Structural capacity, or ultimate strength, is that property of a structural member that serves as a measure of is ability to support all potential loads without severe cracking or excessive deformations. To indicate when the limit on load-carrying usefulness has been reached, design specifications for the various structural materials establish allowable unit stresses or design strengths that may not be exceeded under maximum loading. Structural theory provides methods for calculating unit stresses and for estimating deformations. Many of these methods are presented in the rest of this section.
If a structure and its components are so supported that, after a very small deformation occurs, no further motion is possible, they are said to be in equilibrium.
Under such circumstances, internal forces, or stresses, exactly counteract the loads.
Several useful conclusions may be drawn from the state of static equilibrium:
Since there is no translatory motion, the sum of the external forces must be zero;
and since there is no rotation, the sum of the moments of the external forces about any point must be zero.
For the same reason, if we consider any portion of the structure and the loads on it, the sum of the external and internal forces on the boundaries of that section must be zero. Also, the sum of the moments of these forces must be zero.
In Fig. 5.1, for example, the sum of the forces RL and RR needed to support the roof truss is equal to be the 20-kip load on the truss (1 kip 1 kilopound 1000 lb 0.5 ton). Also, the sum of moments of the external forces is zero about any point. About the right end, for instance, it is 40 x 15 - 30 x 20 = 600 - 600.
In Fig. 5.2 is shown the portion of the truss to the left of section AA. The internal forces at the cut members balance the external load and hold this piece of the truss in equilibrium.
Generally, it is convenient to decompose the forces acting on a structure into components parallel to a set of perpendicular axes that will simplify computations.
For example, for forces in a single plane a condition commonly encountered in building design the most useful technique is to resolve all forces into horizontal and vertical components. Then, for a structure in equilibrium, if H represents the horizontal components, V the vertical components, and M the moments of the components about any point in the plane,
Unit Stress and Strain
To ascertain whether a structural member has adequate load-carrying capacity, the designer generally has to compute the maximum unit stress produced by design loads in the member for each type of internal force tensile, compressive, or shearing and compare it with the corresponding allowable unit stress.
When the loading is such that the unit stress is constant over a section under consideration, the stress may be obtained by dividing the force by the area of the section. But in general, the unit stress varies from point to point. In that case, the unit stress at any point in the section is the limiting value of the ratio of the internal force on any small area to that area, as the area is taken smaller and smaller.
Sometimes in the design of a structure, unit stress may not be the prime consideration.
The designer may be more interested in limiting the deformation or strain.
Deformation in any direction is the total change in the dimension of a member in that direction.
Unit strain in any direction is the deformation per unit of length in that direction.
When the loading is such that the unit strain is constant over a portion of a member, it may be obtained by dividing the deformation by the original length of that portion. In general, however, the unit strain varies from point to point in a member. Like a varying unit stress, it represents the limiting value of a ratio.
For many materials, unit strain is proportional to unit stress, until a certain stress, the proportional limit, is exceeded. Known as Hookes law, this relationship may be written as
Hence, when the unit stress and modulus of elasticity of a material are known, the unit strain can be computed. Conversely, when the unit strain has been found, the unit stress can be calculated.
When a member is loaded and the unit stress does ot exceed the proportional limit, the member will return to its original dimensions when the load is removed.
The elastic limit is the largest unit stress that can be developed without a permanent deformation remaining after removal of the load.
Some materials possess one or two yield points. These are unit stresses in the region of which there appears to be an increase in strain with no increase or a small decrease in stress. Thus, the materials exhibit plastic deformation. For materials that do not have a well-defined yield point, the offset yield strength is used as a measure of the beginning of plastic deformation.
The offset yield strength, or proof stress as it is sometimes referred to, is defined as the unit stress corresponding to a permanent deformation, usually 0.01% (0.0001 in / in) or 0.20% (0.002 in / in).
Constant Unit Stress
The simplest cases of stress and strain are those in which the unit stress and strain are constant. Stresses due to an axial tension or compression load or a centrally applied shearing force are examples; also an evenly applied bearing load. These loading conditions are illustrated in Figs. 5.3 to 5.6.
For the axial tension and compression loadings, we take a section normal to the centroidal axis (and to the applied forces). For the shearing load, the section is taken along a plane of sliding. And for the bearing load, it is chosen through the plane of contact between the two members.
Since for these loading conditions, the unit stress is constant across the section, the equation of equilibrium may be written
P = AÆ’
where P load
Æ’ a tensile, compressive, shearing, or bearing unit stress A cross-sectional area for tensile or compressive forces, or area on which sliding may occur for shearing forces, or contact area for bearing loads
For torsional stresses, see Art. 5.4.2.
The unit strain for the axial tensile and compressive loads is given by the equation
[Since long compression members tend to buckle, Eqs. (5.21) to (5.23) are applicable only to short members.]
While tension and compression strains represent a simple stretching or shortening of a member, shearing strain represents a distortion due to a small rotation.
The load on the small rectangular portion of the member in Fig. 5.5 tends to distort it into a parallelogram. The unit shearing strain is the change in the right angle, measured in radians.
Modulus of rigidity, or shearing modulus of elasticity, is defined by
Within the elastic limit, when a material is subjected to axial loads, it deforms not only longitudinally but also laterally. Under tension, the cross section of a member decreases, and under compression, it increases. The ratio of the unit lateral strain to the unit longitudinal strain is called Poissons ratio.
For many materials, this ratio can be taken equal to 0.25. For structural steel, it is usually assumed to be 0.3.
Assume, for example, that a steel hanger with an area of 2 in2 carries a 40-kip (40,000-lb) load. The unit stress is 40,000/2, or 20,000 psi. The unit tensile strain, taking the modulus of elasticity of the steel as 30,000,000 psi, is 20,000/ 30,000,000, or 0.00067 in / in. With Poissons ratio as 0.3, the unit lateral strain is 0.3 x 0.00067, or a shortening of 0.00020 in / in.
When the temperature of a body changes, its dimensions also change. Forces are required to prevent such dimensional changes, and stresses are set up in the body by these forces.
If is the coefficient of expansion of the material and T the change in temperature, the unit strain in a bar restrained by external forces from expanding or contracting is
When a bar is stressed, energy is stored in it. If a bar supporting a load P undergoes a deformation e the energy stored in it is
This equation assumes the load was applied gradually and the bar is not stressed beyond the proportional limit. It represents the area under the load-deformation curve up to the load P. Applying Eqs. (5.20) and (5.21) to Eq. (5.28) gives another useful equation for energy:
Since AL is the volume of the bar, the term Æ’2/2E indicates the energy stored per unit of volume. It represents the area under the stress-strain curve up to the stress Æ’. Its value when the bar is stressed to the proportional limit is called the modulus of resilience. This modulus is a measure of the capacity of the material to absorb energy without danger of being permanently deformed and is of importance in designing members to resist energy loads.
Equation (5.28) is a general equation that holds true when the principle of superposition applies (the total deformation produced by a system of forces is equal to the sum of the elongations produced by each force). In the general sense, P in Eq. (5.28) represents any group of statically interdependent forces that can be completely defined by one symbol, and e is the corresponding deformation.
The strain-energy equation can be written as a function of either the load or the deformation.
For axial tension or compression:
where M bending moment
@ angle of rotation of one end of the beam with respect to the other
L length of beam
I moment of inertia of the cross section
E modulus of elasticity
For beams carrying transverse loads, the strain energy is the sum of the energy for bending and that for shear.
See also Art. 5.10.4.
very helpfull thanksss