The moment M that will produce a maximum stress of 70 MPa on the cross-section of 177.3 will be 13.533 kN-m.
<h3>What is bending stress?</h3>
Bending stress is the typical stress that an item experiences when it is exposed to a heavy load at a specific spot, causing it to bent and fatigue.
The moment of inertia above the neutral axis will be
Iₓₓ = 99 x 84.7³/3 - (99 - 12 x 3) (84.7 - 12)³/3
Iₓₓ = 11983247 mm⁴
The moment of inertia below the neutral axis will be
Iₓₓ = 2 x 12 x (87 - 84.7)³/3 + 12 x 177.3³/3
Iₓₓ = 22294005 mm⁴
Then the moment of inertia about the neutral axis will be
Iₓₓ = + 11983247 + 22294005
Iₓₓ = 34277252 mm⁴
Then for maximum bending stress, we have
M = σ x Iₓₓ / y(max)
We have
σ = 70 MPa
Iₓₓ = 34277252 mm⁴
y(max) = 177.3
Then we have
M = 70 x 34277252 / 177.3
M = 13.533 kN-m
More about the bending stress link is given below.
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Hope this helps you out :D
It’s 2 equations so in order to solve the previous system, you can use different methods, as for example substitution or addition of equations. In this case, you use the second one, due to the fact you have 7x in one equation and -7x in the other equation. In this way you can easily eliminate variable x and then solve for y. With the value of y you can replace in any of the two equations and solve for x.
7x-y=-1
-7x+3y=-25
Summarizing, you proceed as follow:
- add up the given equations
7x - y = -1
-7x+3y=-25
——————
0 +2y=-26
- solve for y in the previous equation
2y=-26
y=-26/2
y=-13
- replace the obtained value of y in one of the given equations, and solve for x
7x-(-13)=-1
7x+13=-1
7x=-1-13
7x=-14
x=-14
x=-14/7
x=-2
Hence, the solution of the given systems of equation is:
X=-2
Y=-13
You can divide the field into two squares of area 2500 yd^2. The side of each of those will be √2500 = 50 yd.
The dimensions of the field are 50 yd by 100 yd.
Answer:
Step-by-step explanation:
We are told the school sold raffle tickets, and each ticket has a digit either 1, 2, or 3. The school also sold 2 tickets with the number 000.
Therefore we have the following raffle tickets:
123
132
213
231
312
321
000
000
From the given information, we can deduce that the school sold 8 tickets and only one ticket can contain the number arrangement of 123, but 000 appeared twice.
Probability of 123 to be picked=
1/8 => 0.125
Probability of 000 to be picked=
2/8 => 0.25
Since the probability of 000 to be picked is greater than 123, a ticket number of 000 is more likely to be picked
