ENGI 1331代做、代寫R程序語言
時間：2024-04-28 來源： 作者： 我要糾錯
ENGI 1331: Project 3 - Problem 3 Sample Calculations
1
Given:
A simply supported beam of length L subject to a force F. The deflection of
the beam y is characterized by the deflection equation
where I is the moment of inertia of the cross section, R is the reaction force on the beam at the left end, 𝜃 is the
clockwise rotational angle of the beam at the left end, and E is the Young’s modulus of the beam’s material. I, R,
𝜃 and E are all geometry- or material-based constants. E is found in a table (MaterialElasticity.mat), and the
other values are found with the following (already derived) equations
Since all the properties are either given or can be directly calculated, to find deflection at a point simply find the
coefficients I, R, 𝜃 and E and then plug them into Equation 1.
Case 1: Single force
For the beam shown with width w = 0.2 [m], height h = 0.2 [m], and modulus
E = 190 * 10^9 [Pa], calculate the beam deflection at x = 2 [m] and x = 5 [m].
First, calculate the constants:
(2𝐿 − 𝑎)(𝐿 − 𝑎) =
(500)(3)
6(190∗109)(1.33∗10−4)(10)
(17)(7) = 1.682 ∗ 10−5
[rad]
For 𝑥 = 2 (𝑥 ≤ 𝑎), use the first half of Eq. 1:
𝑦(2) = −𝜃(2) +
𝑅(2
3)
6𝐸𝐼
= −(1.682 ∗ 10−5
)(2) +
(350)(2
3)
6(190∗109)(1.33∗10−4)
→ −1.517 ∗ 10−5
[m] 𝑜𝑟 ~0.015 [mm]
For 𝑥 = 5 (𝑥 > 𝑎), use the second half of Eq. 1:
𝑦(2) = −𝜃(5) +
𝑅(5
3
)
6𝐸𝐼
−
𝐹
6𝐸𝐼
(𝑥 − 𝑎)
3 = −(1.682 ∗ 10−5
)(5) +
(350)(5
3
)
6(190∗109)(1.33∗10−4)
−
500
6(190∗109)(1.33∗10−4)
(2)
3 →
−1.781 ∗ 10−4
[m] 𝑜𝑟 ~0.18 [mm]
This application can be generalized with x as a vector instead of a single value to find the deflection at all points
on the beam.
=3 [m]
=500 [N]
=10 [m]
ENGI 1331: Project 3 - Problem 3 Sample Calculations
2
Case 2: Multiple forces
Similar case, but with more than one force. Simply treat the problem as
two instances of Case 1. Calculate the deflection caused by force F at
distance a, then calculate the deflection caused by force F2 at distance a2.
The sum of those deflections will be the total deflection across the beam.
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