**Goal:** Produce a 90-to-120 second visual lesson that rigorously explains how to find the mid-span balancing force **F** on a simply supported beam with two equal point loads **P**.
> **Audience:** Undergraduate structural-mechanics students.
> **Narration tone:** Concise, authoritative, step-by-step.
> **Please include the following visual sequence:**
>
> 1. **Title card (3 s)** – “Moment-Area Method: Cancelling Mid-Span Deflection of a Beam.”
> 2. **Problem sketch (5 s)** – Clean line drawing of beam A-E (length L), supports at A and E, loads **P** at B (L/4) and D (3L/4), unknown upward force **F** at C (L/2). Label all distances.
> 3. **Method choice slide (4 s)** – Quick bullet list: “Linear system → superposition,” “Moment-area theorem (Mohr’s 2nd) relates deflection to 1st moment of M/EI diagram.”
> 4. **Step 1 – M/EI diagram for a single P (15 s)**
>
> * Draw left-half triangle from A to C caused by the left-hand **P**.
> * Highlight area $A_1$ and centroid distance $\bar{x}_1$.
> * Show formula $t_{AC}=\bar{x}_1A_1$.
> * Display computed value $PL^3/(192EI)$.
> * Add subtitle “Right-hand P gives identical value.”
> 5. **Step 2 – Combine the two P loads (4 s)** – Sum the two contributions → $11PL^3/(384EI)$ downward. Use a downward arrow icon.
> 6. **Step 3 – M/EI diagram for the balancing force F (10 s)**
>
> * Draw inverted triangle under C (span A-C).
> * Mark its area and centroid → $FL^3/(48EI)$.
> * Arrow up to stress “opposite direction.”
> 7. **Step 4 – Superposition equation (8 s)**
>
> * Show equation $11PL^3/(384EI) - FL^3/(48EI) = 0$.
> * Animate cancelling terms, solve for $F = 11/8\,P$.
> 8. **Result slide (5 s)** – Big boxed answer $F = 1.375\,P$ (upward), plus a brief note “deflection at C now zero.”
> 9. **Wrap-up (5 s)** – One-line takeaway: “Moment-area + superposition = fast, exact mid-span balancing force.”
> **Technical style:** minimal text on screen, key equations appear beside diagrams, gentle zooms/pans to guide attention, no background music needed.
---**Question 1**
**Question Stem (Chinese):**
有一橫樑 AE 受到兩個集中力 P 作用，如下圖所示，樑斷面之撓曲剛度為常數 EI：
此時若增加集中力 F 欲使樑中 C 點回到中線 (抵銷垂直位移)，則剛集中力大小為何？
**Question Stem (English):**
There is a beam AE with constant flexural rigidity as EI and a series loading.
If the vertical deflection of the beam at midspan (i.e., point C) is to be zero, determine the magnitude of force F.
**Chart/Diagram Description:**
* **Type:** Structural diagram of a beam with applied forces and supports.
* **Main Elements:**
* A horizontal beam labeled AE.
* Supports: Point A is supported by a pin (indicated by a triangle on a base). Point E is supported by a roller (indicated by a circle on a base).
* Points on the beam: A, B, C, D, E. Point C is located at the midspan of the beam AE.
* Applied Forces:
* A downward vertical force labeled P is applied at point B.
* A downward vertical force labeled P is applied at point D.
* An upward vertical force labeled F=? is applied at point C.
* Distances: The beam is divided into four equal segments by points B, C, and D.
* Distance from A to B is L/4.
* Distance from B to C is L/4.
* Distance from C to D is L/4.
* Distance from D to E is L/4.
* The total length of the beam AE is L (sum of the four segments L/4 + L/4 + L/4 + L/4).
* Flexural rigidity of the beam is denoted as EI, which is constant.
视频信息
答案文本
视频字幕
We will use the moment-area method to find the balancing force that cancels mid-span deflection of a simply supported beam.
Here is our simply supported beam A-E with length L. Two equal point loads P are applied at points B and D, located at L/4 and 3L/4 from the left support. We need to find the upward force F at the midspan point C that will cancel the vertical deflection at that point.
We will use superposition and the moment-area theorem. Mohr's second theorem states that the tangential deviation equals the first moment of the M over EI diagram. We start by analyzing the moment diagram for a single point load P.
For the left-hand P load, the M over EI diagram forms a triangle from A to C. The area is P L squared over 32 E I, and the centroid distance is L over 6. This gives a deflection of P L cubed over 192 E I. The right-hand P load contributes the same amount. Combined, the total downward deflection is 11 P L cubed over 384 E I.
The upward force F creates a deflection of F L cubed over 48 E I. Setting the total deflection to zero: 11 P L cubed over 384 E I minus F L cubed over 48 E I equals zero. Solving for F gives us F equals 11 over 8 times P, or 1.375 P upward. This balancing force cancels the mid-span deflection completely.