Draw the load and moment diagram of the beam with the shear diagram as shown: (KN) 2 +¹ 36 $ 1.6 16 2.4 20 -24 —x (m)(
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- -100N/m |20N- h ਸਾਧ P E थे Zm B Am C med Im IR₁ R Draw the shear and moment diagramGet the bending moment diagram. 8kN/m ΕΙ 4m B 10KN EI _|_2m| 2m 8kN/m EI 4m DProblem Ii A beam shown in Fiqure 1 is subected to loads as shown. 38 &N/m 59 N 120 AN.m 68 AN +150m+ -5.50 100 m- a. Dotormine the rosultant of force systom (magnitude and direction from x-axis). b. Compute the total moment about point B (include direction, ccw or ca) c Location of the resultant hrom poirit A
- 4KN 1. FROM THE GIVEN BEAM COMPUTE THE FOUOWING 4) REACNONS nT A DB 725KA/m b) SHEAR AND MOMONT AT D C) DRAN THE SHEAR AND MOMENT DIAGRAM 1-2m 1-8m O.so T ☺ t Q6) Compute the vertical displacement of joint B of the structure shown in the figure. 100 kN Hinge El, constant C 10KN 4 m مرّر للأعلى لاستعراض الفلاتر إضافة شرح. . . mum <4 5 10 kN 0000 20 IN 73 m izd Which of the following most nearly gives the force in member cd due to a downward vertical virtual unit load at e? OO -1 O-1.41 20 AN H (w3 min d 7-3m 1-3 m Which of the following most nearly gives the vertical displacement (mm) at e? O 80/AE O 113.14/AE O183.14/AE O 70/AE 10 KN
- 110% Problem 2: For the step pulley shown, solve for the resultant of the non-concurrent forces applied on it. 2875N 1000N 300 40cm 25cm 5000N :-A.) CREATE A SHEAR DIAGRAM BY AREA MOMENT AND BY PARTS B.) CREATE A M/EI DIAGRAM BY AREA MOMENT AND BY PARTS ITEM 1.0 180 kN 15 kN/m A В C- D -5 m- -5 m- -4 m- El = constant E = 70 GPa %3D I = 2,340 (10°) mm4 %3D10KN/m A В D 4m 2m 3m Draw Shear and Moment Diagram of the Beam
- What is the location of the Maximum moment from A in meters. Givern; w 18, a ad1,6-5. andc 1. w kN/m Ram B bm c m KA FETC 43 FS F6 s44 F7 WeWhich of the following gives the bending moment at point A on the beam shown? (kN m) 80 KN - m 90 KN ✓ 50 kN/m ✓ 4 A D Lamhamd 6 m- 75 KN 4 m. O zero 895 O 1000 375 ŽIN. B 75 kN m-UAB = 1 wac = ? *16-60 The slider block C moves at 8 m/s down the inclined groove. Determine the angular velocities of links AB and BC, at the instant shown. Cotohon around a fixed ³8 = WAR X PAB cand off = _WAB K x (2i) = -2WABj →general plane of motion → √₂ = vc + BCB/C - 2 m -2WABJ = 8 sinus 18 twBck * 25 FB/1C 45 A 45° TAB WAB # B ль BIC 2 m SUBC с 45 Vc = 8 m/s 8 $145 T