Question

Consider the beam ABC of length L[m] in Figure 1 below, with simple supports at both ends. The beam supports a concentrated load P[N] at point B. You may assume the beam to be weightless in your analysis. P A B L/3 2L/3 Figure 1: Schematic of beam ABC. Part (a) [4 marks] Determine the vertical reaction forces at points A and C in terms of P. Part (b) [8 marks] Determine expressions (in terms of Pand L) for the shear force, V(x) and the bending moment, M(x) along the length of the beam. Here, x is defined as the distance [m] along the beam from point A. Part (c) [18 marks] Determine expressions (in terms of P, L, E, and I) for the deflection, v(x) and slope, v'(x) along the length of the beam. Hint: The deflection is continuous at point B (its value is the same whether we are considering segment AB or segment BC). The same holds for the slope.

Answer 1

We can find the support reactions in vertical direction by using the equilibrium conditions. Then we can easily write the shear force and bending moment in every section of the beam across the length. Please note that bending moment and shear force expressions will be different in different sections of beam.

Then we can apply the deflection equations and find the slope and deflection across the beam by finding the constants of integration from boundry conditions. Please find the attached solution.

Prob B 2/3 2L) Rc RA P1 X x Let Reactions are RA & Rc {fy=0 RAIRC P- EMA EO REAL PX 13 Rc =P/ from RA= 21/3 shear force at section x-x (fig. 1) sign convention section BC. tue -ve fx= RA-P for section AB. fa 2P - P Fe = RA Fx = -P/ FE OP 3 constant constant.

Bending moment Sign Convension (rue) (-ve) Section AB Section Be (P-2 P. Mx B RAIR 43 Me = 2.2 2P X Linear Max = 2P a P(x-4) Ma = 2 px _ px + 4 Ma = alm - DX 3 » Į (L-2) 3 Ma = P (2-2) SFD BMD. 21/3 + B P/3 B

To find slope & deflection, Let's write the moment equation Section X-X. (Fig. 1) P3) Mat RAX P(x-22) "Ma = o dě dx² 2 EI. dy da² 2x - P(x-1/4) EI dye PL Px de ² (1 ---- Integrating wortox. El dy PLx P22 tc, 6 Again integrating 8.2.4 = Plze po te ox+6 € Boundry conditione yo) = 0 y (L) = 0

9(0)= 9(4)=0 > 2=0 PL3 6 PL3 18 + ₂ C2 P23 (to = 5) PL² 9 (12-4 Then. from ② Slope erregion PL 8x2 va) dy dae x ЗЕТ 6 EI from Deflection exceation. PL3 *V(x) y = PL. 0²_ P.x² 6 ET 18 ET 9