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Simply supported beam with a single eccentric concentrated load. An illustration of the Macaulay method considers a simply supported beam with a single eccentric concentrated load as shown in the adjacent figure. The first step is to find . The reactions at the supports A and C are determined from the balance of forces and moments as
The deflection at any point, , along the span of a center loaded simply supported beam can be calculated using: [1] = for The special case of elastic deflection at the midpoint C of a beam, loaded at its center, supported by two simple supports is then given by: [ 1 ] δ C = F L 3 48 E I {\displaystyle \delta _{C}={\frac {FL^{3}}{48EI}}} where
Besides deflection, the beam equation describes forces and moments and can thus be used to describe stresses. For this reason, the Euler–Bernoulli beam equation is widely used in engineering, especially civil and mechanical, to determine the strength (as well as deflection) of beams under bending.
Simply supported beam with a constant 10 kN per meter load over a 15m length. Take the beam shown at right supported by a fixed pin at the left and a roller at the right. There are no applied moments, the weight is a constant 10 kN, and - due to symmetry - each support applies a 75 kN vertical force to the beam. Taking x as the distance from ...
The bending moment at a particular cross section varies linearly with the second derivative of the deflected shape at that location. The beam is composed of an isotropic material. The applied load is orthogonal to the beam's neutral axis and acts in a unique plane. A simplified version of Euler–Bernoulli beam equation is:
The resulting equation is of 4th order but, unlike Euler–Bernoulli beam theory, there is also a second-order partial derivative present. Physically, taking into account the added mechanisms of deformation effectively lowers the stiffness of the beam, while the result is a larger deflection under a static load and lower predicted ...
Using the beam sign convention and cutting the beam at B, we can deduce the figure shown. Part (e) of the figure shows the influence line for the bending moment at point B. Again making a cut through the beam at point B and using the beam sign convention, we can deduce the figure shown.
A beam supported at its Airy points has parallel ends. Vertical and angular deflection of a beam supported at its Airy points. Supporting a uniform beam at the Airy points produces zero angular deflection of the ends. [2] [3] The Airy points are symmetrically arranged around the centre of the length standard and are separated by a distance equal to