Mechanics of Materials and Structural Analysis
Structural analysis asks what the forces do inside a determinate structure. The Bifocal paper concentrates on shear force and bending moment diagrams, truss analysis by joints and sections, and simple deflections; determinate analysis supplies the numericals.
Rules for SFD and BMD
- Shear force changes abruptly by the value of a point load, varies linearly under a UDL, and is unchanged by a pure moment.
- Bending moment is maximum where the shear force is zero or changes sign; a point load gives a kink, a UDL gives a parabola, a couple gives a jump.
- Draw the BMD on the tension side; for a simply supported beam with downward load, sagging is positive and plotted below the axis.
Trusses
A perfect truss satisfies m = 2j - 3, where m is the number of members and j the number of joints including supports. The method of joints resolves two unknowns per joint; the method of sections cuts at most three members.
Worked example
A simply supported beam of span 6 m carries a UDL of 10 kN/m over the whole span. Each reaction = 10 x 6 / 2 = 30 kN. Shear force is 30 kN at the support, falling linearly to zero at midspan. Maximum bending moment at midspan = wL squared / 8 = 10 x 36 / 8 = 45 kNm. The BMD is a parabola peaking at 45 kNm under the load.
Deflection
For a simply supported beam with a UDL, the maximum deflection is 5wL to the fourth / (384EI) at midspan; for a central point load it is WL cubed / (48EI). Keep units consistent: w in N/mm and E in N/mm2 with I in mm to the fourth give deflection in mm.
Traps
- Sign errors from taking moments about the wrong end.
- Forgetting that a couple produces no shear but a jump in bending moment.
- Applying m = 2j - 3 without counting support joints.
- Plotting the BMD on the compression side.