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Problem 5.6: for the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
Solution:
Drawing the free body diagram and applying equilibrium are shown in the
following figure:
……(1)
Solved Examples
∫
Now, we want to get values of shear and bending moment at :
Substitute in (1) &(2) &(3) we find that :
It is clear that moment at x=0 ,x=L is zero as there is not concentrated M.
Problem 5.11: Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
Solution:
Free body diagram:
∑ ∑
∑
There is a jump in bending moment at D(x=0.3 m)
There is a jump in bending moment at F (x=0.9 m)
Now we can draw the S.F.D and the B.M.D:
| | | |
Problem 5.55: Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum normal stress due to
bending.
Solution:
Free body diagram:
∑
∑
∫
∫ ∫
∫
Now, we want the values of S,M at A,C,B:
Now we can draw the S.F.D and the B.M.D:
Note: This diagram is not to scale.
Problem 5.67: For the beam and loading shown, design the cross section of
the beam. Knowing that the grade of timber used has an allowable stress of
12 MPa.
Fig. P5.67
Solution:
Free body diagram:
∑
∑
Shear & bending moment diagrams:
From B.M.D, We can see obviously that
Problem 5.79: A steel pipe of 100 mm diameter is to support the loading
shown. Knowing that the stock of the pipes available has thickness varying
from 6 mm to 24 mm in 3-mm increments, and that the allowable normal
stress for the steel used is 165 MPa, determine the minimum wall
thicknesses t that can be used.
Fig. P5.79
Solution:
Shear & bending moment diagrams:
From B.M.D, We can see obviously that
, the min. wall thickness that can be used is t=9 mm.
Problem 5.91: A 240-KN load is to be supported at the center of the 5-m span
shown. Knowing that the allowable normal stress for the steel used is 165
MPa, determine the smallest allowable length of beam if the
beam AB is not to be oversterssed, the most economical
shape that can be used for beam . Neglect the weight of both beams.
Solution:
For the beam
(
) (
)
Substituting in (1), we have :
Shear & bending moment diagrams:
[
]
(
)
(Reqired no. 1)
For Beam AB:
