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Fundamental Mechanics ENGG 1300A
All rights reserved 2σ3 = F3/A = 25 MPa (tension) F=10kN F3 = 10kN F2 = -10kN F=10kN F=10kN 20 kN F1 =
-20kNF=10kN 20 kN 10 kN σ2 = F2/A = - 25 MPa (compression) σ1 = F1/A = -
50 MPa (compression) 3b = 2 m, and g = 10 m/s2 = F1 = ρVg F1 F1 F2 4= F1 d F2 σ + F1 d =
12 m e F2 e = 12/2–1.5= 4.5 m superposition of axial and flexural stresses Flexural stress σf =
6M/(bd 2) = 6F1e/(bd 2) (See slide no. 12, axially loaded members) Axial stress σ = (F1+F2)/A F1 d
F1 σf σf e 5superposition of axial and flexural stresses Weight F2 = 3x12x2ρg = 72ρg Axial stress =
126ρg /(12x2) = 5.25ρg Eccentricity (e) = 4.5 m Moment (M) = eF1= 243ρg σA = axial stress +
flexural stress = 232 kPa σB = axial stress - flexural stress = 4.22 kPa Flexural stress = 6M/(bd2) =
5.0625ρg (See slide no. 12, axially loaded members) Weight F1 = 3x9x2ρg = 54ρg F1 d = 12 m e F2 e =
12/2–1.5= 4.5 m Flexural stress σf = 6M/(bd 2) Axial stress σ = (F1+F2)/A 6F = ρgd(d/2)b = ρgbd2/2 d = 3 m;
b = 2 m and g = 10 m/s2 Take moment at B (d/3)F – (d/2)RA=0 RA = 2/3F = ρgbd2(1/3) =
60 kN RB = F - (1/3)ρgbd 2 = ρgbd2/2 - (1/3)ρgbd2 = (1/6)ρgbd2 = 30 kN F RA RB ρgd 7F1 =
106 = 60 kN; F2 = 100 kN Take moment at A -F16/2 – F2(6+4) + RB6 = 0 RB = [603 + 10010]/6 =
196.7 kN Vertical force equilibrium RA + RB = F1 + F2 RA = 60 + 100 – 196.7 = -36.7 kN F1 F2 = 100 kN RBRA 8Q3(a) Zero-force members Answer: 1, 3, 5, 6, 7, 8 9Q3(b) Zero-force members Answer: ad, db, de, be, eg, cg, lg, fh, hi, il, kl, ki, kh, and jk 10 Q3(c) Degrees of indeterminacy Answer: 3 11 Q3(c) Degrees of indeterminacy Answer: 10 (ii) 12 Solve the reactions first, you can then use the method of joints to obtain all the internal forces. 13 Take moment at F, we can get internal force CE Take moment at C, we can get internal force DF Consider vertical force equilibrium, we can get internal force CF Method of Sections: 14 Take moment at D, you can get VA. HA VA HD VD Cut section through BC, you can get HA. Take moment at A, you can get VD. Cut section through BC, you can get yc. Consider horizontal force equilibrium, HA = HD 15 The end!