Geoscience Reference
In-Depth Information
Solution:
10 tons = 20,000 lb
R
L
=
R
R
=
P
/2 = (20,000 lb)/2 = 10,000 lb
Applying a safety factor of 4 requires that each support be designed to handle at least 32,000 lb, or
16 tons of force.
11.6.11.3 Load at Any Point
The equation below is used to calculate the reactions
R
L
and
R
R
in simple beams of length
L
that are
loaded off-center with a weight of
P
:
R
L
=
Pa
/
L
R
R
=
Pb
/
L
(11.17)
where
R
L
= Left reaction (lb).
R
R
= Right reaction (lb).
P
= Concentrated load on the beam (lb).
a
= Distance from the left support of the beam (ft).
b
= Distance from the right support of the beam (ft).
L
= Length of the beam (ft).
■
EXAMPLE 11.17
Problem:
A 10-ft-long beam supported at each end is loaded with a weight of 1200 lb at a point 2 ft
left of its center. What are the reactions on the supports? (Neglect weight of the beam.)
Solution:
R
L
=
Pa
/
L
= [(1200 lb)(2 ft)]/10 ft = 240 lb
R
R
=
Pb
/
L
= [(1200 lb)(8 ft)]/2 ft = 960 lb
11.6.12 b
uCKling
s
tress
(W
ood
C
olumns
)
Columns that support a load have a tendency to bend. If the bending becomes too great, the column
will become unstable and fail by buckling under the pressure from the load. The equation below is
for columns made of wood, with intermediate length:
P
/
A
= σ[
L
- 1/3(
L
/
KD
)]
4
(11.18)
where
P
= Maximum acceptable load applied to the column (lb, N).
A
= Cross-sectional area of the column (in.
2
, m
2
).
σ = Allowable unit compressive stress parallel to the grain (psi, kPa).
L
= Free unsupported length of the column (in., m).
K
= Minimum value of l/
D
for which Euler column mechanics may be used.
D
= Dimension (width) of the column in the expected direction of buckling (in., m).
11.6.13 F
floors
Environmental engineers commonly spend a considerable amount of time and attention on ensur-
ing the proper maintenance of floors and flooring in general, in ways that range from the mundane
(housekeeping) to the structurally essential (calculating floor load). Housekeeping is always a focal
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