SN=25
- Topic Approximate Analysis Of Indeterminate Frames Sn 25 Full Solution A Kn 42 B Kn 45 C Kn 15 D M 3 60 E M 1 (29.6 KiB) Viewed 80 times
- Topic Approximate Analysis Of Indeterminate Frames Sn 25 Full Solution A Kn 42 B Kn 45 C Kn 15 D M 3 60 E M 2 (38.07 KiB) Viewed 80 times
- Topic Approximate Analysis Of Indeterminate Frames Sn 25 Full Solution A Kn 42 B Kn 45 C Kn 15 D M 3 60 E M 3 (35.75 KiB) Viewed 80 times
- Topic Approximate Analysis Of Indeterminate Frames Sn 25 Full Solution A Kn 42 B Kn 45 C Kn 15 D M 3 60 E M 4 (42.78 KiB) Viewed 80 times
- Topic Approximate Analysis Of Indeterminate Frames Sn 25 Full Solution A Kn 42 B Kn 45 C Kn 15 D M 3 60 E M 5 (24.51 KiB) Viewed 80 times
- Topic Approximate Analysis Of Indeterminate Frames Sn 25 Full Solution A Kn 42 B Kn 45 C Kn 15 D M 3 60 E M 6 (46.76 KiB) Viewed 80 times
a.(KN) 42
b.(KN) 45
c. (KN) 15
d.(m) 3.60
e.(m) 4.10
f.(m) 4.20
g.(m) 4.20
- Topic Approximate Analysis Of Indeterminate Frames Sn 25 Full Solution A Kn 42 B Kn 45 C Kn 15 D M 3 60 E M 7 (67.39 KiB) Viewed 80 times
- Topic Approximate Analysis Of Indeterminate Frames Sn 25 Full Solution A Kn 42 B Kn 45 C Kn 15 D M 3 60 E M 8 (32.27 KiB) Viewed 80 times
Instruction: Solve the following problem completely using portal method and write the solutions neatly in the plate template. Summarize your final answers at the end of each problem. Round off your answers to 3 decimal places. 1. Portal Method: The frame shown is subjected to lateral loads as shown. Supports S, T, U V, W and X are rigid supports. Determine the following: a) The shear forces in columns and girders. b) The axial forces in columns and girders. c) The moment at the end of each member. d) Summarize your results by drawing the frame with the internal forces
5.0 m 5.50 m 6.0 m 7.0 m 5.0 m (150+SN) KN> B 40 m 144 IN E 4.20 ml 120 l H 4.50 m 96 W M N O Q R 5.0 m U w
2 Cantilever Method: The frame shown is subjected to lateral loads as shown. Supports S, T, U V, W and X are rigid supports. Column HMS (400mm x 400 mm) Column ADINT (500 mm x 600 mm) Column BEJOU (600 mm x 600 mm) Column CFKPV (500 mm x 600 mm) Column GLQW (500 mm x 500 mm) Column RX (300mm x 300 mm) Determine the following a) The shear forces in columns and girders. b) The axial forces in columns and girders. c) The moment at the end of each member. d) Summarize your results by drawing the frame with the internal forces.
5.0 m 5.50 m 6.0 m 7.0 m 5.0 m (100+SN) » A B 40 m 144 IN D 4.20 m 120 W Н. 4.50 m 96 KN M 0 P R 5.0 m U On w 1x
3. Factor Method: The frame shown is subjected to lateral loads as shown. Supports S, T, U V, W and X are rigid supports Determine the following a) The shear forces in columns and girders. b) The axial forces in columns and girders. c) The moment at the end of each member. d) Summarize your results by drawing the frame with the internal forces.
5.0 m 5.50 m 6.0 m 7.0 m 5.0 m 1.51 (160+SN) LON 1.51 A B с 1.51 4.0 m 1.51 21 144 KN E F G 1.51 21 21 1.51 4.20 m 1.51 21 120 KN 21 31 H 1 J K 1.51 21 2.51 2.51 21 4.50 m 21 21 31 31 96 kW M N o P Q R 21 31 31 31 21 1.51 5.0 m S T U V w
Instruction: Solve the following problem/s and show your complete solution for each. Input your final answers on the online Form in the link. Variables (a, b, c, and d) can be found on the list on page. You must upload your plate on the last part of the online Form. Problem 1. The indeterminate frame shown below is loaded with lateral forces applied on each level. Using Factor Method, draw the approximate moment and shear diagram of the structure due to the applied loads. Show your complete solution from the determination of girder/column factor, carry over factor, girder/column moment factor up to computation of end-moment for each member. From your computation, provide the following unknowns in the Form; a. Girder factor at joint D, go b. Column factor at joint K, CK C. Girder moment factor at joint F, GF d. Column moment factor at joint E, CE e. Bending moment at column AC, MAC f. Bending moment at girder ED, MED g Bending moment at girder CD, MCD h. Bending moment at column HL, MHL i. Shear at column DG, VG j Shear at beam HI, MHI
a KN 21 B B d m 41 31 b KN C С 21 31 E d m 41 31 51 a C KN F 21 3 ЗІ H 21 E 41 31 51 41 e m J K L M fm gm fm