strength of materialsDerive the formula for the maximum shear and maximum bending moments for Case 7. Shear and Moment FormulaL/2L/2MMPL348 EI5 WL3D.384 EICase ICase 2P PM. PLV - Pv . P23 PL3648 EIPL3D.3EICase 3Case 4

Answered: strength of materials Derive the…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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8:30 Test 3.pdf F1 F2 a CF3 A b a F4 D Consider the following values: - F5 a = 12 m; b = 10 m; c = 10 m; d = 8 m; F1 = 6 kN; F2 = 7 kN; F3 = 8 kN; F4 = 9 kN; F5 = 10 kN; a = 30°, 0= 45° , B = 60° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 111.6 kN.m b) 112.6 kN.m e) - 184.6 kN.m d) - 109.6 kN.m e) - 104.1 kN.m 0 None of them 21 What is the resultant moment of the five forces acting on the rod about point B? a) -95.8 KN.m b) 42.7 kN.m c) 80.9 kN.m d) 10.9 kN.m e) 51.8 kN.m ) None of them 31 What is the resultant moment of the five forces acting on the rod about point C? b) - 21.1 kN.m e) 17.4 kN.m d) - 16.5 kN.m e) 15.6 kN.m ) None of them a) 32.6 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 37.6 kN.m b) 73.8 kN.m c) 71.6 kN.m d) 52.1 kN.m e) 21.1 kN.m ) None of them 5] What is the moment of the force F2 about point E? a) 90.7 kN.m b) - 88.1 kN.m c) 54.6 kN.m d) 103.1 kN.m e) 100.7 kN.m ) None…
9-22. Solve Prob. 9-21 using Castigliano's theorem. Internal Moment Function M(x): The internal moment function in terms of the load P' and couple moment M' and externally applied load are shown on figures (a) and (b), respectively. Castigliano's Second Theorem: The displacement at C can be determined aM(x) ap' aM dx [M(M) +/G-04] ↓ with PL³ 48E1 To determine the slope at B, with setting M = 0. and set P = P. EI PL² 16EI OM dx OM EL ~ ) ) » (¹-2) & aM(x₂) X₁ M(x₂) aM' LaM Ans. and Ans. 12 M MOC+ PMOC-M²+(£- #%₂ Ti *44 6 244 aft 214
1. - solve the nodal displacements and reaction forces: Al 20 kN/m 20 kN/m 20 kN/m 20 kN/m 5 kN ww Nom
Use the graphical method to construct the bending-moment diagram and identify the magnitude of the largest moment (consider both positive and negative). The ground reactions and the shear-force diagram are provided. M = 13 kN-m Ay = 3.46 KN Dy Units: KN = 32.54 kN M 4 m 29.5 kN-m 43.3 kN-m 25.8 kN-m O 22.9 kN-m 31.48 kN-m B Ay 4 m 15 kN C Ay 4 m Ay-15 7 kN/m D 3 m 21 Ay-15 E X
42 F=Fcos ex lok 14k = Fsin By Fy= Fsin Ox = Fuos y F - angles 12k 15k 8k Find Re = 7k TF (hye slopes #3/ 4 tan = 15 16 1316 3 IN 1016 48 63 ISK hyp √√13 = 3.61 Fx-156 Fy = 15 ( 2³ 41 ) = 2516 7th 3lb Find R+O
C8-49 A 500-lb load is supported by a roller on a beam as shown in Fig. Plot |Mmax, the maximum bending moment in the beam, as a function of the roller position b (0 sbs 25 ft). -25 ft- A B 500 lb
In the fig. shown, compute the ff: (11-13) the tension in cable BD 7.5 rt To = (14-15) the components of the support 10 rt At C. Ch Cv = 5 rt 1000 lb 12.5 ft S00 I In the fig. shown, compute the ff: (16-18) the resultant using cosine law (force polygon) 60 R = 35 (19-20) the angle of the R measured CW from the x- axis. 500 lb
am of Mechanics/Firs X E Monthly Exam of Mechanics/Firs x PQLS ZSh4aZJdw9EBqROOfLVx M7KUGF6 9bgBbd-YGOXA/formResponse * For the force shown in figure, find the resultant (R) by component method F=10 kN 0=30° R=5 kN R=10 KN O R=15 kN O R=60 KN For the forces shown in figure, find just (EFx) components F=4 kN F=2 kN 0-30 A=300
750-10UNC-24 100 0.600 4.3875 THRU 2776-2 CRS M BAM 756 -178- 1-8UNC-2A 1.25 LG GENERA S Rockwell STEM TEST MODEL A-2776-2 5-6
:Q81 The moment M for the force the figure shown below is 250 N 153 -200 mm 30 mm 46.5 N.M O 46.6 N.M O 46.4N. M of L is
The linkage shown is used in a vehicle suspension system. Find the forces indicated below when a static force of F = 820 lb is applied to the tire at point C at an angle of 0 = 22° as shown. Assume the connection of the wheel to member AB is rigid. hs hy hi F BY NC SA W4. W, Ws 2013 Michael Swanbom Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value Variable Value hi 10 in Wi 8.8 in h2 15.6 in W2 5.6 in h3 18.1 in W3 10.8 in h4 10.9 in WĄ 7.4 in h5 4.9 in W5 9.3 in For all answers, take x as positive to the right and y as positive upward. At point B, member AB exerts a force of B = j lbs on member BHE. At point A, member AD exerts a force of A + j lbs on member AB. At point H, shock GH exerts a force of H = i + i lbs on member ВНЕ. At point E, the magnitude of the force exerted on the car body by member BHE is lbs. At point D, the magnitude of the force exerted on the car body by member AD is lbs. At point G, the…
8:44 6 A m • O Zoom v Leave DA bar is as shown ig. and is subjected do axial Jensile lood of 5 KN. And the totalelongation am Power Urted Pege Ue Pg Produc ed im the bas Est = 200 ap a, GAL=70GPA Am- 200 Ast- 50 mm mm? 5KN 5 KN .5 m - +1.8m 30 Unmute Start Video Share Participants More II

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strength of materials Derive the formula for the maximum shear and maximum bending moments for Case 7.