5.4. Examples

Example (1):

For the beam shown in the figure below, calculate the deflection of the beam at the mid-span.

Given: E = 200GPa, I = 200×106mm4

Solution:

1m w = 5kN/m = 0.005kN/mm, L = 5m = 5000mm, E = 200GPa = 200kN/mm2

$\Delta = \frac{5wL^4}{384EI} = \frac{5(0.005kN/mm)(5000mm)^4}{384(200kN/mm^2)(200×10^6mm^4)} = 1.017 \text{ mm}$

In this example, we are calculating the deflection of a beam at its mid-span under a uniform load. We have been given specific values for the modulus of elasticity (E) and moment of inertia (I), which are important material properties. The formula used is derived from beam theory and helps in determining how much the beam will bend under the load applied. By substituting the given values into the formula, we calculate the total deflection, resulting in a deflection of approximately 1.017 mm.

Imagine a long, flexible diving board. When someone jumps on it at the center, it bends down. The amount it bends is similar to the deflection we calculated – it depends on how heavy the person is (load), how rigid the board is (modulus of elasticity, or E), and the thickness of the board (moment of inertia, or I).

Example (2):

For the beam shown in the figure below, calculate the deflection of the beam at the free end.

Given: E = 90GPa, I = 100×106mm4

Solution:

P = 10kN, P = 20kN, l = 6m = 6000mm, I = 100×106mm4

x = 2m = 2000 mm, b = 4m = 4000 mm, E = 90GPa = 90kN/mm2

$\Delta_1 = \frac{P_1}{6EI}(2l^3 - 3l^2x + x^3) = \ldots = 41.48 \text{ mm}$ $\Delta_2 = \frac{P_2b^2}{6EI}(3l - 3x - b) = \ldots = 47.41 \text{ mm}$ $\Delta = \Delta_1 + \Delta_2 = 41.48 \text{ mm} + 47.41 \text{ mm} = 88.88 \text{ mm}$

This example involves finding the deflection at the free end of a beam subjected to two different loads. Each load has its formula for calculating deflection based on its properties such as the applied load, length of the beam, and its rigidity. By applying the relevant formulas for each load, we derive two separate deflections. Adding these two values gives us the total deflection at the end of the beam, which in this case is approximately 88.88 mm.

Think of a seesaw in a playground where two kids of different weights sit at different distances from the center. The greater the weight and the distance, the more the seesaw tilts. In our case, each child represents a different load on the beam, causing it to bend at the end. This deflection shows how much the beam bends under loads, similar to how the seesaw tilts with unequal weights.

Example (3):

For the beam shown in the figure below, calculate the deflection of the beam at point C.

Given: E = 100GPa, I = 120×106mm4

Solution:

1m w = 10kN/m = 0.01kN/mm, l = 5m = 5000mm, E = 100GPa = 100kN/mm2

P = 10kN, x = 2m = 2000 mm, a = 2m = 2000 mm, I = 120×106mm4

$\Delta = \frac{Pax^2(l^2 - x^2)}{6EIl} = 2.33 \text{ mm}$ $\Delta = \frac{wax^2(l^2 - x^2)}{2EIl} = 1.17 \text{ mm}$ $\Delta = \Delta_1 + \, \Delta_2 = 2.33 \text{ mm} + 1.17 \text{ mm} = 3.5 \text{ mm}$

In this calculation, we are determining the deflection of a beam at a specific point (point C). It involves two components: one from the concentrated load (P) applied at a distance 'x' from a point, and another from a uniform load (w). We use tailored formulas to derive the deflections resulting from each type of load separately and then sum them to find the total deflection effect at point C, which totals approximately 3.5 mm.

Consider a trampoline with one person jumping towards one side (a concentrated load) while the surface of the trampoline is being pressed down evenly (like a uniform load). The total bending of the trampoline at a certain point is similar to our calculation; combining both effects creates a noticeable deflection at that point.

Example (4):

For the beam shown in the figure below, calculate the deflection of the beam at the mid-span.

Given: E = 95GPa, I = 100×106mm4

Solution:

1m w = 5kN/m = 0.005kN/mm

l = 6m = 6000mm, E = 95GPa = 95kN/mm2

x = 3m = 3000 mm, a = 2m = 2000 mm, I = 100×106 mm4

$\Delta = \frac{5wL^4}{384EI} = 8.88 \text{ mm}$ $\Delta = \frac{wa^2(l - x)}{24EIl} = 2.19 \text{ mm}$ $\Delta = \Delta_1 + \Delta_2 = 8.88 \text{ mm} + 2.19 \text{ mm} = 11 \text{ mm}$

In this example, we calculate the mid-span deflection of a beam considering both a uniform load and an additional point load. Using the well-established formulas, we account for the contributions of each load on the mid-span. The use of the 384EI term in the first part reflects how the bending due to uniform loading affects the overall deflection, while the second formula addresses point effects. Adding together the calculated deflections gives us a total deflection of about 11 mm.

Imagine a heavy blanket stretched across a bed. If you sit in the middle, it dips down due to your weight (uniform load), and if you place a heavy box on one side, that also pulls the corner down, creating an additional dip (point load). The combined dips represent the total deflection as calculated, showing how both loads work together to influence the blanket's position.

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