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    Plate No.6 - Solution

    Uploaded by

    Billie Ian. Salamante Jr
    1. The hydrostatic force acting on the 1m wide gate submerged 2m in water is 19.62 kN. 2. The center of pressure of the gate is located 0.667m above the hinge. 3. A minimum of 0.316m3 of concrete with a unit weight of 23.6 kN/m3 is needed to keep the gate closed.

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    Plate No.6 - Solution

    Uploaded by

    Billie Ian. Salamante Jr
    199 views6 pages
    1. The hydrostatic force acting on the 1m wide gate submerged 2m in water is 19.62 kN. 2. The center of pressure of the gate is located 0.667m above the hinge. 3. A minimum of 0.316m3 of concrete with a unit weight of 23.6 kN/m3 is needed to keep the gate closed.

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    1. The hydrostatic force acting on the 1m wide gate submerged 2m in water is 19.62 kN. 2. The center of pressure of the gate is located 0.667m above the hinge. 3. A minimum of 0.316m3 of concrete with a unit weight of 23.6 kN/m3 is needed to keep the gate closed.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    199 views6 pages

    Plate No.6 - Solution

    Uploaded by

    Billie Ian. Salamante Jr
    1. The hydrostatic force acting on the 1m wide gate submerged 2m in water is 19.62 kN. 2. The center of pressure of the gate is located 0.667m above the hinge. 3. A minimum of 0.316m3 of concrete with a unit weight of 23.6 kN/m3 is needed to keep the gate closed.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 6

    ENGR.

    BON RYAN ANIBAN


    𝑆𝑈𝑀𝑀𝐴𝑅𝑌
    1. 𝑊𝐿 = 8.489 kN
    𝑊𝐿 = 7.732 kN

    2. ℎ = 0.917 ft

    3. 𝜃 = 19.471°

    4. 𝐹 = 19.62 kN
    1
    𝑦= 2 = 0.667 m
    3
    𝑉𝑐𝑜𝑛𝑐 = 0.316 m3
    𝑃𝑅𝑂𝐵𝐿𝐸𝑀 1
    A hollow cylinder 1m in diameter and 2 m high weighs 3825 N.
    (a) How many kN of lead weighing 110 kN/m3 must be fastened to the outside bottom of the cylinder to make it
    float with 1.5 m submerged in water?
    (b) How many kN of lead if it is placed inside the cylinder?
    Solution
    1m
    (a) Lead is fastened outside (b) Lead is fastened inside the cylinder
    𝐵𝐹𝐶 = 𝛾𝑤𝑉𝐷 0.5 m WC [Σ𝐹𝑉 = 0] 1m

    𝐵𝐹𝐶 = 9.81(𝜋/4)(12)(1.5) 𝐵𝐹 = 𝑊𝐶 + 𝑊𝐿
    2m
    𝐵𝐹𝐶 = 11.557 kN 1.5 m 11.557 = 3.825 + 𝑊𝐿 0.5 m WC
    BFC
    𝑊𝐿 = 7.732 kN
    2m
    𝐵𝐹𝐿 = 𝛾𝑤𝑉𝐿 1.5 m
    𝐵𝐹𝐿 = 9.81𝑉𝐿
    WL
    𝑊𝐿 = 𝛾𝑤𝑉𝐿 = 110𝑉𝐿
    WL

    BFL
    [Σ𝐹𝑉 = 0] BF
    𝐵𝐹𝐶 + 𝐵𝐹𝐿 = 𝑊𝐶 + 𝑊𝐿
    11.557 + 9.81𝑉𝐿 = 3.825 + 110𝑉𝐿
    𝑉𝐿 = 0.077 m3

    𝑊𝐿 = 8.489 kN
    𝑃𝑅𝑂𝐵𝐿𝐸𝑀 2
    A cube 2.2 ft on an edge has its lower on an edge has its lower half of sg = 1.6 and upper half of sg = 0.7. It rests
    in a two-layer fluid, with lower sg = 1.4 and upper sg = 0.8. Determine the height h of the top of the cube above
    the interface.
    Solution
    sg = 0.8
    𝐵𝐹 = 62.4 × 1.4 2.2 2 2.2 − ℎ + 62.4 × 0.8 [ 2.2 2 ℎ ]
    𝐵𝐹 = 62.4 × 2.22 (3.08 − 1.4ℎ + 0.8ℎ)
    sg = 0.7
    𝑊 = 62.4 × 1.6 [ 2.2 2 1.1 ] + 62.4 × 0.7 [ 2.2 2 1.1 ] h
    𝑊 = 62.4 × 2.22 (2.53) 2.2 ft

    sg = 1.6
    𝐵𝐹 = 𝑊
    62.4 × 2.22 (3.08 − 1.4ℎ + 0.8ℎ) = 62.4 × 2.22 (2.53)
    sg = 1.4 2.2 ft
    ℎ = 0.917 ft
    𝑃𝑅𝑂𝐵𝐿𝐸𝑀 3
    A wooden beam of specific gravity 0.64 is 150 mm by 150 mm and is hinged at A, as shown in the figure. At what
    angle ϴ will the beam float in water?
    Solution
    A
    𝑊 = 𝛾𝑏𝑒𝑎𝑚𝑉𝑏𝑒𝑎𝑚
    𝑊 = 9810 × 0.64 0.15 × 0.15 × 5 = 706.32 N
    1m
    𝐵𝐹 = 𝛾𝑤𝑉𝐷
    𝐵𝐹 = 9810 0.15 × 0.15 × 𝑥 = 220.725𝑥 N ϴ

    Water
    Σ𝑀𝐴 = 0
    𝑥
    𝐵𝐹 + 5−𝑥 cos 𝜃 − 𝑊(2.5 cos 𝜃) = 0
    2

    𝑥 A
    𝐵𝐹 + 5−𝑥 cos 𝜃 = 𝑊(2.5 cos 𝜃)
    2
    𝑥
    220.725𝑥 + 5−𝑥 = 706.32(2.5) 1m
    2
    𝑥 =2m ϴ
    1 1 Water
    sin 𝜃 = = 𝜃 = 19.471°
    5−𝑥 5−2 W

    BF
    𝑃𝑅𝑂𝐵𝐿𝐸𝑀 4
    From the figure below, it is shown that the gate is 1.0 m wide and is hinge at the bottom of the gate. Compute the
    following:
    (a) the hydrostatic force in kN acting on the gate
    (b) the location of the center of pressure of the gate from the hinge,
    (c) the minimum volume of concrete (unit weight = 23.6 kN/m3) needed to keep the gate closed position.

    Solution
    Pulley
    ത = 9.81(1)(2 × 1)
    𝐹 = 𝛾ℎ𝐴
    𝐹 = 19.62 kN
    1m
    1
    𝑦 = 2 = 0.667 m T
    3

    W Submerged
    Σ𝑀𝐴 = 0 Σ𝐹𝑦 = 0 concrete
    𝐹𝑦 = 𝑇(3) 𝑇 + 𝐵𝐹 = 𝑊 2m block

    (19.62)(0.667) = 𝑇(3) 4.36 + 9.81𝑉𝑐𝑜𝑛𝑐 = 23.6𝑉𝑐𝑜𝑛𝑐


    𝑇 = 4.360 kN 𝑉𝑐𝑜𝑛𝑐 = 0.316 m3 A

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