Example #1
Example #1
• Equilibrium conditions for the whole beam:
Example #1
Example #1
• Equilibrium of the forces & moment
Example #1
Example #2
• The following problem has been considered
and V = 200N and M’ = 2100 Nm. Plot the
shear force and bending moment diagram.
Express the shear force and bending moment
in terms of the Macaulay’s bracket.
Example #2
• Solution:
(a) 0 < x < 4,
SFy = 0 Þ F = 200
SMcut = 0
Þ M + 2100 = 200x
Þ M=200x-2100
Example #2
(b) 4 < x < 10,
SFy = 0 Þ F = 200
SMcut = 0
Þ M + 2100 = 400+200x
Þ M=200x-1700
Example #2
(c) 10 < x < 14,
SFy = 0 Þ F +200= 200 Þ F=0
SMcut = 0
Þ M + 2100 = 400+200(10)
Þ M=300
Example #2
For S.F.D. and B.M.D.,
F(x=0) = 200, F(x=4) = 200
F(x=10-) = 200, F(x=10+) = 0
F(x=14) = 0
Note : wherever there is a point moment, there is step
change in M. E.g
M(x=0) = -2100
M(x=4-) = -1300, M(x=4+) = -900
M(x=10) = 300
M(x=14) = 300.
Example #2
Example #2
• Since the 400 Nm moment would enter the free
body diagram when x > 4 and the 200 N load
would enter the free body diagram when x > 10.
F = 200 - 200 [x - 10]0
M = -2100 + 200 x + 400 [x-4]0 - 200 [x - 10]
Example #3
• For the problem shown in the figure.
a) Calculate the supporting reaction.
b) Plot the shear force diagram and the bending
moment diagram.
Example #3
• Solution:
(a) For vertical force equilibrium,
/
/
=
2 =
/
⇒
(a) Using
=
/
= /
dV
V
V
2
Example #3
(b) The following boundary conditions apply,
Hence, A = B = 0 and
V
The reaction forces can be verified by
V
V
Example #4
• For the following stress
state, determine
a) The principal stress
angles and the principal
stresses,
b) The stress components
after a rotation of 30°
anti-clockwise, and
show these stresses on
a sketch.
Example #4
• Solution:
Example #4
• Solution:
(a)
= 100
= −48
=
;
= 60
;
100 − 60
(
) +48 = 52
2
= 80 + 52 = 132
Hence,
=
and
= 33.69
= 80 − 52 = 28
180 − 67.38 = 56.31
Example #4
(b) After rotating the point (100, -48) by 2X30o=60o in
clockwise direction (opposite to the question)
= 80 −
= 80 − 52cos 180 − 60 − 67.38
= 48.43
= 80 +
= 111.57
= −52
= −41.32
Example #4
Example #5
• Given a Mohr circle for a stress state as shown in the
figure.
a) Determine ,
and
?
Example #5
• If the material element is rotated, determine the
values of the following stress components and the
angles of rotation required for obtaining:
b) Maximum normal stress;
c) Minimum normal stress;
d) Maximum shear stress;
e) Minimum shear stress; and
f) The shear stresses corresponding to zero normal
stress.
Example #5
• Solution:
(a)
= 18 + 25 45 = 35.68 ( ,
= 18 − 25 45 = 0.32
= −25 45 = −17.68
(b)
)
= 18 + 25 = 43
= 22.5
(c)
as it is anti-clockwise on Mohr Circle.
= 18 − 25 = −7
= 67.5
as it is clockwise on Mohr Circle.
(
,
)
Example #5
• Solution:
(d)
= 25 =
= 67.5
as it is anticlockwise on Mohr
Circle.
(e)
= −25 =
= 22.5
as it is clockwise on Mohr
Circle.
(
,
)
Example #5
(f) The normal stress is zero at b and c which are the
intersection of the circle with shear stress axis. They
make an angle a with the normal stress axis.
=
⟹
= 43.95
,
= 90.53
= 45.53
=−
= 25
= 17.35
180 − + 45
2
180 − − 45
2
Example #6
• For an industrial laboratory a pilot unit is to employ a
pressure vessel of the dimensions shown below.
The vessel will operate at
an internal pressure of 0.7
(MPa). If for this unit 20 bolts
are to be used on a 650 (mm)
bolt circle diameter, what is
the required bolt diameter at
the root of the threads?
Example #6
• Set the allowable stress in tension for the bolts at 125
(MPa). However, assume that at the root of the bolt
threads the stress concentration factor is 2.
• Solution:
The vertical force F acting on the cover is given by
600
) = 198 × 10
=
= 0.7 (
2
Assuming that this force is equally distributed among
the 20 bolts, the force P per bolt is given by
198 × 10
= 9.9 × 10
20
Example #6
• The required bolt area is given by
=
⟹
2 × 9.9 × 10
=
=
125 × 10
= 0.1584 × 10
= 158.4
• Hence, the required bolt diameter at the root of the
threads is
= 2 / = 14.2
• Note that initial tightening of the bolts results in a
relatively small increase in total bolt stress when the
vessel is pressurized.
Example #7
• Determine the moments of inertia for the I-profile.
Simplify the results for , ≪ , ℎ.
Example #7
• Solution:
ℎ
ℎ
=
+2
+( + )
12
12
2 2
ℎ
2
ℎ
=
+
+
+ ℎ
12
3
2
ℎ
2
=
+
12
12
Example #7
• In case , ≪ ,ℎ, the terms which contain d, t
quadratically or to the third power may be
neglected as compared with the terms that
are linear in d and t:
Example #8
• The beam of symmetric I-section as shown is simply
supported over a span of 9 m. If the max. permissible
stress is 75 N/mm2, determine the concentrated load that
the beam can carry at a distance 3 m from one supported
end.
all dimensions in mm
Example #8
• Solution: From equilibrium consideration,
Bending
moment
diagram
RA = W/3 and RB = 2W/3
Mmax = 6W/3 kNm = 2 ´ 106 W Nmm
Example #8
For the I-section,
And
Hence,
Example #9
• Solve the deflection for a simply supported beam
that carries a uniform distributed load.
Statically
Indeterminate
Beams
• Solution:
35
Example #9
36
Example #9
37