Question

Part A Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment. El is constant. (Figure 1) Express your answer in terms of L. EVO AED It vec ? Submit Request Answer ure 1 of 1 Provide Feedback Activate Go to Settin

Answer 1

Draw free body diagram of the beam. P MA A B L-a B, Apply equilibrium conditions. Consider forces acting along y-direction F, = 0 4, +B, =P (1) M=0 M,+B, L = Pa (2) Consider the section. A M(x) M M(x) B, At 0 < x <a, M (34) = 4,24-M At0 < x <(L-a), M(x)=B,

d'y Using equation EI = M(x) dx? EI dx =M(x) = 4,4-M (3) dx EXY **** (4) dy EI =M(+5)=B, 32 dx? El dv, B, dx, 2 B -x+ ₂ (5) EI v3 = x <3 +34 +64 -(6) Apply Boundary Conditions. du At x1 = 0, 1 = x = 0 dx From equation (3) Er dn _ A Mata dy EI(0) = (0) -M()+4 q=0 From equation (4) El vi 4-M4x+64 +63 (0)-44(0)+2(0)+c 4 EI (0) C₂ = 0

From Equation (6) At x = 0.12 = 0 B EI vz = 5 x +C3x3 +64 EI(O) = (0) +c,(0)+C. C4 = 0. -dva Apply Continuity Conditions. Atx = a.x = L-a, αχ, αχ, Equate equations (3) and (4) and y = 12 2-M24+9= 44-Apr-1 (2-a)+] 4a B (2-a) Ca = Ma- (7) 2 6 (a) M42 - B.(2-a+c(2-a) +M_a(2-a) - 4. (2-a) - (L-a). L-a!).3 (1.0)(0---)- (a) Mia? - By (L-a) 6 6 a a'(L-a) 4 = 0 (8) 6 MA Solve Equation (1). (2) and (8). Pa (2L? +a? -3al 31 Р 3 Pa? B = 4 = (3€ – 3dL+a") y 2E (3L-a)

Pa 21. (2 +a– 3aL) La P 20 (2L?-34+) (3L-a) 21 Pa? 2L (3L-a)(L-a) Bending moment (M) L - Pa (2L+a- 3al.) 2L From bending moment diagram equation. Pa? M. (3L-a)(L-a) 2 MMAX -Pa (21? + a²-3aL) 21 But required condition is Mmx|=|Main| Pa (3L-a)(L-a) Pa(2Ľ +a? -3aL 2Ľ 2L? 3al-a? = 212 - aL 3al-d=-(2-a) a?-2 al-2Ľ = 0 a=(2-2)