2. Axial Force, shear Force and Bending Moment Diagrams, Study notes of Theory of Structures

Shear and Bending moment diagram for a simply supported beam with a concentrated load at mid-span. Page 3. Theory of Structures. Dr. Mohammed ...

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

margoth
margoth 🇬🇧

4.4

(11)

229 documents

1 / 7

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Theory of Structures
Dr. Mohammed Abdulhussain Radi
17
2. Axial Force, shear Force and Bending Moment Diagrams:
Sign convention:
N: Axial Force (tension +ve, compression ve).
V: Shear Force (turning structure clockwise +ve, counterclockwise ve).
M: Bending Moment (compression outside of structure and tension inside
+ve, otherwise ve).
Positive (+) Negative (-)
There are three interior forces
are generated when any beam
element was cutting:
pf3
pf4
pf5

Partial preview of the text

Download 2. Axial Force, shear Force and Bending Moment Diagrams and more Study notes Theory of Structures in PDF only on Docsity!

Dr. Mohammed Abdulhussain Radi

2. Axial Force, shear Force and Bending Moment Diagrams:

Sign convention: N: Axial Force (tension +ve, compression – ve). V: Shear Force (turning structure clockwise +ve, counterclockwise – ve). M: Bending Moment (compression outside of structure and tension inside +ve, otherwise – ve). Positive (+) Negative (-) There are three interior forces are generated when any beam element was cutting:

Dr. Mohammed Abdulhussain Radi Notes:

  • Axial, shear and bending moment diagrams are usually started from the left side.
  • The value of the interior force (axial, shear or moment) at any location is equal to the cumulative area under the corresponding force diagram from (x = 0) to the considered location.
  • Any concentrated load (force or moment) causes a jump in the corresponding diagram. Shear and Bending moment diagram for a simply supported beam with a concentrated load at mid-span.

Dr. Mohammed Abdulhussain Radi ∑Fy = 0 VBc + 4 – 4 × 2 = 0 V = 12 kN ∑MC = 0 MC + 4 × 4 × 2 – 4 × 4 = 0 MC = - 16 = 16 kN.m

Dr. Mohammed Abdulhussain Radi

Dr. Mohammed Abdulhussain Radi