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د.م يوسف حميضة

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اريد شرح عن deflection

حساب الانتقال والسهم

deflection for various loads and supports

2waysl-1.jpg




Timber_deflection.gif


simplebm01.jpg










461_lecture40_pic1.gif



20.JPG

image003.jpg


Beam deflection for various loads and supports

Beams can vary greatly in their geometry and composition. For instance, a beam may be straight or curved. It may be of constant cross section, or it may taper. It may be made entirely of the same material (homogeneous), or it may be composed of different materials (composite). Some of these things make analysis difficult, but many engineering applications involve cases that are not so complicated. Analysis is simplified if:
-The beam is originally straight, and any taper is slight-The beam experiences only linear elastic deformation-The beam is slender (its length to height ratio is greater than 10)-Only small deflections are considered (max deflection less than 1/10 the span)
In this case, the equation governing the beam's deflection (
f1290186a5d0b1ceab27f4e77c0c5d68.png
) can be approximated as:
4c1f36e87e71fd8e28a73abbd0b6b2d1.png
where the second derivative of its deflected shape with respect to
9dd4e461268c8034f5c8564e155c67a6.png
is interpreted as its curvature,
3a3ea00cfc35332cedf6e5e9a32e94da.png
is the Young's modulus,
dd7536794b63bf90eccfd37f9b147d7f.png
is the area moment of inertia of the cross-section, and
69691c7bdcc3ce6d5d8a1361f22d04ac.png
is the internal bending moment in the beam.
If, in addition, the beam is not tapered and is homogeneous, and is acted upon by a distributed load
7694f4a66316e53c8cdd9d9954bd611d.png
, the above expression can be written as:

7478d35393c67993af00fa5abaa94efb.png

This equation can be solved for a variety of loading and boundary conditions. A number of simple examples are shown below. The formulas expressed are approximations developed for long, slender, homogeneous, prismatic beams with small deflections, and linear elastic properties. Under these restrictions, the approximations should give results within 5% of the actual deflection.
Cantilever beams

Cantilever beams have one end fixed, so that the slope and deflection at that end must be zero.


Schematic of the deflection of a cantilever beam.

End-loaded cantilever beams



Cantilever beam witha force on the free end

The elastic deflection
f10f03c9836c36537d2539196058bfa2.png
and
angle of deflection
7f20aa0b3691b496aec21cf356f63e04.png
(in
radians) at the free end in the example image: A (weightless) cantilever beam, with an end load, can be calculated (at the free end B) using:[SUP][1][/SUP]

c32ad9c31ad56dbb27215044fcaaba64.png
06431808649098106e8a7c70a3f0aff3.png

where
800618943025315f869e4e1f09471012.png
= Force acting on the tip of the beam
d20caec3b48a1eef164cb4ca81ba2587.png
= Length of the beam (span)
3a3ea00cfc35332cedf6e5e9a32e94da.png
= Modulus of elasticity
dd7536794b63bf90eccfd37f9b147d7f.png
= Area moment of inertia of the beam's cross section
Note that if the span doubles, the deflection increases eightfold. The deflection at any point,
9dd4e461268c8034f5c8564e155c67a6.png
, along the span of an end loaded cantilevered beam can be calculated using:[SUP][1][/SUP]

0b9ecf276634f2507010eb5995832788.png
56a5962669aceca67add5e94530b05aa.png

Note that at
3c098f36119f1c2104b704c3c2b84699.png
(the end of the beam), the
80a0e70dc48d34c9687bc691696a7d4e.png
and
7586d0f39aaa12dbd632db78a8c67532.png
equations are identical to the
b47e2db356a037b1828ba2f9df2a4b45.png
and
15a3f29a6b2865710de38972c7d78335.png
equations above.

Uniformly-loaded cantilever beams



Cantilever beam with a uniform distributed load

The deflection, at the free end B, of a cantilevered beam under a uniform load is given by:[SUP][1][/SUP]
ff0c9a0ce2fafc46072992ac3f1c79dd.png
80ee1f59737ddbbeb43c7bace97e04e5.png

where
7694f4a66316e53c8cdd9d9954bd611d.png
= Uniform load on the beam (force per unit length)
d20caec3b48a1eef164cb4ca81ba2587.png
= Length of the beam
3a3ea00cfc35332cedf6e5e9a32e94da.png
= Modulus of elasticity
dd7536794b63bf90eccfd37f9b147d7f.png
= Area moment of inertia of cross section
The deflection at any point,
9dd4e461268c8034f5c8564e155c67a6.png
, along the span of a uniformly loaded cantilevered beam can be calculated using:[SUP][1][/SUP]
22bfc584010e2e7d92680945be20170c.png
243851dd80618307b1ed15d15e6ffa47.png
Simply-supported beam

Simply-supported beams have supports under their ends which allow rotation, but not deflection.


Schematic of the deflection of a simply-supported beam.

Center-loaded simple beams



Simply-supported beam with a force in the center

The elastic deflection (at the midpoint C) of a beam, loaded at its center, supported by two simple supports is given by:[SUP][1][/SUP]
6f407eb3a0a273601ba11063a49f9276.png

where
800618943025315f869e4e1f09471012.png
= Force acting on the center of the beam
d20caec3b48a1eef164cb4ca81ba2587.png
= Length of the beam between the supports
3a3ea00cfc35332cedf6e5e9a32e94da.png
= Modulus of elasticity
dd7536794b63bf90eccfd37f9b147d7f.png
= Area moment of inertia of cross section
The deflection at any point,
9dd4e461268c8034f5c8564e155c67a6.png
, along the span of a center loaded simply supported beam can be calculated using:[SUP][1][/SUP]

for
041ef0cfb37da4775a27fca2f4a10531.png

e425531273bea2c367fdf72b8aad26b0.png
Off-center-loaded simple beams



Simply-supported beam with a force off center

The maximum elastic deflection on a beam supported by two simple supports, loaded at a distance
0cc175b9c0f1b6a831c399e269772661.png
from the closest support, is given by:[SUP][1][/SUP]

682d9ee921193f89a4fcb0ccaaeac5ea.png

where
800618943025315f869e4e1f09471012.png
= Force acting on the beam
d20caec3b48a1eef164cb4ca81ba2587.png
= Length of the beam between the supports
3a3ea00cfc35332cedf6e5e9a32e94da.png
= Modulus of elasticity
dd7536794b63bf90eccfd37f9b147d7f.png
= Area moment of inertia of cross section
0cc175b9c0f1b6a831c399e269772661.png
= Distance from the load to the closest support (i.e.
2a64410903432c708970b34123d5acb5.png
)
This maximum deflection occurs at a distance
f9a3b8e9e501458e8face47cae8826de.png
from the closest support and is given by:[SUP][1][/SUP]

8013ab48daa520e5f46e7a613b2c8b8c.png
Uniformly-loaded simple beams



Simply-supported beam with a uniform distributed load

The elastic deflection (at the midpoint C) on a beam supported by two simple supports, under a uniform load (as pictured) is given by:[SUP][1][/SUP]
d0a031089dc2e47e769612a492cf59e6.png

where
7694f4a66316e53c8cdd9d9954bd611d.png
= Uniform load on the beam (force per unit length)
d20caec3b48a1eef164cb4ca81ba2587.png
= Length of the beam
3a3ea00cfc35332cedf6e5e9a32e94da.png
= Modulus of elasticity
dd7536794b63bf90eccfd37f9b147d7f.png
= Area moment of inertia of cross section
The deflection at any point,
9dd4e461268c8034f5c8564e155c67a6.png
, along the span of a uniformly loaded simply supported beam can be calculated using:[SUP][1][/SUP]

dcc47505df8fe5c59b83c345f4e6f862.png


460098_01327944493.jpg


 
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Mohamed_Osman09

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