40 Important Question of STRUCTURE ANALYSIS -II

 1-4- Analyze using Moment Distribution Method 2 frame 2 inclined frame

 

1-                ABCD is a frame, with AB and CD as columns, and BC as the beam with end A as fixed and D as hinged. AB=4m, BC=4m and CD=3m. BC has a udl of 4kN/m. AB has a point load of 6kN at its mid point.

2-                ABCD is a frame, with AB and CD as columns, and BC as the beam with ends A and D as hinged. AB=CD=4m, BC=3m. A point load of 120kN at 1m from B.

3-                ABCD is a frame, with AB and CD as inclined columns having slant height(original height of column) of AB=13m and CD=5m. BC=6m having a udl of 10 kN/m. A and D are hinged.

4-                ABCD is a frame with AB and CD as inclined columns having slant height (original height of column) of 8m. BC is a beam=8m. A and D are fixed. A point load of 40kN is there on BC from B. A sway force of 20kN is there on B.

5-8 Analyze using Kani’s method- 2 beams, 2 frames

 

5-                ABCD is a beam with A as fixed, B and C as roller with CD as an overhang. AB=4m, BC=3m, CD=1m. A point load of 2kN from A at a distance 1m, a clockwise moment of 2kNm at a distance of 1m from B, a point load of 0.5kN at point D.

6-                ABCD is a beam with A as fixed, B, C and D as roller. AB=4m with I, BC=4m with  I and CD=3m with 1.5I. A udl of 60kN/m on AB, a point load of 80kN at mid span of BC and a point load of 45kN on CD at a distance of 1m from C.

7-                ABCDE is a frame with AB and CD as columns with A and D as fixed. BC and CE are beams, with E as fixed. All members are of 4m. A clockwise moment of 100kNm at mid point of BC is there.

8-                ABCDEF is a frame with BE and CF as columns and AB, BC and CD as beams. A and F are hinged and, D and E are fixed. AB carries a udl of 50kN/m. A point load of 40kN on BC at a distance of 1.5m from B and a point load of 120kN at midpoint of CD is there. AB=3m with I, BC=4m with I, CD=4m with I, BE=4m with 2I and CF=3m with 2I.

9-12 shape factor derivation, 1 ques

 

9-                Derive shape factor for a rectangular section with base B and depth D. 10- Derive shape factor for a circular section with radius R.

11-  Derive shape factor for a triangular section with base B and height H.

12-  Find out the shape factor for a Plus section ‘+’ having horizontal length and vertical height of 400mm. The thickness of each flange is 50mm.

13-16 Using Plastic analysis 2 beams 2 frames

 

13-  A beam AB of span 6m is fixed at both ends and carries an idle of 100kN/m over a half span from A. Allow a load factor of 1.75. Yield stress for steel is 250N/mm².

 

14-  Determine the collapse load in the fixed beam AB. C is the midpoint of AB, AB=L. Plastic moment carrying capacity of 2Mp for AC and Mp for CB. A point load of W at L/3 from A is there.

15-  Determine the plastic moment capacity of the frame ABCD with A and D as fixed, AB and CD as columns and BC as beam. AB= 4m, BC=4m and CD=6m. A point load on AB at B of 20kN is there and a point load of 50kN on BC at the midpoint is there.

16-  Determine the plastic moment capacity of the frame ABCD with A as hinged D as fixed, AB and CD as columns and BC as beam. AB= 8m with 2Mp, BC=7m with 1.5Mp and CD=6m with Mp. A udl on AB of 40kN/m is there and a point load of 200kN on BC at a distance of 5m from B is there.

17-20 2 portal 2 cantilever

 

17-  Analyze using portal method. ABC, DEF, GHI are beams of a frame with J, K and L as fixed. Beams=6m and Columns=4m. A horizontal force of 12kN is there on A, a horizontal force of 24kN is there on D and a horizontal force of 24kN is there on G.

18-  Analyze using portal method. ABCD, EFGH, IJKL are beams of a frame with M, N, O and P as fixed. Beams=6m and columns=3m. Wind load at A=6kN, E=12kN and  I=12kN.

19-  Analyze, using the cantilever method. ABCD, EFGH are beams of a frame with I, J, K and L as fixed. AB=EF=4m, BC=FG=5m and CD=GH=6m. AE=BF=CG=DH=3m and EI=FJ=GK=HL=4m. A horizontal force of 12kN at A and 24kN at E is there.

20-  Analyze using cantilever method. ABCD, EFGH, IJKL are beams of a frame with M, N, O and P as fixed. Beams=6m and columns=3m. Wind load at A=6kN, E=12kN and I=12kN.

