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FromTheMoon [43]
1 year ago
14

A rectangular beam 200 mm deep and 300 mm wide is simply supported over a span of 8 m. What uniformly distributed load per metre

the beam may carry, if the bending stress is not to exceed 120N/mm2.
Physics
1 answer:
Alexandra [31]1 year ago
5 0

The uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

<h3>What is bending stress?</h3>

When an object is subjected to a heavy load at a specific spot, it often experiences bending stress, which causes the object to bend and tire.

The given data in the problem is;  

Bending stress, σ = 120 N/mm2  

Moment of inertia, I = 8.5 × 106 mm⁴

Depth of beam, y = d/2 = 200/2 = 100 mm  

Length of beam, L = 8 m = 8000 mm  

Width of beam, W = 300 mm

The maximum bending moment of the beam with UDL;

\rm W = \frac{wL^2}{8}

From the bending equation;

\rm M = \frac{\sigma I}{y_{max}} \\\\ M = \frac{120 \ N / mm^2 \times 8.5 \times 10^6 }{100 \ mm } \\\\ M = 10.2 \times 10^6 \ N - mm

The maximum bending moment of the beam with UDL;

\rm M = \frac{wL^2}{8} \\\\  10.2 \times 10^6 = \frac{w \times 8^2}{8} \\\\ w = 1.275 \times 10^6 \  N/mm

Hence the uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

To learn more about the bending stress, refer;

brainly.com/question/24227487

#SPJ1

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3 years ago
(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en
V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

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height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

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2 years ago
A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un
Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

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G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

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3 years ago
a radio controlled car rolls 10 m south then reverses direction and rolls 8 m north the car traveled a distance of 18 m and has
EastWind [94]

The displacement is 2 m south

Explanation:

Distance and displacement are two different quantities:

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For the car in this problem, the motion is:

10 m south

8 m north

Taking north as positive direction, we can describe the two parts of the motion as

d_1 = -10 m

d_2 = +8 m

Therefore,  the final position of the car with respect to the original position is

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Learn more about  distance and displacement:

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