To close a gate valve, it is necessary to exert two forces of 6kN at opposite sides of a handwheel, 1m in diameter. Through an accident, the wheel is broken and the valve must be closed by thrusting a bar through a slot in the valve stem and exerting a force at 1.2m out from the center. Determine the force required and draw a free body diagram of the bar.

Q: I KN Im 2-50 KN /m ROUER B HINGE 2.25m 0.757, 1m 5.0 m FIGURE 003

A: GIVEN BEAM

Q: determine the magnitude and direction of the force reaction at support C

A: To determine the support reactions at C.   To determine the support reactions at C, the free body…

Q: 2. A wheel of 10-in radius carries a load of 1000 Ib, as shown in Fig. P-324. (a) Determine the…

A: Given data,  Wheel radius = 10 in Load = 1000 lb Block height = 5 in   To find out the horizontal P…

Q: 75 KN/M 75KN/m A W 3.5m B 3.5m 35KN/m с 35 KN/M T om am 2m

A: In triangle BDO, BD=32+3.52=4.61 m sinθ=3/24.61/2=0.651 cosθ=3.5/24.61/2=0.759

Q: Q2) Three forces act on the piping. Determine the sum of the moments of the three forces about point…

A:

Q: 1. Determine the value of reactions at the supports A & B. see Figure1. Figure1 15 AN/m AA 2m 1-5m…

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Q: 2. Calculate the support reaction for the beam shown. 60 KN A -2m -4m 20 kN/m E 'm C

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Q: Determine the reactions and level of determinancy

A:

Q: Determine the Reactions.

A: Degree of static determinacy, Ds = 0  (3 -3) thus, reactions can be found out by using only…

Q: 10m B 100 KN 1000 KU-m

A: The magnitude of the point load The magnitude of the concentrated moment We have to determine the…

Q: Determine the value of Support Reaction B 400 N/m 1m t 3m 200 N 2m +

A: Given Data: UVL= 400N/m Point load = 200N

Q: . Determine the magnitude of the reaction in the structure below due to loading. 3KN/M 37° A (m 4.m…

A: Given  inclined load  8 KN at a 37-degree angle from horizontal  Horizontal component of 8 KN force…

Q: 30 kN/ m A C 6 m 5 m

A:

Q: Compute the support reaction

A:

Q: The three Portal Structure Itself is focused on joints A and D and connected with hinges S as seen…

A:

Q: Two beams at 6 meters in length each are arranged so that the reactions at A, B, and D are the…

A: Given data  Ay = By = Dy (equal reaction at A, B and C)  Length of each member (AB and CD ) = 6m…

Q: 90 k 3 20 ft- 25 ft 75 k 3 20 ft

A:

Q: Calculate the reactions in the given beam 15 kN 4.8m 3.4m 67 7.2m B 9 kN/m Vertical Reaction on B,…

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Q: For the continuous bcam shown dotermine the reactions over the supports

A: As per the Fixed end moments formula, the diagram can be modified as shown in figure below

Q: Problem No. 2: Determine the horizontal and vertical components of reaction at the supports A and C.…

A: Please note that the Force F BC will be at slope of 3 - 4 - 5

Q: The simply supported beam shown is supported by a pin at  A and a roller at C. Draw a free body…

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Q: Problem 8: Replace the loading by an equivalent resultant force and couple moment acting at point 0.…

A:

Q: 18 Klft A 20 ft B 22 f+ 12 ft * Determine the reactions at the sup ports for the Structure Shown…

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Q: 20kN/m 50KN 10KN/m 2n 3m F3 F2 30KN/m 4m 1.5m F1

A: In UVlLoad=wL2=30×32=45kNIn UDl1Load=wL=20×4=80kNIn UDL2Load=wL=10×2.5=25kNLet x be distance of…

Q: Calculate the reactions in the given beam 15 kN 4.8m 3.4m 7.2m A В 9 kN/m Vertical Reaction on B, kN…

A:

Q: 90 kips 60 kips 6 kips/ft C D E F 6 ft 6 ft 6 ft 6 ft 12 ft 6 ft

A:

Q: 5 kN/m 150 kN-m B 5m 15m

A: Find the moment at A

Q: Determine the magnitude of the reaction acting at the support.

A: Given: A simply supported beam carrying a uniformly varying load (UVL). Magnitude of UVL at support…

Q: w = wo + kx². 4 kN 6 kN/m w 0.5 m = kN/m Bx Ay By 0.6 m 1.4 m 0.4 m 0.6 m

A: To find support reaction

Q: 5. The solid shaft that is fixed connected at both ends has a diameter of 20 mm and a total length…

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Q: calculate the reactions using static equilibrium Please show all steps + FBD 1.0 kN/m 7 kN B 3.0 m

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Q: 350KN 18KN/m 36KN/m 25 kN/m 3.00 meters 1.80 meters 3.50 meters en

A:

Q: The system below shows a solid shaft composed of 2 circular cross sections (H15 mm.r30 mm). It is…

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Q: 5 kN 20 kN/m B C A D 0000 -2 m² 3 m 1 m

A:

Q: F 450 N +1.5 m 25m 2.0 m B 60° D 1.5 m 3.0 m 100 N 85 N + 1.5 m +

A:

Q: 22.4 KN 20 kN 6 m THalino MATHalino com 6 m 6 m 6 m 6 m 10 kN 20 kN 30 KN

A: find the perpendicular distance for the force 22.4 KN  and then use equilibrium conditions to find…

Open Channel Flow

Open channel flow is a branch of fluid mechanics. It deals with the flow of fluid in an open conduit or free surface. In an open channel flow, gravity induces the flow of water whose surface is open to the air. This flow is free-falling rather than confined in rigid body conduits. The open channel flow depends upon two parameters that are the cross-section of the conduit and the velocity profile. Practical examples of open channel flow are streams, rivers, canals, and other drainage systems.In open channel flow, the cross-sectional area of at least one of the sides is exposed to air, and the surface of water faces atmospheric pressure. This flow is also known as free gravity flow. The cross-sectional area of the conduit can be either man-made or a natural channel. For instance, the flow in rivers and streams are natural conduits, and the flow in a canal and sanitary systems are man-made or engineered channels.