![](\"resources/question-resources/LoadedBeam.png\"/)
Euler-Bernoulli simply supported beam bending example with point force and moment.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "A beam is simply supported at two points from below, and loaded from above with force $F$. A moment $M$ is applied also. The beam is in static equilibrium, has constant cross-section with second moment $I$ and uniform material properties with Young's elastic modulus $E$. It can be treated like an Euler-Bernoulli beam with bending equation:
\n$-{\\sigma_x \\over y} = {m(x) \\over I} = {E \\over R(x)} = E v''(x)$
\nwhere $m(x)$ is the bending moment, $R(x)$ is the bend radius and $v''(x) = 1/R(x)$ is the corresponding curvature; the coordinate $x$ is zero at the left end of the beam.
\n
The equation of force per unit length can be written for this case using the Dirac delta function, $\\delta(x)$:
\n$f(x) = R_L \\delta(x-L_1) - F \\delta(x-(L_1+L_2)) + R_R \\delta(x-(L_1+L_2 + L_3 + L_4))$
\nwhere we are ignoring the applied moment for now. This can be integrated to get the shear force function:
\n$V(x) = R_L \\langle x-L_1 \\rangle^0 - F \\langle x-(L_1+L_2)\\rangle^0 + R_R \\langle x-(L_1+L_2 + L_3 + L_4)\\rangle^0$
\nwhere $\\langle x - a \\rangle^0$ is a step-function that steps up from zero to one at $x=a$. In general, these Macauley brackets are defined as:
\n$\\langle x - a \\rangle^n = \\left\\{ \\begin{array}{ll} (x-a)^n & x > a\\\\ 0 & x < a\\end{array} \\right.$
\nIn these equations, positive forces are upwards, and positive moments are anticlockwise. When integrating to get the bending moment function, we need to subtract any applied moments as we pass them. In this case, $M$ is clockwise, so is technically already negative and ends up getting added:
\n$m(x) = R_L \\langle x-L_1 \\rangle^1 - F \\langle x-(L_1+L_2)\\rangle^1 + R_R \\langle x-(L_1+L_2 + L_3 + L_4)\\rangle^1 + M \\langle x-(L_1+L_2 + L_3)\\rangle^0$
\nNote: The maximum bending moment (of either sign) will be either where the force is applied or immediately one side or the other of the applied moment.
\nUsing the bending equation, this can be integrated to give us the deflection gradient of the beam, $v'(x)$:
\n$EIv'(x) = {1 \\over 2} R_L \\langle x-L_1 \\rangle^2 -{1 \\over 2} F \\langle x-(L_1+L_2)\\rangle^2 +{1 \\over 2} R_R \\langle x-(L_1+L_2 + L_3 + L_4)\\rangle^2 + M \\langle x-(L_1+L_2 + L_3)\\rangle^1 + A$
\nwhere $A$ is a constant of integration. A final integration step gives the vertical deflection of the beam, $v(x)$:
\n$EIv(x) = {1 \\over 6} R_L \\langle x-L_1 \\rangle^3 -{1 \\over 6} F \\langle x-(L_1+L_2)\\rangle^3 +{1 \\over 6} R_R \\langle x-(L_1+L_2 + L_3 + L_4)\\rangle^3 + {1 \\over 2} M \\langle x-(L_1+L_2 + L_3)\\rangle^2 + Ax + B$
\nwhere $B$ is a constant of integration. The constants of integration are determined by the requirement that the vertical deflection is zero at the supports, i.e., $v(L_1) = 0$ and $v(L_1+L_2+L_3+L_4) = 0$. The first of these is trivial because the Macauley brackets all disappear for the first part of the beam:
\n$0 = EIv(L_1) = A L_1 + B$
\ni.e., $B = -A L_1$, so there is only one constant still to be fixed. At the other support, none of the Macauley brackets disappear, although one of them is technically zero:
\n$\\begin{array}{rcl}0 = EIv(L_1+L_2+L_3+L_4) &=& {1 \\over 6} R_L \\langle L_1+L_2+L_3+L_4-L_1 \\rangle^3 -{1 \\over 6} F \\langle L_1+L_2+L_3+L_4-(L_1+L_2)\\rangle^3 + \\\\ &&{1 \\over 6} R_R \\langle L_1+L_2+L_3+L_4-(L_1+L_2 + L_3 + L_4)\\rangle^3 + \\\\ && {1 \\over 2} M \\langle L_1+L_2+L_3+L_4-(L_1+L_2 + L_3)\\rangle^2 + Ax + B \\\\ &=& {1 \\over 6} R_L ( L_2+L_3+L_4 )^3 -{1 \\over 6} F ( L_3+L_4)^3 +{1 \\over 6} R_R ( 0 )^3 + {1 \\over 2} M ( L_4 )^2 + A(L_2+L_3+L_4)\\end{array}$
\nwhere you can see that $L_1$ and $R_R$ have vanished from the expression, and the constant $A$ is given by:
\n$A(L_2+L_3+L_4) = -{1 \\over 6} R_L ( L_2+L_3+L_4 )^3 +{1 \\over 6} F ( L_3+L_4)^3 - {1 \\over 2} M ( L_4 )^2$
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\nThe dimensions shown in the illustration are: $L_1$={L1}cm, $L_2$={L2}cm, $L_3$={L3}cm, $L_4$={L4}cm and $L_5$={L5}cm.
\nDetermine:
\n