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: - \( A_{\mathrm{s}} = \frac{265}{kN \cdot m} \) - \( A_{M} = \frac{135}{kN \cdot m} \) -

Answered: Image /qna-images/answer/ca45bd8e-7186-415f-9164-516c970dc0a0.jpg

Answered: moment diagrama. A 265KN A = 135 kNm 5…

Engineering

Civil Engineering

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

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

ENGINEERING MECHANICS (UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS AND FBD LEGIBLY. NO LONG EXPLANATION NEEDED. ANSWER IN 2 DECIMAL PLACES. BOX THE FINAL ANSWER.)
Beam PORS as in Figure Q1 is pin connected at P and S. (a) Develop the shear force (V) and bending moment (M) as function of x for section RS, taking point P as the origin. (b) Draw the shear force and free body diagrams of the entire beam and indicate all important points on the diagrams. [Note: All calculations must be clearly shown] 40 kN 10 kN/m P S 20 kN.m 4 m 2 т 4 m Figure Q1
Answer only number 5 - need answer in 20 mins
Please answer attached image.thank you
AB, BC, CD AND DE are 2m in length. all members act at right angles sketch the bending and torsional moment diagrams. please could you explain the steps in the diagrams or highlight which force causes which moment (use a different coloured pen for each force maybe ? thankyou!
Solve the problem and answer the questions that follow. - If the supports were swapped such that the left support were a roller and the right support a pin, how would your solution change? - If the value of “w” were doubled, what would be the bending moment at the left most point of the beam?
1 ft 3 ft B 3 ft 3 ft (-9002 + 800f + 1000k) Ib.ft 6 ft 3 ft E (-300î – 400ĵ + 100k) Ibs The applied force at point E and the applied couple moment at point D are shown in the figure. Complete/calculate the following: (Where applicable, express the results as the (x, y, z) components of a Cartesian vector) 1. Draw the free-body diagram (FBD), showing the appropriate reaction forces and/or moments at the supports. (You must show the reactions in their component form i.e. show RAx, MRAX etc. on the FBD). Note: The support at point A is not able to resist a moment about the 'X' axis but is able to resist a force in this direction. 2. The unit vector ÅBc 3. The position vector Tɛ/a 4. The position vector īB/a
ANSWER TYPEWRITTEN FOR UPVOTE. SKIP IF YOU ALREADY DID THIS OR ELSE DOWNVOTE
please answer correctly