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myOKState-Stillwater/Tuls x emView?assignmentProblemID=16663783&attemptNo=1 posite Beams) - Attempt 1 Segment Attributes Segment type: decreasing linear x Mastering Engineering Co X Left point Right point -12.00 0.00 87 hp * Draw the shear diagram for the beam. Begin by placing vertical lines. Place the appropriate function between the vertical lines, ensuring the endpoints have the correct values. Note - Make sure you place only one vertical line at places that require a vertical line. If you inadvertently place two vertical lines at the same place, it will appear visually correct because the lines overlap, but the system will mark it wrong. 8 00 Mastering Engineering Mas X 8 V (kip) 30 20 10 00 -10 ( 9 5 9 2 kip/ft f10 O 6 ft De O A beam is shown in the fig X 10 kip P 4 ft O 1 8 kip 1. 40 kip ft ft 61°F Mostly cloudy ? num lik + x insert 12 of 18 4 0 X prt scr backspace 12:12 PM 11/11/2023 Review Help : > do delete home Do up

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William

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Step 1/2

To create the shear diagram for the given beam, we need to follow a systematic approach. The beam in question is subjected to various loads, and we'll use these to compute the shear forces at different points along the beam. Let's break down the process:

1. Understanding the Beam and Loads:

- The beam has different loads applied: point loads of 10 kip, 8 kip, and a uniformly distributed load (UDL) of 9.2 kip/ft over a certain length.

- There's also a moment of 40 kip-ft applied.

Explanation:

2. Calculating Reactions at Supports:

Explanation:

- First, we need to calculate the `reactions` at the supports (assuming the beam is `statically` determinate). This involves summing moments about a point (usually one of the supports) and using equilibrium equations:

Explanation:

\[ \Sigma M = 0, \; \Sigma V = 0 \]

Explanation:

- With these equations, we can find the vertical reactions at the supports.

Step 2/2

Explanation:

3. Constructing the Shear Diagram:

Explanation:

- Start from one end of the beam (usually the left).

Explanation:

- At the first point load or reaction, the shear force will jump up or down by the `magnitude` of that load.

Explanation:

- For the UDL, the shear force will `decrease` linearly as we move along the beam.

Explanation:

- At any point load along the beam, the shear force will again jump.

4. Drawing the Diagram:

- Plot these calculated shear forces on the Y-axis against the beam length on the X-axis.

- Between loads, the shear force is constant; under a UDL, it's a straight line sloping up or down.

- At each point load or support reaction, there will be a vertical jump in the shear diagram.

5. Checking for Errors:

- Ensure continuity in the diagram: it should start and end at zero (if the beam is simply supported).

- Check the diagram against the loading diagram for consistency.

Final Answer

6. Conclusion:

- The resulting shear diagram will visually represent how shear force varies along the length of the beam.

- This diagram is crucial for understanding the internal shear forces and designing the beam for strength and stability.

Remember, each beam and loading condition is unique, so the exact shape and values of the shear diagram will depend on the specific loads and support conditions of your beam. This explanation provides a general method for constructing a shear force diagram.

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