Recent content by yellowbird321

Hi! Thanks for the reply. However, if Q is 75 m3/hr and the flow is pumped through a 0,01 m diameter pipe then: A = PI D2 / 4 = PI x (0,01)2 / 4 = 7,85 x 10-5 m2 and u = (750 / 3600) x 1 / (7,85 x 10-5) = 2,65 x 10-8 m/s. With the numbers given in the exersice, how can I get 2 m/s.

Allright: What about this: Rod AB. ∑Fx = 0: Ax – T cos 60 = 0 ∑Fy = 0: Ay + sin 60 – 75 = 0 ∑Mx = 0: T sin 60 - 75 = 0 ∑Mx = 0: 75 = T sin 60 T = 75 / sin 60 T = 86,6 N∑Fx = 0...

Hi! Thanks for the input. You mean, instead of using Q = u Ac u = 4Q / PI p2 = 4 x 75 m3/h / PI x (0,01 m)2 = 0,0095 m/s, I should instead do this: A = PI D2 / 4 = PI x (0,01)2 / 4 = 7,85 x 10-2 m2 u = (75 / 3600) x 1 / (7,85 x 10-2) = 2,65 x 10-4 m/s.

Homework Statement Explain with basis in the figure how the necessary pumping effect is determined for a pump in a pipe system. We are given the following data: T = 15 C 75 m3/h vand H = 11 m L = 70 m rørledning D = 100 mm R = έ = 0,1 mm ή = 80 % 9 of 90 % standard elbows R/D =...

Hi! Thanks for the comments. I have attemped the following: Rod AB. ∑Fx = 0: Ax – T cos 60 = 0 ∑Fy = 0: Ay + sin 60 – 75 = 0 ∑Mx = 0: T sin 60 - 75 = 0 ∑Mx = 0: 75 = T sin...

Thanks for the comments. So we have: I = 22,2*106 mm4 h = 162 mm. M = -22.0 kN.m σ = - (Mr y / I) y = D/2. Distance from the centroid. Stress at the beams neutral axis is (the beam is symmetrical): σneutral = 0. We are asked for the normal stress distribution in the cross...

As far as I can see, this is the solution to the question: I = 22,2*106 mm4 D = 162 mm. M = -22.0 kN.m σ = - (Mr y / I) y = D/2.σneutral = 0. σtop = - (Mr y / I) = - (-22.0 kN.m) (81 mm) / 22,2 x 106 mm4 = 80,3 N/mm2 .σdown = - (Mr y / I) = - (-22.0 kN.m) (81 mm) / 22,2 x 106 mm4 =...

Hi! Thanks for your reply. That explained a lot, and I think that that I got it now. Then only part 2 remain. 2. Explain how one calculate size and distribution normal stress of the cut line. The beam is a I profile with moment of inerti I= 22,2*106 mm4 and height = 162 mm. We are given the...

Hi! Thanks for the comments. I have made the following: ∑Fy = 0: Ay + Cy – Dy = 0 → -20(2) + Cy(2 + 3) – 15(2 + 3 + 2.5) = 0 → Cy(5) = 40 + 112,5 = 0 → Cy = 30,5 kN ∑Fy = 0: Ay + Cy – By – Dy = 0 → Ay + 30,5 – 20 –...

Hi! Thanks for the reply. I thought I needed to multiply the force 15 kN with the distance 2,5 m.

Hi Chestermiller! Thanks for the reply. This might be the answer (at last!): ∑Fy = 0: -Fa + Vx + Dy = 0 -80 + Vx + 64,25 = 0 Vx = 15,75 kN ∑Fx = 0 Nx = 0. + ∑Ma = 0: Mx + Fb(2) – F (5) = 0 Mx + 80(2) – 37,5(5) = 0 Mc = 27,5 kN.

Hi PhantomJay! Thanks for your suggestion. I have followed your approach and made the following (I can currently not scan my hand drawings of the beam, but hope it make sense anyway): + ∑Ma = 0: -20(2) + RCy (5) – 37,5(5 + 2.5) = 0 → RCy (5) = 40 + 37,5(7,5) = 0 → RCy (5) =...

Homework Statement 1.Explain how one with the help of the free body diagram can calculate forces and moments on the vertical cut line through the beam in a distance 1 m to the right from the point C. 2. Explain how one calculate size and distribution normal stress of the cut line. The beam...

Hi! Thanks for the reply. So will this be the correct and complete solution to the problem? Rod AB. 75 = T sin 60 T = 75 / sin 60 T = 86,6 NAx = T cos 60 Ax = 86,6 cos 60 Ax = 43,3 NAy = 0 Rod BC. 75 = T sin 60 T = 75 / sin 60 T = 86,6 NBx = T cos 60 Bx = 86,6 cos 60 Bx = 43,3...

Homework Statement Determine the forces acting on the two rods in the figure. Explain which stresses that act on the rods and the bolts holding the rods to the brackets in point A and C. Homework Equations ∑Fx = 0 ∑Fy = 0 ∑Fz = 0...