21-24 2 factor 2 substitute frame

 

21-  Analyze using factor method. ABCD, EFGH are beams of a frame with I, J, K and L as fixed. AB=EF=3m with stiffness 2K, BC=FG=4m with 3K and CD=GH=5m with 4K. AE=BF=CG=DH=3m and EI=FJ=GK=HL=4m with stiffness for AE, DH, EI, HL as stiffness K and for BF, FJ, CG, GK as stiffness 2K. A horizontal force of 12kN at A and 24kN at E is there.

22-  Analyze using factor method. ABC, DEF are beams of a frame with G, H and I as fixed. AB=DE=3m with stiffness 2K, BC=EF=4m with 3K, AD=BE=CF=2m with stiffness for AD and CF as K and BE as 2K. DG=EH=FI=3m with stiffness for DG and FI as 2K and EH as 3K.

23-  Analyze using substitute frame method ABCD, EFGH, IJKL, MNOP are fixed ends. AB=DE=IJ=9m, BC=EF=JK=3m, CD=FG=KL=6m. spacing between frames is 3.6m, DL on floors=4kN/m², LL on floors=3kN/m², self wt of beams=5kN/m for 9m, 3kN/m for 3m, 4kN/m for 6m, I for 9m=6I, I for 3m=2I, I for 6m=4I, I for columns is I.

 

24-  Analyze using substitute frame method ABCD, EFGH, IJKL, MNOP are fixed ends. AB=DE=IJ=6m, BC=EF=JK=3m, CD=FG=KL=6m. spacing between frames is 3m, DL on floors=4kN/m², LL on floors=2kN/m², self wt of beams=5kN/m for 6m, 3kN/m for 3m, 4kN/m for 6m, I for 6m=6I, I for 3m=2I, I for 6m=4I, I for columns is I.

25-28 flexibility 2 beam, 2 frame

 

25-  Analyze using flexibility matrix method.A beam ABC with A as fixed, B and C as rollers. AB=6m, BC=8m. A point load of 12t on AB at 4m from A and a point load of 8t on BC at 3m from B is there.

26-  Analyze using flexibility matrix method. A fixed beam AB of span 6m with a udl of 2t/m and a point load of 12t at the mid span along with an anticlockwise moment of 24tm at the mid span is there.

27-  Analyze the portal frame ABCD by flexibility matrix method. A and D are fixed. AB=5m, BC=4m and CD=3m. Horizontal load at joint B=20kN and udl at BC=50kN/m.

28-  Analyze the portal frame ABCD by flexibility matrix method. A and D are fixed. AB=6m with 2I, BC=10m with 3I and CD=4m with I. Horizontal load at joint B=6t and a point load of 20t on BC at 6m from C.

 

29-32 stiffness 2 beam, 2 frame

29-  Analyze using stiffness matrix method. A beam ABC with A as fixed, B and C as rollers. AB=6m, BC=8m. A point load of 12t on AB at 4m from A and a point load of 8t on BC at 3m from B is there.

30-  Analyze using stiffness matrix method. A fixed beam AB of span 6m with a udl of 2t/m and a point load of 12t at the mid span along with an anticlockwise moment of 24tm at the mid span is there.

31-  Analyze the portal frame ABCD by stiffness matrix method. A and D are fixed. AB=5m, BC=4m and CD=3m. Horizontal load at joint B=20kN and udl at BC=50kN/m.

32-  Analyze the portal frame ABCD by stiffness matrix method. A and D are fixed. AB=6m with 2I, BC=10m with 3I and CD=4m with I. Horizontal load at joint B=6t and a point load of 20t on BC at 6m from C.

 

 

33-40 3 parabolic arch, 3 semicircular arch, 2 fixed arch

 

33-  Find the horizontal thrust for the two-hinged semicircular arch.

34-  A two-hinged semicircular arch of radius R carries a udl of w/m over the whole span.

Determine the horizontal thrust at each support.

35-  A two-hinged semicircular arch of radius 20m carries a load system. Determine the horizontal thrust at each support.

 

36-  A two-hinged parabolic arch of span 20m and rise 4m carries a udl of 50kN/m on left half of span. Find the reactions at the supports and the position and amount of maximum bending moment.

37-  A two-hinged parabolic arch of span 40m and rise 8m carries a point load of 80kN at a distance of 10m from the left support. Find the horizontal thrust at each support.

38-  A two-hinged parabolic arch of span 25m and rise 5m carries a udl of 40kN/m over the left half of the span and a concentrated load of 100kN at the crown. Find the horizontal thrust at each support.