Using the simply supported beam from Prob. 8.24, the moment along the beam,
Use a numerical method to find the point
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- F1 F2 a F3 A F4 F5 Consider the following values: - a = 25 m; b = F1 = 7 kN; F2 = 8 kN; F3 = 9 kN; F4 = 10 kN; F5 = 11 kN; 5 m; c 25 m; d =5 m; a = 30° , 8 = 60° , B = 30° Tor a) ou. . 4] What is the resultant moment of the five forces acting on the rod about point D? a) - 117.8 kN.m b) 82.7 kN.m c) 121.9 kN.m d) - 259.9 kN.m e) 127.1 kN.m f) 300.1 kN.m Activate 5] What is the moment of the force F2 about point E? a) 151.7 kN.m b) 118.1 kN.m c) 214.6 kN.m d) 158.6 kN.m e) 101.1 kN.m boarrow_forwardPlease do this carefully.arrow_forwardO 1001, 1:16 docs.google.com/forms 37 M=144 Ib.in The lower lumbar region A of the spine is the part of the spinal column most susceptible to abuse while resisting exco by the moment about A of a force F =10KN. For given values of F, b = 50cm , and h = 10 cm, determine the moment at point (A) from the following Answers which of them is bending caused correct .if the angle 0 = 0° : O M=0 O M=500 N.m M=500000 N.m O M=100 N.m O M=100000 N.m Page 3 of 3 Вack Submitarrow_forward
- 3 A) A(force vector)= A(x)i + A(y)j +A(z)k and z component of Moment (by support A) is zeroB) A(force vector)= A(y)j +A(z)k , A(x)=0 and z component of Moment (by support of A) is zeroC) A(force vector)= A(x)i + A(z)k, A(y)=0 and x component of Moment (by support A) is zeroD) A(force vector)= A(x)i + A(y)j, A(z)=0 and x component of Moment (by support A) is zeroE) A(force vector)= A(x)i + A(y)j +A(z)k and y component of Moment (by support A) is zeroarrow_forwardF1 F2 ba C F3 A B d. b F4 Consider the following values: - F5 a = 7 m; b = 3 m; c = 7 m; d = 3 m; F1 = 3 kN; F2 = 11 kN; F3 = 2 kN; F4 = 20 kN; F5 = 2 kN; a = 30° e = 60° , B = 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) 90.8 kN.m b) 72.7 kN.m c) 14.6 kN.m d) 55.9 kN.m e) 21.6 kN.m f) 61.3 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 77.8 kN.m b) 25.7 kN.m c) 38.6 kN.m d) 18.9 kN.m e) 37.9 kN.m f) 84.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 35.7 kN.m b) 21.1 kN.m c) 714.6 kN.m d) 69.1 kN.m e) 51.4 kN.m f) 85.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 17.8 kN.m b) 55.7 kN.m c) 44.6 kN.m d) 18.9 kN.m e) 21.5 kN.m f) 30.1 kN.m 5] What is the moment of the force F2 about point E? a) 81.7 kN.m b) 78.1 kN.m c) 64.6 kN.m d) 73.1 kN.m e) 54.6 kN.m f) 61.3 kN.marrow_forwardF1 F2 a F3 A d. F4 Consider the following values: F5 a = 25 m;b = 5 m; c = 25 m; d = 5 m; F1 = 3 kN; F2 = 11 kN; F3 = 2 kN; F4 = 20 kN; F5 = 2 kN; a = 30°, 0 = 60° , B = 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 15.8 kN.m b) 12.7 kN.m c) - 74.6 kN.m d) 52.9 kN.m e) 24.1 kN.m f)- 47.1 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 8.8 kN.m b) 2.7 kN.m c) 31.6 kN.m d) 8.9 kN.m e) 5.4 kN.m ) 24.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 86.7 kN.m b) 51.1 kN.m c) 14.6 kN.m d) 27.9 kN.m e) 74.1 kN.m ) 35.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) - 17.8 kN.m b) 82.7 kN.m c) - 21.9 kN.m d) 36.9 kN.m e) - 27.1 kN.m f) 30.1 kN.m Activate V 5] What is the moment of the force F2 about point E? Go to Setting ) 321.3 kN.m a) 111.7 kN.m b) 218.1 kN.m c) 254.6 kN.m d) 93.1 kN.m e) 101.1 kN.m boarrow_forward
- F1 F2 a C F3 A d b F4 E o Consider the following values: - F5 a = 25 m; b = 5 m; c = 25 m; d = 5 m; F1 = 4 kN; F2 = 2 kN; F3 = 5 kN; F4 = 7 kN; F5 = 3 kN; a = 30° , 0= 60° , B 30° 2] What is the resultant moment of the five forces acting on the rod about point B?arrow_forwardThe figure below shows a moving vehicle represented by two moving loads. The two loads are x = 2 m apart and they are to pass across a simple beam of length L = 13 m. If P1 = 46 kN and P2 = 93 kN, how far will the car travel when the moment induced is maximum. Initial position: P1 X P2 Position at maximum moment: Simply put, what is the distance D (m) when the moment is maximum? D P1 L P2 +arrow_forwardFind the moment on intertia of the area bounded by the parabola y^2=4x and the line x=2 about the x-axis. Answer: 12.07 units^4arrow_forward
- 2 / 5 78% + 1-2 2) Find the magnitude and direction of the resultant force and moment at point A 250lb 150lb A JA ++ 350lb 400lb 1" 1 \150lb FORCE e F, =F.Cos(e) F, =F.Sin( e) (Ib) 5M, =Fxd FORCES COUPLES R= R = M= +M, -M, RESULTANT FORCE AND MOMENT DIAGRAM R= OR = ++ M.= -Y AY 6. étv A Aaarrow_forwardNeed help pleasearrow_forwardNeed neat and clean handwritten solution, I repeat need handwritten solution. I will upvote if correct solution is provided. THANKS ?arrow_forward
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