Question

For questions 33 through 36, refer to the frame loaded with a concentrated load at A and a couple at C, as shown. It is supported by a hinge at B and by a rocker at C. 6AN kN-m 33 Determine the horizontal reaction at B, in kilonewtons. (b) 4.000 (a) 3.333 (c) 6.000 (d) 6.667 34. Determine the vertical reaction at B, in kilonewtons. (b)4.000 (c) 6.000 (d) 6.667 (a) 3.333 Determine the horizontal reaction at C, in kilonewtons. (b) 4.000 35 (d) 6.667 (e) 6.000 (a) 3.333 Calculate the vertical deflection at A, in millimeters. E 200 GPa.I-90 X 10 mm (a) 15.4 36. (d) 41.0 (c) 34.2 (b) 18.7

Answer 1

FSy Fex A Fen 4 kN-m Horizontal re action at B (F): 33 0. EMC 6 + 0 - 6x4 t F8 = 20 B 20 11 Bn KN Ans 3.333 FBL (Fay) 24 Vertical reachan at B E M - G 4+ 4 + Fe F 3.333 kN. CY

Now moment about A, taking Fo By 6 +4 Fr.r F 3.333 x 6 +4 KN. F By Ams F Horizontal re achin od 3 5 This already cal culated jo the Previons Part KN. 3.3 33 Ans. Cn

A : VerHical deflection at 26 apply principle of virtual work Let KN A AB A J X - X - C . KN-m. Due to virtual load, Real Load case, first find horizontal reachom at Let us unit load at A. dne to virtnal B for virtual leading moment about c Taking Case, KBn

Now Let Moment at sechon X-x due to real load M = due to vistual Moment a sechon m Load Pay hian A B B C Origin A Limit o to O to 6 6 x 4+ F - 6 n M 24 + 3.333 -1x4 K Bn m1 - 1 n=- -4 + 24 + 3.333)(-4+- ) Mm +6n 2. 2.22 29.333 + 96 Verhical deflechon at A is Now M m dn. E 1 2-2.22 29.3331+96 6 2 dn. E 1

-*W .29.333*- 3 2 222 x 3 1 2. A E1 -(0-40.020-099pe) 1 +96 6 - 29.333x - + 2.222 E1 336 A E 1 Now, 10 N/mt 2=200 Pa 2.00 x 200 x l0 KN/mt - 6 9 0x l0 m4. I 9 0x 1o mm lo x9 O x lo- 200 10 KN m2 . EI kN-m 18000 336 : 4 A m 9810 O I 8.67 81 18.7 AA Ans