ENGI1331 Computing and Problem Solving for Engineers: Project 3 - Problem 3 Sample Calculations
ENGI 1331: Project 3 - Problem 3 Sample Calculations 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
−𝜃𝑥 + 𝑅𝑥3 , 𝑦(𝑥) = { 6𝐸𝐼
𝑥 ≤ 𝑎
𝑥>𝑎
Eq. 1
−𝜃𝑥+𝑅𝑥3−𝐹 (𝑥−𝑎)3, 6𝐸𝐼 6𝐸𝐼
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
𝐼 = 𝑤h3 12
𝑅 = 𝐹 (𝐿 − 𝑎) 𝐿
𝜃 = 𝐹𝑎 (2𝐿 − 𝑎)(𝐿 − 𝑎) 6𝐸𝐼𝐿
Eq. 2 Eq. 3 Eq. 4
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
𝜃 =
=3 [m]
=500 [N]
=10 [m]
𝐹𝑎 (2𝐿 − 𝑎)(𝐿 − 𝑎) = 6𝐸𝐼𝐿
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:
𝐼 = 𝑤h3 = 0.2∗0.23 = 1.33 ∗ 10−4 [m4] 12 12
𝑅= 𝐹(𝐿−𝑎)=500(10−3)=350[N] 𝐿 10
(500)(3) (17)(7) = 1.682 ∗ 10−5 [rad] 6(190∗109)(1.33∗10−4)(10)
For 𝑥 = 2 (𝑥 ≤ 𝑎), use the first half of Eq. 1:
𝑦(2) = −𝜃(2) + 𝑅(23) = −(1.682 ∗ 10−5)(2) + (350)(23) → −1.517 ∗ 10−5[m] 𝑜𝑟 ~0.015 [mm]
6𝐸𝐼 6(190∗109)(1.33∗10−4) For 𝑥 = 5 (𝑥 > 𝑎), use the second half of Eq. 1:
𝑦(2) = −𝜃(5) + 𝑅(53) − 𝐹 (𝑥 − 𝑎)3 = −(1.682 ∗ 10−5)(5) + (350)(53) − 500 (2)3 → 6𝐸𝐼 6𝐸𝐼 6(190∗109)(1.33∗10−4) 6(190∗109)(1.33∗10−4)
−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.
1
ENGI 1331: Project 3 - Problem 3 Sample Calculations
Case 2: Multiple forces
2
2
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.
