1. Statically Determine of Beams and Frames

3. Axial Force
The axial force at any transverse cross section of a straight beam is the algebraic sum of the component acting parallel to the axis of the member of all the loads and reactions applied to the portion of the member on either side of that cross section.
For curved members the summation is done of the of the force components parallel to the tangent to the curve at the selected cross section.

4. Shearing Force:
The shearing force at any transverse cross section of a straight member is the algebraic sum of the components acting transverse to the axis of the member of all the loads and reactions applied to the portion of the member on either side of that cross section.
For curved members the summation is done of the of the force components transverse to the tangent to the curve at the selected cross section.

5. Bending Moment:
The bending moment at any transverse cross section is the algabric sum of the moment taken about on axis normal to the plane of loading and passing through the cross section of all the loads and reactions applied to the portion of member on either side of the cross section.

6. Sign Conventions

7. Internal Forces in a System: General Procedure
The general method of obtaining internal forces at certain cross-section of a system under a given loading (and support) condition is by applying the concepts of equilibrium. To illustrate, let us consider the beam-column AB in figure below.

8. Solution:
Consider F. B. D. of whole structure as shown below

9. If a system is in static equilibrium condition, then every segment of it is also in equilibrium. Let us consider the equilibrium of part AC , and draw its free body diagram.

10. Figure below shows the free body diagram of CB .

11. Example:-
Find the axial force, shearing force and bending moment at the section (1-1) located at 3 m from the left side of simply supported beam shown below.

12. Solution:-
Draw F. B. D. of the beam

14. Axial force = - 30 kN (compression) Shear force = 40 – 20 = 20 kN Bending moment= 40*3 – 20*1 =100 kN.m

15. Relationships Between Loads, Shearing Force, and Bending Moment

16. Figure (b)
𝑭𝒚 =𝟎 →𝑽− 𝑽+𝒅𝑽 −𝒘.𝒅𝒙=𝟎
𝒅𝑽=−𝒘.𝒅𝒙 (𝟏)
𝒐𝒓 𝒅𝑽 𝒅𝒙 =−𝒘 (𝟐)
Eq. (1): - States that the change in the value of shearing force between two points along the member is equal to the area of the load between these points.
𝑴 =𝟎 (𝒂𝒃𝒐𝒖𝒕 𝒕𝒉𝒆 𝒍𝒆𝒇𝒕 𝒇𝒂𝒄𝒆)

17. 𝑽+𝒅𝑽 𝒅𝒙+𝒘.𝒅𝒙. 𝒅𝒘 𝟐 − 𝑴+𝒅𝑴 +𝑴=𝟎
Neglecting higher order terms
𝒅𝑴=𝑽.𝒅𝒙 (𝟑)
𝒐𝒓 𝒅𝑴 𝒅𝒙 =𝑽 (𝟒)
Eq.(3) :- The change in bending moment between two points along a member is equal to the area of the shearing force between these points.
Eq.(4):- States that the slope of the bending moment diagram at any point is equal to the shearing at that point.

18. For the case of a concentrated load Fig.(c): -
Vertical equilibrium
𝒅𝑽=−𝑷𝒐
For the case of a concentrated moment Fig.(d): -
𝒅𝑴=𝑴𝒐

19. Example: Draw S.F.D , and B.M.D for the beam shown in figure below.

20. Example: Draw A.F.D , S.F.D , and B.M.D for the frame
shown in Fig.

21. Example: Draw A.F.D , S.F.D , and B.M.D for the frame
shown in Fig.