Strength 1st Sem 2015-2016

STRENGTH OF MATERIALS

W

P = ?

STATICS SOLUTION: STRENGTH OF MATERIALS SOLUTION:- external effects of forces on rigid bodies

∑M about the pin support

- the solution is extended further

- investigate the bar itself to be sure that it will neither break nor be so flexible that it bends without supporting the load!

- also known as mechanics of materials

- it deals with the elastic behavior of loading materials, relationships between externally applied loads and internal resisting forces associated deformations

F3

F4

F1

F2

STATICS SOLUTION:

ARBITRARY SHAPED BODY

ANALYSIS OF INTERNAL FORCES

- start by determining the resultant of the applied forces to determine whether or not the body remains at

rest

If the resultant is

zero we have static

equilibrium!

ANALYSIS OF INTERNAL FORCES

ARBITRARY SHAPED BODY

F1

F2

F3

F4

STRENGTH OF MATERIALS SOLUTION:

- additional investigation on the internal distribution of the forces must be done

F2

1

1

cut an exploratory section 1-1 through

the body and exposing the internal forces

that are necessary to maintain the equilibrium

F1

F3

F4

F1

F2

O

x

y

z

Mxy

Mxx

Mxz

Pxy

Pxz

Pxx

Pxx = Axial Force, P - this component measures the pulling (or pushing) action perpendicular to the section

a. pull – tensile force that tends to elongate the member

b. push – compressive force that tends to shorten the member

Pxy , Pxz = Shear Forces, V- these are components of the total resistance to sliding to one side of the exploratory section past the other

2 COMPONENTS: 1.Vx 2.Vy

Mxx = Torque, T- this component measures the resistance to twisting the member

Mxy , Mxz = Bending Moments- these components measures the resistance to bending the member about the y or x axis

2 COMPONENTS: 1.Mx 2.My

a. Equation of equilibrium

TYPE OF CONNECTION EQUIVALENT REACTION OR SUPPORT

1. CABLE

2. ROLLER

3. EXTERNAL PIN/HINGE

θ

TYPE OF CONNECTION EQUIVALENT REACTION OR SUPPORT

4. INTERNAL PIN

5. FIXED SUPPORT

θ

B. Simple stresses

σ(sigma) = unit strength of a material or stressP = axial force (compressive or tensile)A = cross sectional Area (uniform)

TYPES OF SIMPLE STRESS: 1. Axial Stress

a. Compressive Stressb. Tensile Stress

2. Shearing Stress

3. Bearing Stress

4. Stresses in Thin-walled Cylinder

B. Simple stresses ( axial stress)AXIAL STRESS is the stress due to axial force or force acting at the centroid of

resisting cross-sectional area.

Tensile Stress: (σt )

At

Ac

Compressive Stress: (σt )

P

P

tensile area normal to the applied load

compressive area normal to the applied load

B. Simple stressesSelection of proper material and to proportion to enable the structure or

machine to perform its function efficiently is on of the basic problem facing an engineer.

For this purpose, it is essential to determine the strength, stiffness and other properties of materials.

Example:

Consider two bars of equal length but different materials, suspended from a common support as shown in the figure:

BAR 1 BAR 2

500 n 5000 n

Which of the material is stronger?

BAR 1 BAR 2

500 n 5000 n

Cross-sectional Area = 10 mm2

Cross-sectional Area = 1000 mm2

SOLUTION:

COMPARE THEIR STRENGTH BY REDUCING THE DATA TO LOAD CAPACITY PER UNIT AREA

Cross-sectional Area = 10 mm2

Cross-sectional Area = 1000 mm2

BAR 1 BAR 2

BAR 1 BAR 2

500 n 5000 n

Cross-sectional Area = 10 mm2

Cross-sectional Area = 1000 mm2

Therefore:

the material of BAR 1 is ten (10) times as strong as the material of BAR 2

B. Illustrative problems

1. A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section. Axial loads are applied at the positions indicated. Determine the stress in each section.

BRONZEA = 1.2 in2

ALUMINUMA = 1.8 in2 STEEL

A = 1.6 in2

1.3 ft 1.6 ft 1.7 ft

4000 lb 9000 lb 2000 lb 7000 lb

1

1

2

2

3

3

4000 lb

1

1

Pbr

9000 lb

SOLUTION:

After cutting and making the appropriate free body diagrams, we must determine the axial loads in each section!

BRONZE (tension)

4000 lb

Pal

2

2

2000 lb9000 lb

ALUMINUM

(compression)

STEEL

4000 lb 9000 lb 2000 lb 7000 lb

3

3

Pal

(compression)

B. Illustrative problems

2. For the truss shown, determine the stress in member AC and BD. The cross-sectional area of each member is 900 mm2.

A

B D

E

F

GH

C

30 kN

Hx

70 kNHYAY

4 panels at 4m = 16 m

3 m

1

1

2

2

SOLUTION:

Make three assumptions for the elementary analysis of the trusses.

1. Weights of the members are neglected.

2. All connections are smooth pin.

3. All external loads are applied directly to the pins.

+

Solve for the values of the external reactions of the entire truss.

A

B D

E

F

GH

C

30 kN

Hx

70 kNHYAY

3 m

4 m 4 m 4 m 4 m

A

AY

AC

AB

1

1

+

+

3

4

5

B

C

3 m

(tension)

4

A

B

C

30 kNAY

2

2

3

E

BD

BE

CE

4m 4m

5

3 m

+

(compression)SOLVING FOR STRESSES IN MEMBERS AC AND BD:

(tension)

(compression)

D

B. Illustrative problems3. The bars of the pin-connected frame shown are each 30 mm by 60 mm in section. Determine the maximum load P that can be applied so that the stresses will not exceed 100 MN/m2 in tension or 80 MN/m2 in compression.

A

B

C

P

8 m

10 m

6 m

θ

Consider the FBD and force polygon for joint B.

θθ

P

AB

B

BC P

AB

BC

FBD @ B

FORCE POLYGON

A θ

Consider the FBD of joint B.

AB

AC

RA

= 6/10 P

C

A

SOLVING FOR THE MAXIMUM LOAD P THAT CAN BE APPLIED ACCORDING TO THE REQUIRED LIMITS…

LIMITING FACTORS:

for AB: (compression)

0.6 P = (30 x 60) (80)

PAB = 240,000 N

for BC: (compression)

0.8 P = (30 x 60) (80)

PBC = 180,000 Nfor AC: (tension)

0.48 P = (30 x 60) (100)

PAC = 375,000 N

Therefore, the maximum safe load:

P = 180,000 N = 180 kN

250 kN (1000 N/kN) = (π/4) ((40,000 mm2 - Di2) (50 MN/m2)

((1x 106 N/m2) / (MN/m2)) ( 1x 10-6 m2/mm2)

3. A cast-iron column supports an axial compressive load of 250 kN. Determine the inside diameter of the column if its outside diameter is 200 mm and limiting compressive stress is 50 MPa.

B. Illustrative problems

D0= 200 mm

Di = ?

A = (π/4) D2

Uniform Cross-sectional Area Normal to the Applied Load:

P = 250 kN

for Solid Cylinders: for Hollow Cylinders:

A = (π/4) (Do – Di)2 A = (π/4) (200 – Di)2 A = (π/4) (40,000 – Di2) P = Aσ - load formula

5000 mm2 = (π/4) (40,000 mm2 – Di2)

(40,000 mm2 – Di2) = 6366.2 mm2

Di = 183.4 mm

for Hollow Cylinders: