Mechanics of Materials(ME-294)

Lecture 2

Statics and Strength of Materials

Statics is the study of forces acting in equilibrium on rigid bodies

• “Bodies” are solid objects, like steel cables, gear teeth, timber beams, and axle shafts (no liquids or gases);

• “rigid” means the bodies do not stretch, bend, or twist; • “equilibrium” means the rigid bodies are not accelerating.

In Strength of Materials, we keep the assumptions of bodies in equilibrium, but we drop the “rigid” assumption.

Real cables stretch under tension and real axle shafts twist under torsional load.

Strength of Materials Statics is the study of forces acting in equilibrium

on rigid bodies.• “Bodies” are solid objects, like steel cables, gear teeth,

timber beams, and axle shafts (no liquids or gases);

• “rigid” means the bodies do not stretch, bend, or twist;

• “equilibrium” means the rigid bodies are not accelerating.

In Strength of Materials, we keep the assumptions of bodies in equilibrium, but we drop the “rigid” assumption.

▫Real cables stretch under tension and real axle shafts twist under torsional load.

▫The most fundamental concepts in mechanics of materials are stress and strain.

Stress and Strain• The words “stress” and “strain” are used interchangeably:

“I’m feeling stressed” or “I’m under a lot of strain.”

• In engineering, these words have specific, technical meanings.

If you tie a steel wire to a hook in the ceiling and hang a weight on the lower end, the wire will stretch.

•Divide the change in length by the original length, and you have the strain in the wire.

•Divide the weight hanging from the wire by the wire’s cross sectional area, and you have the tensile stress in the wire.

Stress and strain are ratios. ▫The symbol for stress is σ, the lower case

Greek letter sigma. Stress has units of force per unit area

When SI units are used, force is expressed in newtons (N) and area in square meters (m2). Consequently, stress has units of newtons per square meter (N/m2), that is, pascals (Pa).

However, the pascal is such a small unit of

stress that it is necessary to work with large multiples, usually the megapascal (MPa).

1 MPa = 106 Pa = 106 N/ m2 = 1 N/mm2

1N/m2 = 1 Pa

The symbol for strain is ε, the lower case Greek letter epsilon.

Because normal strain is the ratio of two lengths, it is a dimension- less quantity, that is, it has no units. Therefore, strain is expressed simply as a number, independent of any system of units.

Numerical values of strain are usually very small, because bars made of structural materials undergo only small changes in length when loaded.

A prismatic bar is a straight structural member having the same cross section throughout its length, and an axial force is a load directed along the axis of the member, resulting in either tension or compression in the bar

The internal actions in the bar are exposed if we make an imaginary cut through the bar at section mn (Fig. 1-2c). Because this section is taken perpendicular to the longitudinal axis of the bar, it is called a cross section.

we will consider the tow bar of Fig. 1-1 and isolate a segment of it as a free body.

This action consists of continuously distributed stresses acting over the entire cross section, and the axial force P acting at the cross section is the resultant of those stresses. (The resultant force is shown with a dashed line in Fig. 1-2d.)

The force per unit area, or intensity of the forces distributed over a given section, is called the stress on that section.

Tensile and Compressive Stress and strain• A load that pulls is called a tensile

load and stress induced is tensile stress.

• If the load pushes, we call it a compressive load and stress is called compressive stress.

• The equations are the same.• Stresses act in a direction

perpendicular to Cross section, they are called normal stresses.

• Normal stresses may be either tensile or compressive.

• Another type of stress, called shear stress, that acts parallel to the surface.

• sign convention ▫ tensile stresses as positive(+)▫ compressive stresses as negative(-)

Stress-Strain Curve stress-strain curve illustrates the elastic and plastic zones. • If you hang a light weight to the

wire hanging from the ceiling, the wire stretches elastically; remove the weight and the wire returns to its original length.

• Apply a heavier weight to the wire, and the wire will stretch beyond the elastic limit and begins to plastically deform, which means it stretches permanently. Remove the weight and the wire will be a little longer (and a little skinnier) than it was originally.

• Hang a sufficiently heavy weight, and the wire will break.

• Two stress values are important in engineering design. ▫ The yield strength, σys, is the limit

of elastic deformation; beyond this point, the material “yields,” or permanently deforms.

▫ The ultimate tensile strength, σUTS (also called tensile strength, σTS) is the highest stress value on the stress-strain curve.

• The rupture strength is the stress at final fracture; this value is not particularly useful, because once the tensile strength is exceeded, the metal will break soon after.

• Young’s modulus, E, is the slope of the stress-strain curve before the test specimen starts to yield.

• The strain when the test specimen breaks is also called the elongation.

Many manufacturing operations on metals are performed at stress levels between the yield strength and the tensile strength.

▫Bending a steel wire into a paperclip, deep-drawing sheet metal to make an aluminum can, or rolling steel into wide-flange structural beams are three processes that permanently deform the metal, so σYS<σApplied .

▫ During each forming operation, the metal must not be stressed beyond its tensile strength, otherwise it would break, so σYS<σApplied<σUTS .

▫Manufacturers need to know the values of yield and tensile strength in order to stay within these limits.

Expression for elongation In this Strength of Materials course, almost all of the problems are

elastic, so there is a linear relationship between stress and strainTake an aluminum rod of length L, cross-sectional area A, and pull on it with a load P. The rod will lengthen an amount δ. We can calculate δ in three separate equations, or we can use algebra to find a simple equation to calculate δ directly.

Direct equation for calculating the change in length of the rod.

Deformation of a body due to self weight In Fig determine the total increase of length of a bar of constant cross

section hanging vertically and subject to its own weight as the only load.

The normal stress (tensile) over any horizontal cross section is caused by the weight of the material below that section. The elongation of the element of thickness dy shown is

where A denotes the cross-sectional area of the bar and γ its specific weight (weight/unit volume).

Integrating, the total elongation of the bar is

where W denotes the total weight of the bar. Note that the total elongation produced by the weight of the bar is equal to that produced by a load of half its weight applied at the end.

Principle of Superposition

Assignment#1

1. In 1989, Jason, a research-type submersible with remote TV monitoring capabilities and weighing 35 200 N, was lowered to a depth of 646 m in an effort to send back to the attending surface vessel photographs of a sunken Roman ship offshore from Italy. The submersible was lowered at the end of a hollow steel cable having an area of 452 × 10–6 m2 and E= 200 GPa. Determine the extension of the steel cable. Due to the small volume of the entire system, buoyancy may be neglected. (Note: Jason was the system that took the first photographs of the sunken Titanic in 1986.) weight of steel per unit volume is 77 kN/m3.

2. Two prismatic bars are rigidly fastened together and support a vertical load of 45 kN, as shown in Figure. The upper bar is steel having length 10 m and cross-sectional area 60 cm2. The lower bar is brass having length 6 m and cross-sectional area 50 cm2. For steel E = 200 GPa, for brass E = 100 GPa. Determine the maximum stress in each material. Specific weights of brass and steel are 84kN and 77kN.

3. A 70 kN compressive load is applied to a 5 cm diameter, 3 cm tall, steel cylinder. Calculate stress, strain, and deflection.

4. What tensile stress is required to produce a strain of 8×10−5 in aluminum? Report the answer in MPa. Aluminum has a Young’s modulus of E = 70 GPa.

5. Stress-strain curves for materials in compression differ from those in tension. Explain.

Quiz#1

•Chapter 1 and 2 –SOM by R. S. Khurmi •Assignment#1