Effects of Stress and Shear on Materials

The materials used in the construction of a house must be strong enough to withstand the normal stresses of the environment. Of course, here in the Caribbean, the meaning of "normal" is stretched. Structures here must be able to withstand hurricane force winds and earthquakes as well. We divide the house into two components, the foundation and the frame.

The Foundation

The foundation of a house is generally considered to be the interface between the ground and floor of the house. The house rests on the foundation. The importance of a stable foundation is obvious. The typical choice of materials for a foundation is concrete and concrete block. It is strong, impervious to insects and rot, and has the ability to rest on an uneven surface while providing an even top surface on which the rest of the house can sit. Concrete is a mixture of cement and aggregates, such as sand, gravel, or stone. ("Blue bit" is a common stone used here in the VI). Cement is a compound comprised mostly of calcium and silicon. It is the "glue" that holds the aggregate together. There are over 200 cement plants in the U.S. and 4,000 to 5,000 ready mix plants that mix cement with aggregate and water to produce concrete. There is one plant here on St. Thomas, Devcon. They blast and grind sand and gravel for the aggregate, but the cement is shipped from Venezuela or from the continental US. The chemical composition and curing of cement is addressed in the chemistry section of this module.

The foundation must support the weight of the house without cracking or deforming. Below, we will examine how this relates to compressive stress and the elastic limit or yield point.

The Frame

The frame of the house refers to the support structure which rests upon the foundation, namely the walls and roof. There are two different basic building materials use for the frame. Concrete and concrete block walls provide great strength and durablilty, but at significant cost. Wood on the other hand, is easier to work with and generally cheaper. The walls must support the weight of the roof, so compressive stress is of some importance. But the main concern for both the walls and roof is the ability to withstand twisting and lateral deformation. Below, we discuss how this is related to shear stress.

Definition of Compressive Stress

Stress is defined as the perpendicular force per unit area applied to an object, in a way that compresses (compressive stress) or stretches (tensile stress) the object. Let's focus on compressive stress. If we stack a weight on a pillar that is standing on a solid surface, it will under compressive stress and the length will decrease by some distance Dl. If we double the weight, it will compress twice as much, 2 Dl. This linear property is common to almost all materials, provided you do exceed the elastic limit of the material. (The elastic limit is also called the ultimate strength or the breaking stress.) This is important information to know before you pour concrete pillars to support an entire house. The units for stress are exactly the same as for pressure. In fact, compressive stress is a pressure. The units are N/m² (Pascal = Pa) or lb/in² (psi).

The relative change in the length, defined as the change in length divided by the original length, is called the strain. If we do not exceed the elastic limit, the ratio of stress to strain is a constant defined to be the elastic modulus of the material. Since strain has no units, the units for the elastic modulus is the same as for pressure.

Of course, the picture shown is greatly exaggerated for clarity. Neither wood nor concrete will compress to this degree before exceeding the elastic limit, resulting in complete structural failure. A table of elastic moduli can be found below.

Problem: Calculate the change in length of a square concrete pillar with a cross section of 6 in by 6 in and 3 ft long that is place under a load of 10,000 lb.

Solution: Using the table below, the elastic modulus for concrete is about 21 GPa (G means giga = 10⁹) or 3.1 Mpsi (M means mega = 10⁶ and psi means lb/in²). Since the dimensions are in British units, we use the British system value for the elastic modulus, 3.1 Mpsi = 3.1 x 10⁶ lb/in². Using the definition of the elastic modulus, we have

E = 3.1 x 10⁶ lb/in² = ^{10,000 lb / _{6 in x 6 in}} / _{^Dl / 36 in}

Solving for Dl we get Dl = 3.2 x 10^-3 in (or about .08 mm). You would hardly notice such a small compression.

Ultimate Strength

The ultimate strength of the material is critical to determining how big the cross sectional area of the pillar should be. If you exceed this limit, you are courting disaster! The ultimate strength is defined by the maximum stress that can be applied. Below is a table with elastic moduli and average compressive ultimate strengths for a few of the standard building materials.
Note that the unit in the SI system is Pascals (N/m²) and in the British system is lb/in² or psi. G means giga = 10⁹ and M means mega = 10⁶. Do you need a review of prefixes?

Note that the value for the elastic limit is smaller than the elastic modulus. Although the units are the same, they are not the same quantity. If you apply 3,000 lb to a one inch square pillar, you are just at the elastic limit. The elastic modulus is E = 21 x 10⁹ Pa = 3,000,000 lb/in². Since E = stress/strain, strain = stress/E. A stress of 3,000 lb on the one square inch pillar yields a strain of .001. So a 1 m long pillar would be compressed by about 1 mm just before it fractured.

It is clear from the table that concrete is a bit more rigid than wood, but its elastic limit is actually less! Steel is more rigid and much stronger than either. Steel reinforcement (usually in the form of steel rods called "rebar") is very common in concete walls and foundations.

Shear

Shear stress, or just shear, is similar to stress, except that the force is applied such that the material is sheared or twisted. Look at the diagram. The block is "glued" to the table and a force F is applied to the top parallel to the table. (Note that table must be supplying an equal but opposite force -F if the block remains stationary.) The block distorts, the top surface moving a distance Dl. If h is the height of the block and A is the area of the top surface, then the shear modulus is defined as

Note that the distortion (Dl), length (h) and area (A) are defined differently here than for the elastic modulus. In general, the shear modulus is less than the elastic modulus. Usually a material will distort more than it will compress, but that is not always true. The foundation and walls of a house will suffer shear forces during earthquakes and high winds, but rarely do failures occur due to the wall or foundation materials themselves. Typically, it is the "fasteners" that fail from shear. The fasteners are the bolts that hold the sill to the foundation, or the nails that hold studs to the head plate, or the nails in the the hurricane clips that hold the roof to the walls.

Problem: Let's find the force necessary to shear a common 1" nail used to attach a hurricane clip to the rafter and header board. The nail is made of a low grade steel, with an elastic limit of about .3 GPa. The diameter of the nail is about 2 mm.

Solution: We start with,
Shear = .3 x 10⁹ Pa = F/A.
A = pr² for the cross sectional area of the nail
F/A = F/p.001².
Solving, we get F = 940 N (210 lb).

This may seem like a great deal of force. But in the Bernoulli Effect section we will see that during hurricanes, a 10 meter square roof can be subjected to lift forces in the range of 150,000 N (34,000 lb). So we would need a minimum of 150,000 / 940 = 160 nails! Don't skimp on the nails!

Here are a couple more examples from Broome Community College in NY. (Note that the elastic modulus is also called Young's modulus, and the symbol Y is used instead of E in these examples.)

• Compression of Concrete Pillar

• Shearing a Pin