Intermani ENGINEERING Fina ae Programs-Freshman Student Name and ID: Fall 20; wed: 3 Hrs. GG The Exam i Statles 1/2 Consists of 0 Questions in Two Fagor ‘antmum Marks: 40 Marks se (MODEL D] - eee Nectar an OT ay os soph hele Sh apsSagh Sosa el Open Becket i Is as ge pid bn Nya aS hed i Ss Choose the correct answer in each of the followin} a EMz=0) of any If the general equilibrium conditions are (SFX=0, ZFy=0, DF2=0, ZMx=0, © “hitfon) in each case: object. Find the condition which is already satisfied (cancelled or useless cont 60N 200N Case (1): Rod supported by 2] [Case (2): Disk rests| [Case (3): Rod supported cables & ball-socket support | |on smooth planes | {by pin support Case (4): Plate supported by 3 cables| =i a i a 7 1) [Case1 EMy=0. 2) | Case 2 3) [ Case 3 4) | Case 4 The shown rod is in equilibrium at the shown po: support O and vertical force P, the magnitude of: a b c d 1826 | 2438 | 2384 | 319.6 128.6 | 2384 | 3916 | 1826 238.4 | 1286 | 1396 | 3916 5) Gi 7) The shown Uniform rod AB with mass m= 200 kg and length 7 m is in equilibrium with ball-socket joint at A and the end 8 rests on the shown smooth vertical planes, The x, y, z components of the next vectors are: a b ¢ d 8) [we | i962 = 200k 200k | ~a96% | 9) {Res [Zak | Xuevaltzak | Xie¥ah 10) | Re= | Xeit Yel Yes Yaj*Zak For Equilibrium, The magnitude of a b c 11) | Ras ..N | 2850.7 | 25807 2068.1 be 12)|Re=..N] 2568.1 2068.1 2750.7 Ca Scanned with CamScanner2/2 AIN: 7 a SHAMSUNIVERSiy, FACULTY OF ENGINEERING 019 Smational programs Freshmen Time Allowed:3 Hrs. Course code: PHMO3 MODEL D. ‘Statics The Exam Consists of 40 Questions in Two Pages, E, the The shown fi Magnitude o} “ame is in equilibrium by pin supports at A and D # force’s components at the next pin equal roller support at 13) | Xa= 56 42 14) [250 16) 17) 18) 19) 2\2/z/z\z/2/2/> The shown Truss is in equilibrium with pin support A and roller support G, next member's force is: a b ¢ | a 21) [Fox=. KN] 28.3C] 20€ | 283T{ 207 22) | Fu=-.kN| 20C | 28.37 | 28.3C 20T 23) | Focz.kN | 20c | 10¢ | 107 | 207 7a) | Fee=.kN | 207 [ 107 | 10¢ | 283¢ wing cases, 10 KN 20 kN if thi ‘e some forces are removed in the follo ber of zero force members in each case is: = el —___—__—-—7_@ | the num a 6 ¢ | [25) [removed 10 kN at E [6 5 4 None 26) | removed 10 kN at E, C 2 a_{ 8 None 37) [ removed 10 kN at E and 20 kNat 0 8 ole a3 None 1) of density 10 kg/ m* added to The shown shai .e (Part 2) of density 4 ki shaded plate (Part ded area consists of parabolic /m? with removed semicircular hole (Part 3) lower rectangular plat! a c 28) [ centroid of part 2: x1=..™ 2.25 18 39) | Centroid of part 1:y: 18 as 30) | Centroid of part 2: x2¥: =3,15 3,3 31) [ centroid of part 3:x3- 15 54 32) | Centroid of part 3: vs a 0 [-33) | Centroid of total shape: 5-07] 34,0 | 43,02 [-3a) | Center of mass of total shape: | 3.8,0.7| 3,0.4 | 5,-0.2 The second moment of area of the shapes equal: as s : 61.7 30.8 88.3 None 129.6 aol: 259.2 None 2 216 54 None ze 54 261 None 2 9.8 18 313 32 None 101.2 None Examination Committee END of Exam, Good Luck £xam. Date : 15/1/2029 r. Amol Abdel Hafez, Dr. Ahmed Ezzat, Dr. Hussein Elsayed Scanned with CamScanner