solve question no 93 with proper exaplaination---Here is the extracted content from the image:
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**General Information:**
* **Topic:** Structural Analysis
* **Section:** 3.14 (implied from top left corner)
* **Year Markers:** 2015, 2016
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**Question 92:**
**Question Stem:**
92. For the beam shown below, the stiffness coefficient K₂₂ can be written as
**Other Relevant Text:**
Note: 1, 2 and 3 are the d.o.f.
**Diagram Description:**
* **Type:** Beam diagram with degrees of freedom (d.o.f.) indicated.
* **Main Elements:**
* A horizontal beam is depicted.
* The left end of the beam is fixed to a vertical support (indicated by hatching).
* The length of the beam is labeled as 'L'.
* The beam properties are denoted as A, E, I (Area, Young's Modulus, Moment of Inertia).
* At the free (right) end of the beam, three degrees of freedom (d.o.f.) are shown:
* A horizontal arrow pointing right, labeled '1'.
* A vertical arrow pointing up, labeled '2'.
* A clockwise curved arrow representing rotation, labeled '3'.
**Options:**
(a) 6EI / L²
(b) 12EI / L³
(c) 3EI / L
(d) EI / 6L²
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**Question 93:**
**Question Stem:**
93. A fixed end beam is subjected to a load, W at 1/3rd span from the left support as shown in the figure. The collapse load of the beam is
**Diagram Description:**
* **Type:** Fixed-end beam diagram with a point load and varying plastic moment capacity.
* **Main Elements:**
* A horizontal beam is shown.
* Both the left end (labeled X) and the right end (labeled Z) are fixed supports (indicated by vertical hatching).
* A downward point load 'W' is applied at a distance of L/3 from the left support X.
* The beam has a total span of L, divided into two halves by point Y.
* The segment from X to Y has a plastic moment capacity of 2Mₚ.
* The segment from Y to Z has a plastic moment capacity of Mₚ.
* Horizontal dimensions are indicated below the beam: L/2 for the left segment (X to Y) and L/2 for the right segment (Y to Z).
**Options:**
(a) 16.5 Mₚ/L
(b) 15.5 Mₚ/L
(c) 15.0 Mₚ/L
(d) 16.0 Mₚ/L
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**Question 94:**
**Question Stem:**
94. Consider the plane truss with load P as shown in the figure. Let the horizontal and vertical reactions at the joint B be H_B and V_B, respectively and V_C be the vertical reaction at the joint C. Which one of the following
**Diagram Description:**
* **Type:** Plane truss diagram with supports and an applied load.
* **Main Elements:**
* A truss structure composed of members forming a series of triangles.
* Joints are labeled alphabetically: A, B, C, D, E, F, G.
* **Supports:**
* Joint B is a pin support (indicated by a hinged support symbol).
* Joint C is a roller support (indicated by a roller symbol).
* **Load:** A downward point load 'P' is applied at joint G.
* **Dimensions:**
* Horizontal length L between A and G.
* Horizontal length L between B and D.
* Horizontal length L between D and C.
* Slanted member lengths from A to E, and E to B are marked as 'L' each.
* **Angles:** Several internal angles of the triangular elements are marked as 60 degrees. Specifically, angles at A, G, B, and C (within their respective triangles) are shown as 60°. This suggests the truss is made of equilateral triangles.
**Options:**
* (Options are not fully visible in the provided image.)
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We have a fixed-end beam with supports at points X and Z. A point load W is applied at one-third span from the left support. The beam has varying plastic moment capacity: 2Mp from X to Y and Mp from Y to Z. We need to find the collapse load using plastic analysis principles.
Plastic hinges form when the bending moment at a section reaches the plastic moment capacity. In this fixed-end beam, critical locations are at the fixed supports X and Z, and under the point load. These sections experience maximum moments and will yield first during collapse, forming plastic hinges that allow rotation.
We have a fixed-end beam with varying plastic moment capacity. The left segment from X to Y has plastic moment capacity 2Mp, while the right segment from Y to Z has capacity Mp. A point load W is applied at one-third span from the left support.
In plastic analysis, plastic hinges form at points of maximum moment. At collapse, the beam becomes a mechanism. We use the virtual work principle: external work equals internal work. For this beam, plastic hinges will form at support X, under the load, and at support Z.
At collapse, the beam forms a mechanism with plastic hinges at X, under the load, and at Z. The beam deflects with rotation angles at each hinge. Using the virtual work principle, external work done by load W equals internal work done by plastic moments: W times delta equals sum of Mp times theta at each hinge.
From the collapse mechanism geometry, we establish relationships between deflection and rotation angles. The deflection delta at the load point relates to rotation angles through geometric compatibility. For the left segment of length L over 3, delta equals L over 3 times theta 1. From geometric constraints, theta 1 equals theta 2, and theta 3 equals 1.5 times theta 1.
Applying the virtual work principle, external work equals W times delta, which equals W times L over 3 times theta 1. Internal work at plastic hinges: at X is 2Mp times theta 1, under the load is 2Mp times theta 1, and at Z is Mp times 1.5 theta 1. Total internal work is 5.5 Mp times theta 1. Equating external and internal work gives W equals 16.5 Mp over L. The answer is option A.
Now we solve using the virtual work principle. External work equals W times delta, which is W times L over 3 times theta 1. Internal work occurs at three plastic hinges: at X is 2Mp times theta 1, under the load is 2Mp times theta 2, and at Z is Mp times theta 3. Using geometric relations, theta 2 equals theta 1 and theta 3 equals 1.5 theta 1. Total internal work is 5.5 Mp theta 1. Setting external work equal to internal work and solving gives W equals 16.5 Mp over L. The answer is option A.
Let's complete the final calculation. From the virtual work equation, W times L over 3 times theta 1 equals 5.5 Mp theta 1. Canceling theta 1 from both sides gives W times L over 3 equals 5.5 Mp. Solving for W gives W equals 5.5 Mp times 3 over L, which simplifies to 16.5 Mp over L. Comparing with the given options, this matches option A exactly. The answer is A: 16.5 Mp over L. This plastic analysis method involves identifying plastic hinge locations, applying virtual work principle, accounting for varying plastic moment values, and solving the equilibrium equation.