Strength of Materials: Lecture 1 1-Refrences:

• Introduction to Mechanics of Solid, By: E. Popov • Elements of Strength of Materials, By: Timishinko • Strength of Materials, By: Singer

2-Units: British Metric S.I. 1. Force Ib, kip, Ton g, kg, N, kN 1 kip = 1000 Ib 1 kg = 1000 g 1 kN = 1000 N 1 ton = 2240 Ib Ton = 1000 kg 1 kg = 10 N 2. Long in, ft m, cm, mm m, cm, mm 1 f = 12 in 1 m = 100 cm 1 m = 100 cm 1 cm = 10 mm 1 cm = 10 mm 1 m =1000 mm 1 m =1000 mm 1 in = 2.54 cm 1 in = 2.54 cm 3. Stress psi, ksi Pa (

2mmN ), MPa, GPa

2inp ,

2inkip

MPa = 106 Pa = 106 N/mm2 ×

2

221000

1

mmm

MPa = 2mm

N

GPa = 109 Pa = 109 N/mm2 × 2

221000

1

mmm

= 1032mm

N ×

kNN1000

1

GPa = kN/mm2

1

3-Supports Reactions: 1-One Reaction Supports (Roller Supports):

Signs used to represent the ROLLER support

The Roller or a Link is capable of

resisting a force in only one specific

line of action.

One Reaction Supports (Link Supports)

A RA RA

2

2-Two Reaction Supports (Hinge Supports): Signs used to represent the HING support A Pinned support is capable of resisting a force, the

reaction at such a support may have two components

one in the horizontal and one in the vertical direction.

Two Reaction Supports (2-Link Supports) Link

Rx Link

Ry 3-Three Reaction Supports (Fixed Supports): The third support is a Fixed support, Three forces can exist at such support Two components of forces and a moment, the physical such a suport is obtained by building a beam into a brick wall

3

Ex1: 10 kN 10 kN 10 kN 10 kN

R1 + R2 + R3 + M = 4 UKNOWNS > 3 EQUATIONS THEN (Intermediate) Ex2: Ry = 10-10 Ry = 0 Rx = 0 M = 20 + 10×3 -10×5 M = 0 4-Classification of Beams:

1) Simple Beam

2) Cantilever Beam

2 m 3 m 4 m

M

R3

R2 R1 9 m

10 kN 10 kN

1.5 m 1.5 m 2 m

20 kN

Ry =0

Rx =0

M =0

4

3) Simple Beam with Overhanging OR "Overhanging Beam"

4) Compound Beam

Moment at (Internal Hinge) = ZERO

5) Indeterminate Beam

R4

R5 R6 R1 R2

R3

R7

R3 R4

M = 0

5

3 3

6

3 1

Calculations of Beam Reactions Ex3: --- (1) 0 =∑ xF RAx = 0 + --- (2) ∑ = 0 @A M

250 + 80 × 2.5 + 80 × 3.75 – RB × 5 = 0

∴ RBy = +135 N --- (3) 0 =∑Fy RAy = -5 N RAy = 5 N

1 1 1 2

1.25 m 1.25 m1.25 m 1.25 m

80 N 50 N 250 N

A B

RAy RBy

RAx

×

RAy

4 > 3 6 > 3

5 > 3

Note: The number of equations for a suppose structure at the x-y plane are (3) equations.

0 =∑ xF 0 =∑Fy ∑ = 0 axis- x@M , ,

Strength of Materials: Lecture 2 1-Axial Force, Shear Force, and Bending Moment Diagrams:

The three components ofmoment which can occur at a section of a member act around the three coordinate axes Fig. (c).

The positive sense of the force components on the cut section viewed toward the origin coincides with the positive direction of the coordinate axes as shown in Fig. (b).

The quantities shown in Fig.(c) will be represented alternatively by double headed vectors as in fig. (d). The sense of these vectors follows the right-hand screw rule. *Mx is the torque = T

For planar problems the notation for

and the diagrammatic representationof the forces components are shown in Fig. (e and f).

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

M (moment).

7

From the drawing shown before (P = F): Fx: Axial Force σ (normal or direct stress) =

AP --- (1)

Fy, Fz: Shear Force τ (shear force) = It

VQ --- (2)

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

My, Mz: Bending Moment σ = I

cM --- (4)

Q Q

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 portion of the beam. In The Clock Wise Direction about an axis through @ point in side 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 the x-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 beam such as shown in Fig. (1-b) in equilibrium there must be an internal vertical force Fy at the cut to satisfy the equation ∑ = 0 Fy . This force is called; "Shear force, v". The shear is numerically equal to the algebraic sum of all the vertical components of the external forces the external forces acting on the isolated segment, but it opposite in direction. By apply: + ∑ = 0 Fy Fy = P/2

Figure (1) Simply Supported Beam

P/2 X

Fx

Fy Mz

(b) Section 1-1

P/2 X

Fy =V

(c)

v

v (+ v) Shear

v

v (- v) Shear

L/2 L/2

X

Y 1

1

P

X Z

P/2 P/2

(a)

2

2

Assumption

9

P

P/2 P/2

V V

(d) Section 2-2

2) BENDING MOMENT is considered Positive at section when it tends to bend the member Con @ + ve upward; otherwise it is negative.

1

1

P

P/2 P/2

+ M + M

( + ) Moment

Ex1:

3) AXIAL FORCE T=Tension (+), Increase in LENGTH C = Compression (-), decrease in LENGTH

+M -M

- M -M

( - ) Moment

P P1

1

-M

T T

C C

10

1- Methods of Sections

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

3 m 2 m

4 3

10 kN

A

+ ∑ --- (1) = 0 Fx RAX -6 =0 ∴ RAX = 6 kN B + --- (2) ∑ = 0 AM 8 × 3 – RBY × 5 = 0 ∴ RBY = 4.8 kN 10 kN 10×4/5 = 8 kN + ∑ --- (3) = 0 Fy + RAY - 8 + 4.8 = 0 ∴RAY = 3.2 kN For Section 1-1: (0 ≤ X 3) ≤ + ∑ = 0 Fx 6- N1 = 0 ∴ N1= 6 kN + ∑ = 0 Fy + 3.2 – V1 = 0 ∴ V1= 3.2 kN

N1

3.2 kN

6 kN

V1

M1

1

1

X

3 m 2 m

4 3

1

RAx = 10×3/5 = 6 kN 6 kN A B

RAY = 3.2 kN

RBY = 4.8 kN

¯

A.S.F.

1 2

X

2

- 6 kN

Constant

+

¯

3.2kN S.F.D.

-4.8kNLinear 1st Degree

+9.6kN

B.M.D. + +

11

+ ∑ = 0 AM + 3.2 * X – M1 = 0 ∴ M = + 3.2X kN Variable to X-1st Degree For Section 2-2: (3 ≤ X 5) ≤

12

+ ∑ = 0 Fx 6-6- N2 = 0 ∴ N2= 0 + ∑ = 0 Fy + 3.2 – 8 +V2 = 0 ∴ V2= 4.8 kN + ∑ = 0 AM + 3.2 * X2 – 8 (X2 – 3) – M2 = 0 ∴ M2 = - 4.8X2 +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

dxdM

= , PdxdV

=

N2

3.2 kN

6 kN

V2

M2

2

2

8 kN

6 kN

3

X