moment diagrama. A 265KN A = 135 kNm 5 kN 4m 3m eğn (35kNm tasskn

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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find the reactions. Write the equations. Use decimals.Fill in the blanks. Draw the shear and bending moment diagrams

**Understanding Shear Force and Bending Moment Diagrams**

The provided image details the process of calculating shear force and bending moment diagrams for a beam subjected to a combination of point and distributed loads. Below is a transcription and explanation of the key components shown in the image.

**Diagrams:**

1. **Beam Load Diagram:**
   - **Beam Support:** The beam is fixed at one end (A).
   - **Loads:**
     - A point load of \( 12 \mathrm{kN} \) is applied 1 meter from the support A.
     - A uniformly distributed load (UDL) of \( 5 \mathrm{kN/m} \) is applied over the length of 3 meters starting from the point 4 meters from support A.
   - **Dimensions:** The total length of the beam is shown in two segments, 4 meters and 3 meters, for a total of 7 meters.

2. **Triangular Load Equivalent:**
   - A triangular load is indicated equivalent to the UDL.
   - The calculation shows the total load due to this triangular load as the area under the load curve: \( (\frac{1}{2} \times 7 \times 5 = 35 \mathrm{kN}) \).
   - The centroid of the triangular load acts at \( \frac{1}{3} \) from the larger end, shown with the moment arm calculations.

3. **Effective Shear Force and Bending Moment Equations:**
   - **Notations:**
     - \( \bar{q}_{eq1} = \) symbol representing the equivalent distributed load.
     - \( \bar{V}_{eq1} = \) symbol representing the equivalent shear force.
     - \( \bar{M}_{eq1} = \) symbol representing the equivalent bending moment.

4. **Shear Force Diagram (SFD):**
   - No detailed values are indicated, but the general shape of the SFD is drawn with segments.

5. **Bending Moment Diagram (BMD):**
   - Likewise, the general shape of the BMD is shown with indicated segments.

**Key Calculations:**

- The moment and force values indicated:
  - \( A_{\mathrm{s}} =  \frac{265}{kN \cdot m} \)
  - \( A_{M} = \frac{135}{kN \cdot m} \)
  -
Transcribed Image Text:**Understanding Shear Force and Bending Moment Diagrams** The provided image details the process of calculating shear force and bending moment diagrams for a beam subjected to a combination of point and distributed loads. Below is a transcription and explanation of the key components shown in the image. **Diagrams:** 1. **Beam Load Diagram:** - **Beam Support:** The beam is fixed at one end (A). - **Loads:** - A point load of \( 12 \mathrm{kN} \) is applied 1 meter from the support A. - A uniformly distributed load (UDL) of \( 5 \mathrm{kN/m} \) is applied over the length of 3 meters starting from the point 4 meters from support A. - **Dimensions:** The total length of the beam is shown in two segments, 4 meters and 3 meters, for a total of 7 meters. 2. **Triangular Load Equivalent:** - A triangular load is indicated equivalent to the UDL. - The calculation shows the total load due to this triangular load as the area under the load curve: \( (\frac{1}{2} \times 7 \times 5 = 35 \mathrm{kN}) \). - The centroid of the triangular load acts at \( \frac{1}{3} \) from the larger end, shown with the moment arm calculations. 3. **Effective Shear Force and Bending Moment Equations:** - **Notations:** - \( \bar{q}_{eq1} = \) symbol representing the equivalent distributed load. - \( \bar{V}_{eq1} = \) symbol representing the equivalent shear force. - \( \bar{M}_{eq1} = \) symbol representing the equivalent bending moment. 4. **Shear Force Diagram (SFD):** - No detailed values are indicated, but the general shape of the SFD is drawn with segments. 5. **Bending Moment Diagram (BMD):** - Likewise, the general shape of the BMD is shown with indicated segments. **Key Calculations:** - The moment and force values indicated: - \( A_{\mathrm{s}} = \frac{265}{kN \cdot m} \) - \( A_{M} = \frac{135}{kN \cdot m} \) -
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