Determine the maximum intensity ww of the uniform distributed load that can be applied to the beam without risk of causing the s
trut to buckle.
Take F.S.F.S.x = 2.1 against buckling.
Take F.S.F.S.x = 2.1 against buckling.
1 answer:
You might be interested in
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.
Data Given:
x = 27 , 44 , 32 , 47, 23 , 40, 34, 52
y = 30, 19, 24, 13 , 29, 19, 21, 14
It is given that,
∑x = 299
∑y = 167
∑ = 11887
∑ = 3773
We are asked to verify the above values manually in this question.
So,
1. ∑x = 299
Let's verify it:
∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52
∑x = 299
Yes, it is equal to the given value. Hence, verified.
2. ∑y = 167
Let's verify it:
∑y = 30 + 19 + 24 + 13 + 29 + 19 + 21 + 14
∑y = 169
No, it is not equal to the given value.
3. ∑ = 11887
Let's verify it:
For this to find, first we need to square all the value of x individually and then add them together to verify.
∑ =
+
+
+
+
+
+
+
∑ = 11,887
Yes, it is equal to the given value. Hence, verified.
4. ∑ = 3773
Let's verify it:
Again, for this we need to find the squares of all the y values and then add them together to verify it.
∑ =
+
+
+
+
+
+
+
∑ = 3,845
No, it is not equal to the given value.