Question

A beam is subjected to a triangular distributed load whose value at right end of the beam is w=8.1 kN/m. Draw the free- body

0 0
Add a comment Improve this question Transcribed image text
Answer #1

solution : Given: We 8.1 kN/m 30 m We have to draw : FBD of beam & determine vertical reaction @ B Free Body diagram : w38.1

Add a comment
Know the answer?
Add Answer to:
A beam is subjected to a triangular distributed load whose value at right end of the...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A beam is subjected to a triangular distributed load whose value at right end of the...

    A beam is subjected to a triangular distributed load whose value at right end of the beam is w=8.1 kN/m. Draw the free- body diagram of the beam and determine the vertical reaction at A (in kN). Sign: Upward is positive A B 30 m

  • The cantilever beam shown is subjected to a moment at A and a distributed load that...

    The cantilever beam shown is subjected to a moment at A and a distributed load that acts over segment BC, and is fixed at C. Determine the reactions at the support located at C. Then write expressions for shear and bending moment as a function of their positions along the beam. Finally, use these expressions to construct shear and bending moment diagrams Draw a free-body diagram of the beam on paper. Use your free-body diagram to determine the reactions at...

  • The Beam shown will be subjected to a concentrated live load of 100kN, a uniformly distributed...

    The Beam shown will be subjected to a concentrated live load of 100kN, a uniformly distributed live load of 50kN/m and a uniformly distributed dead load of 20kN/m. 45.) determine the maximum reaction at B 46.) determine the maximum positive shear at C 47.) determine the maximum negative moment at B The beam shown will be subjected to a concentrated live load of 100 KN, a uniformly distributed live load of 50 kN/m and a uniformly distributed dead load of...

  • Q2 The 10 m long simply supported beam is subjected to a uniformly distributed load w...

    Q2 The 10 m long simply supported beam is subjected to a uniformly distributed load w = 10 kN/m throughout and a point load P =10 kN at the midspan of the beam, as shown in Figure Q2 (a). The cross section of this beam is depicted in Figure Q2 (b), which consists of three equal rectangular steel members. Self-weight of the beam is neglected. 30 mm P= 10 KN W = 10 kN/m 200 mm 5 m 5 m...

  • Q2 The 10 m long simply supported beam is subjected to a uniformly distributed load w...

    Q2 The 10 m long simply supported beam is subjected to a uniformly distributed load w = 10 kN/m throughout and a point load P =10 kN at the midspan of the beam, as shown in Figure Q2 (a). The cross section of this beam is depicted in Figure Q2 (b), which consists of three equal rectangular steel members. Self-weight of the beam is neglected. 30 mm P = 10 kN W = 10 kN/m 200 mm 5 m 5...

  • Q2 The 10 m long simply supported beam is subjected to a uniformly distributed load w...

    Q2 The 10 m long simply supported beam is subjected to a uniformly distributed load w = 10 kN/m throughout and a point load P =10 kN at the midspan of the beam, as shown in Figure Q2 (a). The cross section of this beam is depicted in Figure Q2 (b), which consists of three equal rectangular steel members. Self-weight of the beam is neglected. 30 mm P = 10 kN w = 10 kN/m 200 mm 5 m 5...

  • The cantilever beam shown is subjected to a moment at A and a distributed load that...

    The cantilever beam shown is subjected to a moment at A and a distributed load that acts over segment BC, and is fixed at C. Determine the reactions at the support located at C. Then write expressions for shear and bending moment as a function of their positions along the beam. Finally, use these expressions to construct shear and bending moment diagrams. Part A - Reactions at support C Draw a free-body diagram of the beam on paper. Use your...

  • HW16.11. Cantilever beam with distributed load Consider a cantilever beam subjected to a uniform distributed load...

    HW16.11. Cantilever beam with distributed load Consider a cantilever beam subjected to a uniform distributed load as indicated below. ty L/4 L/2 Draw the free-body diagram and corresponding shear force and bending moment diagrams. To draw the shear force and bending moment diagrams, you MUST use the minimum number of lines (straight or curved), i.e., the minimum number of objects created by clicking the two buttons under "V and M lines" FBD FBD Concentrated forces: FBD Distributed loads: ttt ???

  • A propped cantilever beam with length an is subjected to a trapezoidal load with intensities, 2018...

    A propped cantilever beam with length an is subjected to a trapezoidal load with intensities, 2018 and 9,- 30 kN/m Find the reactions at A and B. Hint: The loading is the sum of uniform and triangular loads. (Enter your reaction forces in kN and your reaction moments in KN · m. Solve this problem by the method of superposition. This beam has constant flexural rigidity El. Assume that the +x-axis is to the right, the +y-axis is up along...

  • 4. (25 pt.) The beam subjected to a uniform distributed load as shown in Figure 4(a)...

    4. (25 pt.) The beam subjected to a uniform distributed load as shown in Figure 4(a) has a triangular cross-section as shown in Figure 4(b). 1) (6 pt.) Determine mathematical descriptions of the shear force function V(x) and the moment function M(x). 2) (6 pt.) Draw the shear and moment diagrams for the beam. 3) (5 pt.) What is the maximum internal moment Mmar in the beam? Where on the beam does it occur? 4) (8 pt.) Determine the absolute...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT