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Design tools · Beams

Fixed-pinned beam under point load

Instant results: deflection, bending moment and shear force. The formulas used are given below the calculator.

The calculator needs JavaScript. The formulas are given below.

Formulas

Maximum deflection
if α<0.586L\alpha < 0.586 L
δmax=P (L−α) (2 L α−α2)33 E I (2 L2+2 L α−α2)2\delta_{\mathrm{max}} = \frac{P\ (L - \alpha)\ (2\ L\ \alpha - \alpha^2)^3}{3\ E\ I\ (2\ L^2 + 2\ L\ \alpha - \alpha^2)^2}
otherwise
δmax=P (L−α) α26 E I L−α3 L−α\delta_{\mathrm{max}} = \frac{P\ (L - \alpha)\ \alpha^2}{6\ E\ I}\ \sqrt{\frac{L - \alpha}{3\ L - \alpha}}
Maximum bending moment
Mmax=P (L−α) (2 L α−α2)2 L2M_{\mathrm{max}} = \frac{P\ (L - \alpha)\ (2\ L\ \alpha - \alpha^2)}{2\ L^2}
Minimum bending moment
Mmin=−P α2 (L−α) (3 L−α)2 L3M_{\mathrm{min}} = - \frac{P\ \alpha^2\ (L - \alpha)\ (3\ L - \alpha)}{2\ L^3}
Maximum shear force
Vmax=P (L−α) (2 L2+2 L α−α2)2 L3V_{\mathrm{max}} = \frac{P\ (L - \alpha)\ (2\ L^2 + 2\ L\ \alpha - \alpha^2)}{2\ L^3}
Minimum shear force
Vmin=−P α2 (3 L−α)2 L3V_{\mathrm{min}} = - \frac{P\ \alpha^2\ (3\ L - \alpha)}{2\ L^3}

Notations

  • EEmodulus of elasticity, in GPa
  • IIsecond moment of area of the cross-section, in cm⁴
  • LLeffective span, in cm
  • α\alpha position of the point load, in cm
  • PPpoint load, in kN