CE 201 - Statics

Chapter 7 – Lecture 1

INTERNAL FORCES

Internal Forces Developed in Structural Members (2-D)

Internal Forces Developed in Structural Members (3-D)

Internal Forces Developed in Structural Members (2-D)

To design a member, the forces acting within the member need to be determined. These forces are known as the internal forces and can be found by using the method of sections.

The method can be illustrated by the following example.

A B

C

F1 F2

A B

C

F1 F2

Ax

Ay By

The reactive forces at the supports (A) and (B) can be determined by applying the Equilibrium Equations (Fx = 0; Fy = 0; Mo = 0) on the entire structure.

Suppose that the forces acting at (C) are to be found,

then it is necessary to make a section at (C).

A B

C

F1 F2

Ax

Ay By

A

C

F1

Ax

Ay

B

F2

ByC

NC

VC

MC

These external forces and couple moment will develop at the cut section in order to prevent the segments from translating or rotating

N is called the normal forceV is called the shear forceM is called the bending moment or couple moment

A

C

F1

Ax

Ay

B

F2

ByC

VC VC

NC NC

MC MC

Note:VC, NC, and MC on both segments are equal in magnitude and opposite in direction. VC, NC, and MC can be determined by applying the equilibrium equations (Fx = 0; Fy = 0; Mo = 0) on one of the segments.

Internal Forces developed in Structural Membranes (3-D)

A

C

F1

Ax

Ay

B

F2

ByC

VC VC

NC NC

MC MC

x

Cy

z

Vx Mx

Vz

Mz

Ny

My

•Ny is normal to the cross-sectional area.

•Vx and Vz are components of the shear force and they are acting tangent to the section.

•My is the torsional moment or twisting moment.

•Mx and Mz are components of the bending moment.

•The resultant loadings (forces and moments) act at the centroid of the cross-sectional area.

Procedure for Analysis

Support Reactions (before cutting)

Free-body Diagram (of the segment that has the least number of loadings)

Equilibrium Equations