The Question is in the image below along diagrams. The missing values of \" á\"

Question

The Question is in the image below along diagrams.
The missing values of " á" is "18" which is "serial number of student". The value of "á" is missing at 3 spots in the question, so kindly take value of a=18.

I already got a wrong and incomplete solution here. So please solve it completely according to the instructions of question. I shall be thankful to you.

Homework No 1 Mobile flood control wall consists of horizontal beams which slip into vertical posts (see figures). Determine the maximum hydrostatic force acting on a beam (the down- most one - this is the force for which all the beams must be designed as they are interchangeable) and the total hydrostatic force for which the vertical posts are to be designed Also determine minimum mass of the horizontal beams for which they will not flow up (do not account for the friction in slots of the posts) 60-? 30 ml (a [m], their height is c-150 + 2 . ammi The depth of water is h serial No of the student). Length (span) of the horizontal beams is b = 2+- VIEW B SECTION A-A VERTICAL SUPPORT HORIZONTAL BLOCK ELEMENT DETAIL A DETAIL A 15 35 20 35 JOINTING 15 15 HORIZONTAL BLOCK ELEMENT

Explanation / Answer

Depth of water, h = (60 - 18)/30 = 1.4 m

Length of beam, b = 2 + (18/10) = 3.8 m

Here the hydrostatic pressure variation will be triangular

So total pressure force = (1/2) × gamma × H2 × b

= 0.5 × 9810 × 1.42 × 3.8 = 36532.44 N

It is acting at (2/3)h from top but for bottom most beam we have to take entire h. So force acting at a distance of h from top will be calculated by linear interpolation

For (2/3)h ?36532.44 N

h ? x

X = 54798.68 N it is max hydrostatic force acting on beam

For hydrostatic force for vertical support we have to take unit width (b = 1 m)

Hydrostatic force = 0.5 × gamma × h2 × b

= 0.5 × 9810 × 1.42 ×1 = 9613.8 N

Act at a distance of (2/3)h from top

To find the min mass of beam we have to dicide the hydrostatic force of beam by g value

F = W = mg

m = W/g = (54798.68/9.81) = 5586 kgs

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