Statics

Chapter 7

Internal Forces

Eng. Iqbal Marie

iqbal@hu.edu.jo

Hibbeler, Engineering Mechanics: Statics,12e, Prentice Hall

Beams

A beam is defined as a

structural member designed

primarily to support forces

acting perpendicular to the

axis of the member.

Beams – Types

A beam can be classified as

statically determinate beam,

which means that it can be

solved using equilibrium

equations, or it is Statically

indeterminate

Internal pin or hinge

7.1 Internal Forces Developed in Structural

Members

Internal Force in Structures

– Shear Forces ( V)

– Bending Moment ( M)

– Normal Forces (N Tension or compression)

N

N

V

For each cross section, there is a shear force V and a bending moment M and

a normal force N.

Internal Forces

N

– Shear Forces

– Bending Moment

– Normal Forces

7.1 Internal Forces Developed in Structural

Members

Sign Convention

N+

N+

Negative Moment

Positive Moment

Internal forces

at point C

OR

Internal: moments, shear, and

normal forces at point C

Solution Procedure

The general scheme for finding the internal set of forces at

certain point is:

a) Draw the free-body diagram

b) Determine the support reactions by Applying the

equations of equilibrium

d) Make section passing through the point

e) Draw free body diagram for either part of the beam ( the

one to the right or left of the section) showing the positive

internal forces on it ( N, V, M)

f) Apply equilibrium equations for that part and find the

internal forces

Or

8.8 kN 5.2 kN

Force Diagrams

Force diagrams are plots for the internal

forces along the axis of the beam

1) Axial Force Diagram

2) Shear Diagram

3) Bending Moment Diagram

y By D

By D

B D

D

By

0 20 kN 40 kN

60 kN

0 20 kN 2.5 m 40 kN 3.0 m 5.0 m

14 kN

46 kN

F R R

R R

M R

R

R

sections 1-1.

section 2-2

section at 3-3.

section 4-4.

Shear and Bending moment Diagrams

• Draw the free-body diagram

• Solve for reactions

• Solve for the internal forces (shear, V,

and bending moment, M)

In order to generate a shear and bending

moment diagram one needs to

Location (m) Shear (kN) Moment (kN-m)

1 0 -20 0

2 2.5 -20 -50

3 2.5 26 -50

4 5.5 26 28

5 5.5 -14 28

6 7.5 -14 0

7 7.5 0 0

600lb

7000 lb.ft

600 lb

7000lb. ft

4000 lb.ft MD ( lb.ft)

SD ( lb)

V (k)

M (k.ft)

V (k)

M (k.ft)

w (k/ft)

Concentrated Loads:

Shear forces are consistent in magnitude. Therefore, shear diagrams are flat lines (no slope; horizontal). Moment vary linearly between concentrated loads. Therefore, moment diagrams are composed of sloping lines for concentrated loads.

Uniformly Distributed Loads:

UDLs produce linearly varying shear forces—shear

diagrams consist of sloped lines.

UDLs produce parabollically varying moments;

therefore, moment diagrams are curves.

d

d

MV

x

21

2 2 2 8

wL L wLM

Note that the slope of the

moment diagram is equal

to the shear.

w=2 k/ft

15 ft 15 ft

w=30 k

45

V (k)

M (k.ft)

15

15

45

450 450

450

lbRR

PwlRR

BA

BA

452/302/302

2/2/

For the beam shown draw the shear and moment diagram:

4m 2m 2m

Draw shear and moment Diagrams

1.167 32.167

For the beam shown here draw the shear and moment diagram:

24

w=1.5 k/ft

3k

12 ft 4 ft

7 k 20 k

7

11

9 3

X=4.67

12-x =7.33

16.3

40.3

24 16.3

M(k.ft)

v(k)

RAx

RAy RC

x Ax

y Ay C

Ay C

A C

C Ay

0

0 20 kN/m 6 m

120 kN

0 20 kN/m 6 m 3 m 9 m

40 kN & 80 kN

F R

F R R

R R

M R

R R