Question

lili I AYA till Il Emphas hedullileuny. Lung Paragraph Styles 2.1.[35 pts) For the question below, a-) compute the unknown reactions. b-) draw the normal force, shear force and bending moment diagrams. Show the critical values. Joint B is rigid connected 100 KN 200 kN.m hinge В 3 m 100 KN 2 m 10 m

solution

Answer 1

200 kNm 100 KN ✓ B с 3m E 100KN 2m A * HA Ho VA Vp 41 10m X * C No. of unknown reactions = 4. No. of equations = 3 equilibrium + 1 internal hinge. Degree of static indeterminacy = 4 - (3+1) = 0. The given frame is statically determinate. Notation: Anti-clockwise moment = +ve Clockwise moment = -ve (Mz), = 0 (sum of moment of forces about z axis at pivot point A = 0) Vp:10 + 100-2 - 100.10+ 200 = 0 VD = 60 KN Fy = 0 (sum of forces in y direction = 0) Vp-100 + VA = 0 60 - 100 + VA = 0 VA = 40 KN Bending moment at internal hinge at C = 0 E (M2) Right = 0) (sum of moment of forces to the right of C about z axis = 0) Hp: 5 - 100.3 = 0 Hp = 60 KN

Fx = 0 (sum of forces in x direction = 0) HA- 100+ Hp = 0 HA - 100 + 60 = 0 HA = 40 kN To find Bending moment, Shear and axial force in the members: Consider equilibrium of individual members. Member AB: Cut the frame at an arbitary location y on AB and consider FBD of bottom section. Let y be a positive distance from end A. E (Mz)y = 0 (sum of moment of forces about z axis at pivot point Y = 0) 40-y + My = 0 My ↑ My M(y) = -40y 0 Sy < 5m at A, y=0 M(0) = 0 > at B, y=5 M(5) = -200 kNm 4 Fy = 0 (sum of forces in y direction = 0) 0 N(y) = -40 kN 0 < y < 5m 40+ Ny = A 0) Fx = 0 (sum of forces in x direction = 40 – Vy = 0 140KM V (y) = 40 kN 0 Sy < 5m 40KN Member BC: Cut the frame at an arbitary location x on BC and consider FBD of left section. Let x be a positive distance from end B. E (Mz)x= 0 (sum of moment of forces about z axis at pivot point X = 0)

40.5 - 40-x + 200 + Mx = 0 X 200KNm X M(x) = 40.x - 400 0<x< 10m Mx f Nx at B, x=0 at C, x=10 M(0) = -400 kNm M(10) = 0 B X 15m Fy = 0 (sum of forces in y direction = 0) 40 + Vx = 0 V (x) = -40 KN 0<x< 10m 40KN Fx = 0 (sum of forces in x direction = 0) 40+ Nx = 0 N(x) = -40 KN 0<x< 10m 40KN Member CD: CD has an intemediate load of 100kN and FBD of member, above and below the load, should be considered seperate. Cut the frame at an arbitary Y between D and E, and consider FBD of bottom section. Let y be a positive distance from end D. (Mz)y = 0 (sum of moment of forces about z axis at pivot point Y = 0) 60-y - My = 0 M(y) = 60y 0 Sy s 2m ^ a at A, y=0 M(0) = 0 at point load, y=2 M(2) = 120 kNm Fy = 0 (sum of forces in y direction = 0) 60+ Ny = 0 N(y) = -60 KN 0 Sy s 2m боку * Ex = 0 (sum of forces in x direction = 0) 60 - Vy = 0 V (y) = 60 KN 0 <y s 2m 6 OKN Cut the frame at an arbitary location Y between E and C. consider FBD of bottom section. (M2)y = 0 (sum of moment of forces about z axis at pivot point Y = 0)

60-y – 100-(y – 2) – My = 0) M(y) 200 – 40.y 2m sy s 5m any - Mч at point load, y=2 M(2) = 120 kNm fy at C, y=5 M(5) = 0 E TOOKN EFy = 0 (sum of forces in y direction 0) 60+ Ny = 0 N(y) = -60 kN 2m = y < 5m 4 2m 0) Fx = 0 (sum of forces in x direction 60 – 100 – Vy = 0 D → Goku V(y) = -40 kN 2m sy s 5m tookm With calculated value of bending moment, shear and axial force the diagrams are drawn. Note: Bending moment is drawn on the tension side.

400 BMD 7 200 in Kim. B Notation: 75 the Drown on 62 - ve Tension sidh. 120 +1 E А SED 40 B to C Notation: in KN 40 40 A the -ve + 60 40 + A 60 40 D AFD 40 40 in 20 60 B с KN Natations Tensile tue Compressive-ve “А 40 60