9.49.9 Compression Lap Splices Minimum lap-splice lengths of rebars in compression Ls vary with nominal bar diameter db and yield strength y of the bars. For bar sizes No. 11 or less, the compression lap-splice length is the largest of 12 in or the values computed from Eqs. (9.65a) and (9.65b): L
0.0005 d  60,000 psi (9.65a) s yb y L
(0.0009  24)d  60,000 psi (9.65b) s y b y When is less than 3000 psi, the length of lap should be one-third greater than
c the values computed from the preceding equations. When the bars are enclosed by a spiral, the lap length may be reduced by 25%. For general practice, use 30 bar diameters for compression lap splices (Table 9.11). Spiral should conform to requirements of the ACI 318 Building Code: Spirals should extend from top of footing or slab in any story to the level of the lowest horizontal reinforcement in members supported above. The ratio of volume of spiral reinforcement to the total volume of the concrete core (out-to-out of spirals) should be at least that given in Art. 9.83. Minimum spiral diameter in cast-in-place con- TABLE 9.10 Tension Lap Splice Lengths for Grade 60 Uncoated Bars (Inches)

Upward heat flow 1.6 Downward heat flow 1.1 o for outdoor air film, 15-mi/hr wind (winter) 6.0 o for outdoor air film, 7.5-mi /hr wind (summer) 4.0 C for vertical air gap, 3/4 in or more wide 1.1 C for horizontal air gap, 3/4 in or more wide Upward heat flow 1.2 Downward heat flow 1.0 Q  CA (t
t ) (13.17) 2 1 where t2
t1 is the temperature difference causing the heat flow Q, and A is the cross-sectional area normal to the heat flow. Values of k and C for many building materials are given in tables in ASHRAE HandbookFundamentals and other publications on air conditioning. Thermal resistance, the resistance to flow of heat through a material or an assembly of materials, equals the reciprocal of the conductance: 1 C Thermal resistance R is used in HVAC calculations for determining the rate of heat flow per unit area through a nonhomogeneous material or a group of materials. Air Films. In addition to its dependence on the thermal conductivity or conductance of a given wall section, roof, or other enclosure, the flow of heat is also dependent on the surface air films on each side of the constructions. These air films are very thin and cling to the exposed surface on each side of the enclosures. Each of the air films possesses thermal conductance, which should always be considered in HVAC calculations. The indoor air film is denoted by i and the outdoor film by o . Values are given in Table 13.3 for these air films and for interior or enclosed air spaces of assemblies. In this table, the effects of air films along both enclosure surfaces have been taken into account in developing the air-film coefficients. Additional data may be obtained from the ASHRAE HandbookFundamentals. Air-to-Air Heat Transfer. In the study of heat flow through an assembly of building materials, it is always assumed that the rate of heat flow is constant and continues without change. In other words, a steady-state condition exists. For such a condition, the rate of heat flow in Btu per hour per unit area can be calculated from Q  UA(t
t ) (13.19) 2 1 where U  coefficient of thermal transmittance. Coefficient of Thermal Transmittance U. The coefficient of thermal transmittance U, also known as the overall coefficient of heat transfer, is the rate of heat flow under steady-state conditions from a unit area from the air on one side to the air on the other side of a material or an assembly when a steady temperature difference exists between the air on both sides. In calculation of the heat flow through a series of different materials, their individual resistances should be determined and totaled to obtain the total resistance Rt . The coefficient of thermal transmittance is then given by the reciprocal of the

The diagram indicates that R
Z cos  (15.15) L In a similar way, capacitive reactance in the circuit is indicated by and the phasor sum is shown as follows: The diagrams indicate that E
IZ cos  (15.16) C R
Z cos  (15.17) C Note that the phasors for inductance and capacitance are in opposite directions. Thus, when a circuit contains both inductance and capacitance, they can be added algebraically. If they are equal, they cancel each other, and the Z value is the same as R. If the inductance is greater, Z will be in the upper quadrant. A greater value of capacitance will throw Z into the lower quadrant. The diagram indicates that tan 
(15.18) R Kirchhoffs laws are applicable to alternating current circuits containing any combinations of resistance, inductance, and capacitance by means of phasor analysis: In a series circuit, the current I is equal in all parts of the circuit, but the total voltage drop is the phasor sum of the voltage drops in the parts. If the circuit has resistance R, inductance L, and capacitance C, the voltage drops must be added phasorially as described in the preceding. Equations (15.14) to (15.18) hold for ac series circuits. To find the voltage drop in each part of the circuit, compute E
E  E  E (phasorially) (15.19) Z R L C In a parallel circuit, the voltage E across each part is the same and the total current IZ is the vector sum of the currents in the branches, For parallel circuits, it is convenient to use the reciprocals of the resistance and reactances, or susceptances, respectively SR, SL, and SC. To find the current in each branch then, compute I
I  I  I (phasorially) (15.20) Z R L C Power in AC Circuits. Pure inductance or capacitance circuits store energy in either electric or magnetic fields and, when the field declines to zero, this energy is restored to the electric circuit. Power is consumed in an ac circuit only in the resistance part of the circuit and equals ER, the effective voltage across the resistance, times IR the effective current. ER and IR are in phase. In a circuit with impedance, however, the total circuit voltage EZ is out of phase with the current by the phase angle . In a series circuit, the current I is in phase with ER; the voltage EZ, on the other hand, is out of phase with ER by the angle . In parallel circuits, the voltage E is in phase with ER, but the current IZ is out of phase with ER. In both circuits, the power P is given by In series circuits, Er
E cos  and P
(E cos )IR. In parallel circuits, IR
I cos  and P
EI cos . In any circuit with impedance angle , therefore, the power