downkNV
mx
=
Solve itSolve it
Draw the shear and moment diagrams for
simply supported beam.
11
12
Distributed LoadDistributed Load
For calculation purposes, distributed load can be represented as a single load acting on the center point of the distributed area.
Total force = area of distributed load (W : height and L: length)Point of action: center point of the area
13
ExampleExample
14
ExampleExample
15
Solve itSolve it
Draw the shear and moment diagrams the
beam:
16
Solving all the external loads
kN
WlF
48)6(8 ==
=
Distributed load will be
kN48)6(8 ==
Solving the FBD
012
364
)3(48
0)3(4
0
==
==
=
=
xy
y
x
A
AkNA
kNB
FB
M
Line NA: neutral axis Red Line: max normal stress
c = 60 mmYellow Line: max compressive stress
c = 60mm
I
Mc=max
I
Mc=max
Line NA: neutral axis Red Line: Compressive stress
y1 = 30 mmYellow Line: Normal stress
y2 = 50mm
I
My 11 =
I
My 22 =
Refer to Example 6.11 pp 289
I: moment of inertial of the cross I: moment of inertial of the cross
sectional areasectional area
12
3bh
I xx =
644
44Dr
I xx
==
Find the stresses at A and B
I: moment of inertial of the cross I: moment of inertial of the cross
sectional areasectional area
Locate the centroid (coincide with neutral axis)
Ay
yn
n
i
ii
1=
=
mm
AA
AyAy
An
i
i
5.237
)300)(50()300)(50(
)300)(50(325)300)(50(150
21
2211
1
=
+
+=
+
+=
=
I: moment of inertial of the cross I: moment of inertial of the cross
sectional areasectional area
Profile I
4633
)10(5.11212
)300(50
12mm
bhI I ===
A A
I about Centroidal axis
I about Axis A-A using parallel axis theoremA A
46
2623
)10(344.227
)5.87)(300)(50()10(5.11212
)(
mm
Adbh
I AAI
=
+=+=
theorem
Profile II
46
23
23
)10(969.117
)5.87)(50)(300(12
)50)(300(
12)(
mm
Adbh
I AAII
=
+=+=
Total I
46
466
)10(313.345
)10(969.117)10(344.227
)()(
mm
mm
III AAIIAAIAA
=
+=
+=
* Example 6-12 to 6-14 (pp 290-292)
Solve itSolve it
If the moment acting on the cross section of the beam is M = 6 kNm, determine the maximum bending stress on in the beam. Sketch a three dimensional of the stress distribution acting over the cross sectionIf M = 6 kNm, determine the resultant force the bending stress produces on the top board A of the beamproduces on the top board A of the beam
46
32
3
)10(8.786
12
)300(40])170)(40)(300(
12
)40(300[2
mm
I I
=
++=
Total Moment of Inertia
Max Bending Stress at the top and bottom
MPaII
McM top 45.1
)190()10(60003
=
==
MPaM bottom 45.1=1.45MPa
bottom
Bottom of the flange
MPaII
McM topf 14.1
)150()10(60003
_ =
==
MPaM bottomf 14.1_ =
1.14MPa
6kNm
Resultant F = volume of the trapezoid
1.45MPa
1.14MPa
40 mm
300 mm
kN
NFR
54.15
15540)300)(40(2
)14.145.1(
=
=+
=
Solve itSolve it
The shaft is supported by a smooth thrust load at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress on the shaft
Draw the shear and moment diagram
kNF
kNF
F
M
A
D
D
A
3
3
)25.2(3)75.0(3)3(
0
=
=
+=
=
External Forces
Absolute Bending Stress
M = 2.25kNmMmax = 2.25kNm
MPa
I
Mc
8.52
)2540(4
)40()10(2250
44
3
max
=
==