Statically determinate/indeterminate; beam SFD and BMD

Statically determinate and indeterminate structures

A structure is statically determinate if all support reactions and internal forces can be found from the equations of static equilibrium alone (ΣFx = 0, ΣFy = 0, ΣM = 0 in 2D). It is statically indeterminate when the unknowns exceed the available equations, so extra equations from deformation (compatibility) are needed.

Degree of indeterminacy of a beam: D = r − 3 − (number of internal hinges). A propped cantilever has r = 3 + 1 = 4, so D = 1.

Reading of the figure

Origin at the free left end O; roller at A (x = 1.8 m), pin at B (x = 7.8 m). The triangular load rises from 0 at O to w = 15 kN/m at x = 5.4 m (intensity 2.778x kN/m); a 65 kN point load acts at x = 6.6 m.

• Total triangular load = ½ × 15 × 5.4 = 40.5 kN, acting at 2/3 × 5.4 = 3.6 m from O.

   w=15 kN/m               65 kN
  /||||||||||||||           ↓
 O/_______________________________
 ^  A(roller)                      B(pin)
 |<-1.8->|<--3.6-->|<1.2>|<1.2>|

Reactions

• ΣM_B = 0: R_A(7.8 − 1.8) = 40.5(7.8 − 3.6) + 65(1.2), so 6R_A = 170.1 + 78 = 248.1 and R_A = 41.35 kN ↑.
• ΣFy = 0: R_B = 40.5 + 65 − 41.35 = 64.15 kN ↑. (r = 3, so the beam is determinate.)

Equations of V and M (x measured from O)

Load intensity w(x) = (15/5.4)x = 2.778x kN/m, load up to x = ½·2.778x·x = 1.389x².

(Sagging moment and the usual sign convention: upward force on the left of the section gives positive shear.)

Ordinates

Diagrams and salient features

 SFD (kN):  +36.85 just right of A
            decreases (parabola) to +0.85 at x = 5.4
            stays +0.85 to x = 6.6, jumps to −64.15
            constant −64.15 to B

 BMD (kN·m): 0 at O, −2.70 at A (hogging),
            0 at x = 1.87, +75.96 at x = 5.4,
            +76.98 at x = 6.6 (maximum), 0 at B

• Shear changes abruptly by +41.35 kN at the roller A, and by −65 kN at the point load (from +0.85 to −64.15 kN).
• The shear is 0.85 kN, almost zero, between x = 5.4 m and 6.6 m, so the bending moment is nearly flat there: the maximum moment is 76.98 kN·m at x = 6.6 m (under the 65 kN load), where the shear changes sign.
• A small hogging moment of 2.70 kN·m occurs over the support A (cantilever overhang); the moment changes sign at x ≈ 1.87 m (point of contraflexure, from 41.35(x − 1.8) = 0.463x³).
• Shear is a parabola over the triangular load (V falls by 1.389x²); the moment is a cubic curve there, a straight line under no load.
• The moment is zero at the pin B (end of the simply supported span).