output_mds/page_21.md

(a) Singly reinforced section. (b) Strain distribution. (c) Stress distribution. (d) Internal forces.

Fig. 4-18 Steps in analysis of $M_n$ for singly reinforced rectangular sections.

The assumed stress distribution is given in Fig. 4-18c. Above the neutral axis, the stress-block model from Fig. 4-16 is used to replace the actual concrete stress distribution. The coefficient $\beta_1$ is multiplied by the depth to the neutral axis, $c$, to get the depth of the stress block, $a$. The concrete is assumed to carry no tension, so there is no concrete stress distribution below the neutral axis. At the level of the steel, the stress, $f_s$, is assumed to be equal to the steel yield stress, $f_y$. This corresponds to the assumptions that the steel strain exceeds the yield strain and that the steel stress remains constant after yielding occurs (Fig. 4-7).

The final step is to go from the stress distributions to the equivalent section forces shown in Fig. 4-18d. The concrete compression force, $C_c$, is equal to the volume under the stress block. For the rectangular section used here,

$$\begin{array}{cccc}& C_c=0.85f_c^{\mathrm{\prime }}b{\beta }_{1}c=0.85f_c^{\mathrm{\prime }}ba& & \text{(4-13b)}\end{array}C\_c\; =\; 0.85\; f\_c\text{'}\; b\; \backslash beta\_1\; c\; =\; 0.85\; f\_c\text{'}\; b\; a\; \backslash tag\{4-13b\}$$

The compression force in the concrete cannot be evaluated at this stage, because the depth to the neutral axis is still unknown. The tension force shown in Fig. 4-18d is equal to the tension steel area, $A_s$, multiplied by the yield stress, $f_y$. Based on the assumption that the steel has yielded, this force is known.

A key step in section analysis is to enforce section equilibrium. For this section, which is assumed to be subject to only bending (no axial force), the sum of the compression forces must be equal to the sum of the tension forces. So,

$$\begin{array}{cccc}& C_c=T& & \text{(4-2)}\end{array}C\_c\; =\; T\; \backslash tag\{4-2\}$$

or

$$0.85f_c^{\mathrm{\prime }}b{\beta }_{1}c=0.85f_c^{\mathrm{\prime }}ba=A_s{f}_{y}0.85\; f\_c\text{'}\; b\; \backslash beta\_1\; c\; =\; 0.85\; f\_c\text{'}\; b\; a\; =\; A\_s\; f\_y$$

The only unknown in this equilibrium equation is the depth of the stress block. So, solving for the unknown value of $a$,

$$\begin{array}{cccc}& a={\beta }_{1}c=\frac{A_s{f}_y}{0.85f_c^{\mathrm{\prime }}b}& & \text{(4-16)}\end{array}a\; =\; \backslash beta\_1\; c\; =\; \backslash frac\{A\_s\; f\_y\}\{0.85\; f\_c\text{'}\; b\}\; \backslash tag\{4-16\}$$

and

$$\begin{array}{cccc}& c=\frac{a}{{\beta }_1}& & \text{(4-17)}\end{array}c\; =\; \backslash frac\{a\}\{\backslash beta\_1\}\; \backslash tag\{4-17\}$$