lecure2_Strength of Materials.pdf

8/9/2019 lecure2_Strength of Materials.pdf

http://slidepdf.com/reader/full/lecure2strength-of-materialspdf 1/6

Strength of Materials:Lecture 2

1-Axial Force, Shear Force, and Bending Moment Diagrams:

The three components of moment which can occurat a section of a memberact around the threecoordinate axes Fig. (c).

The positive sense of theforce components on thecut section viewed towardthe origin coincides withthe positive direction of thecoordinate axes as shownin Fi g. (b).

The quantities shown inFig.(c) will be representedalternatively by doubleheaded vectors as in fig.(d). The sense of thesevectors follows the right-hand screw rule.

*M x is the torque = T

For planar problems the notation forand the diagrammatic representationof the forces components are shown inFig. (e and f).

Side view of forces acts on the body;P (axial force), V (shear force), and theM (moment).

7



From the drawing shown before (P = F):

Fx: Axial Force σ (normal or direct stress) = AP --- (1)

Fy, F z: Shear Force τ (shear force) = It VQ

--- (2)

Mx: Twisting Moment or Torsion T =JR .T --- (3)

My, M z: Bending Moment σ = I

c M --- (4)

QQ

a) Axial Force b) Shear Force c) Torsion Force

P

d) Bending Moment

Sign Conventions:

1) SHEAR is considered Positive at section when it tends to Rotate the portionof the beam. In The Clock Wise Direction about an axis through @ point inside the force and normal to the plane of loading, otherwise it is negative.

For concreteness consider a beam, such as shown in Fig. (1-a). Any part of this

beam to either side of an imaginary cut, as (1-1), which is made perpendicular to thex-axis of the member, can treated as a free body.

8



Simply supported beam with

Concentrated force in the middle.

To maintain a segment of a beamsuch as shown in Fig. (1-b) in equilibriumthere must be an internal vertical force

Fy at the cut to satisfy the equation∑ = 0Fy . This force is called;"Shear force , v". The shear is numericallyequal to the algebraic sum of all thevertical components of the externalforces the external forces acting on theisolated segment, but it opposite indirection.

By apply: + ∑ =

0Fy

Fy = P/2

Figure (1) Simply Supported Beam

P/2X

Fx

Fy

Mz

(b) Section 1-1

P/2X

Fy =V

(c)

v

v(+ v) Shear

v

v(- v) Shear

L/2L/2

X

Y P

XZ

P/2 P/2

(a)

2

2

Assumption

9

P

P/2 P/2

V V

(d) Section 2-2



2) BENDING M OMENT is considered Positive at section when it tends to bend themember Con @ + ve upward ; otherwise it is negative.

1

1

P

P/2P/2

+ M + M

( + ) Moment

Ex1:

3) AXIAL FORCE

T=Tension (+) , Increase in LENGTH

C = Compression (-) , decrease in LENGTH

+M -M

- M -M

( - ) Moment

P P

-M

TT

CC

10



1- Methods of Sections

Ex2: Draw A.F., S.F & B.M. Diagrams?

3 m 2 m

4

3

10 kN

A

+ ∑ --- (1) = 0Fx R AX -6 =0∴ R AX = 6 kN B

+ --- (2) ∑ = 0 A M 8 × 3 – R BY × 5 = 0 ∴ R BY = 4.8 kN

10 kN10×4/5 = 8 kN + ∑ --- (3) = 0Fy

+ R AY - 8 + 4.8 = 0∴ R AY = 3.2 kN

For Section 1-1: (0 ≤ X 3)≤

+ ∑ = 0Fx

6- N 1 = 0 ∴ N1= 6 kN

+ ∑ = 0Fy

+ 3.2 – V 1 = 0 ∴ V 1= 3.2 kN

N1

3.2 kN

6 kN

V1

M1

1

1

X

3 m 2 m

4

3

1

R Ax = 10×3/5 = 6 kN 6 kN A B

R AY =3.2 kN

R BY =4.8 kN

¯

A.S.F.

1 2

X

2

- 6 kN

Constant

+

¯

3.2kN S.F.D.

-4.8kN

Linear 1st Degree+9.6kN

B.M.D.++

11



+ ∑ = 0 A M

+ 3.2 * X – M 1 = 0 ∴ M = + 3.2X kN Variable to X-1 st Degree

For Section 2-2: (3 ≤ X 5)≤

12

+ ∑ = 0Fx

6-6- N 2 = 0 ∴ N2= 0

+ ∑ = 0Fy

+ 3.2 – 8 +V 2 = 0 ∴ V 2= 4.8 kN

+ ∑ = 0 A M + 3.2 * X 2 – 8 (X 2 – 3) – M 2 = 0 ∴ M 2 = - 4.8X 2 +24

Just for Checking:

Where X = 5m M = - 4.8 × 5 +24 = 0

Where X = 3m M = - 4.8 × 3 +24 = 9.6

Note : V dx

dM = , PdxdV =

N2

3.2 kN

6 kN

V2

M2

2

2

8 kN

6 kN

3

X

Documents

lecure2_Strength of Materials.pdf