output_mds/page_22.md

With the depth to the neutral axis known, the assumption of yielding of the tension steel can be checked. From similar triangles in the linear strain distribution in Fig. 4-18b, the following expression can be derived:

$$\frac{{ϵ}_s}{d-c}=\frac{{ϵ}_{cu}}{c}\backslash frac\{\backslash epsilon\_s\}\{d\; -\; c\}\; =\; \backslash frac\{\backslash epsilon\_\{cu\}\}\{c\}$$

$$\begin{array}{cccc}& {ϵ}_s=\left(\frac{d-c}{c}\right){ϵ}_{cu}& & \text{(4-18)}\end{array}\backslash epsilon\_s\; =\; \backslash left(\; \backslash frac\{d\; -\; c\}\{c\}\; \backslash right)\; \backslash epsilon\_\{cu\}\; \backslash tag\{4-18\}$$

To confirm the assumption that the section is under-reinforced and the steel is yielding, show

$$\begin{array}{cccc}& {ϵ}_s\ge {ϵ}_y=\frac{f_y}{E_s}=\frac{f_y\text{ (ksi)}}{29,000\text{ ksi}}& & \text{(4-19)}\end{array}\backslash epsilon\_s\; \backslash ge\; \backslash epsilon\_y\; =\; \backslash frac\{f\_y\}\{E\_s\}\; =\; \backslash frac\{f\_y\; \backslash text\{\; (ksi)\}\}\{29,000\; \backslash text\{\; ksi\}\}\; \backslash tag\{4-19\}$$

Once this assumption is confirmed, the nominal-section moment capacity can be calculated by referring back to the section forces in Fig. 4-18d. The compression force is acting at the middepth of the stress block, and the tension force is acting at a distance $d$ from the extreme compression fiber. Thus, the nominal moment strength can be expressed as either the tension force or the compression force multiplied by the moment arm, $d - a/2$:

$$\begin{array}{cccc}& M_n=T(d-\frac{a}{2})=C_c(d-\frac{a}{2})& & \text{(4-20)}\end{array}M\_n\; =\; T\; \backslash left(\; d\; -\; \backslash frac\{a\}\{2\}\; \backslash right)\; =\; C\_c\; \backslash left(\; d\; -\; \backslash frac\{a\}\{2\}\; \backslash right)\; \backslash tag\{4-20\}$$

For singly reinforced sections, it is more common to express the nominal moment strength using the definition of the tension force as

$$\begin{array}{cccc}& M_n=A_s{f}_y(d-\frac{a}{2})& & \text{(4-21)}\end{array}M\_n\; =\; A\_s\; f\_y\; \backslash left(\; d\; -\; \backslash frac\{a\}\{2\}\; \backslash right)\; \backslash tag\{4-21\}$$

This simple expression can be used for all singly reinforced sections with a rectangular (constant width) compression zone after it has been confirmed that the tension steel is yielding. The same fundamental process as used here to determine $M_n$ for singly reinforced rectangular sections will be applied to other types of beam sections in the following parts of this chapter. However, the reader is urged to concentrate on the process rather than the resulting equations. If the process is understood, it can be applied to any beam section that may be encountered.

Example 4-1 Calculation of $M_n$ for a Singly Reinforced Rectangular Section

For the beam shown in Fig. 4-19a, calculate $M_n$ and confirm that the area of tension steel exceeds the required minimum steel area given by Eq. (4-11). The beam section is made of concrete with a compressive strength, $f_c' = 4000$ psi, and has four No. 8 bars with a yield strength of $f_y = 60$ ksi.

For this beam with a single layer of tension reinforcement, it is reasonable to assume that the effective flexural depth, $d$, is approximately equal to the total beam depth minus 2.5 in. This accounts for a typical concrete clear cover of 1.5 in., the diameter of the stirrup (typically a No. 3 or No. 4 bar) and half the diameter of the beam longitudinal reinforcement. Depending on the sizes of the stirrup and longitudinal bar, the dimension to the center of the steel layer will vary slightly, but the use of 2.5 in. will be accurate enough for most design work unless adjustments in reinforcement location are required to avoid rebar interference at connections with other members. Small bars are often used in the compression zone to hold the stirrups in position, but these bars normally are ignored unless they were specifically designed to serve as compression-zone reinforcement.