structure analysis chapter 1
DESCRIPTION
Structure analysis chapter 1TRANSCRIPT
-
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
- Boundary Condition 40
- Boundary Condition 64
- 40
-
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
=
==