Singly Reinforced Concrete Beam Strength Limits

Last updated January 5, 2024
By Ian Story

Moment Capacity

Given: axial force, P (tension positive), which may include axial tension from torsion

Assumptions: one layer of rebar near tension face is resisting all tension; ignore rebar layer near compression face

Variables:

$P =$ applied axial force (tension positive)
$F_C =$ resultant force at concrete compression block (tension positive / compression negative)
$F_T =$ resultant force at tension rebar (tension positive)
$d =$ distance from extreme compression fiber to center of tension rebar
$d_t =$ distance from extreme compression fiber to center of top rebar (typically in the compression zone)
$A_c =$ area of compression block
$A_s =$ cross-sectional area of tension rebar
$\epsilon_c = -0.003$ concrete strain at failure
$\epsilon_s =$ steel strain at failure
$b =$ width of concrete at extreme compression fiber
$a =$ height of compression block
$c =$ distance from extreme compression fiber to neutral axis
$\beta_1 = 0.85 =$ ratio between height of compression block and distance from extreme compression fiber to neutral axis
$d_m =$ moment arm / distance between compression resultant and tension resultant
$M_n =$ nominal moment capacity of beam
$E_s = 29,000,000\text{ psi}=$ modulus of elasticity of steel

Scenario 1 – Typical Beam

Scenario criteria:
$F_C > 0$
$\epsilon_s >= \epsilon_y$

$\Sigma F = P - F_C - F_T = 0$

$F_C=P-F_T$

$F_T=A_sf_y$

$A_c = -\frac{F_C}{0.85f_c'} = a*b$

$a = -\frac{F_C}{0.85f_c'b} = \beta_1 c$

$c = -\frac{F_C}{0.85f_c'b\beta_1}$

$d_m = d - \frac{a}{2}$

$M_n = F_Td_m$

$\epsilon_s = \epsilon_c\left(\frac{d}{c}-1\right)$

Putting all the pieces together gives:

$M_n = A_sf_y\left(d + \frac{P-A_sf_y}{2*0.85f_c'b}\right)$

$\epsilon_s = -\epsilon_c\left(\frac{0.85f_c'b\beta_1d}{P-A_sf_y}+1\right)>\epsilon_y$

Check assumptions:

$P < A_sf_y$

$\frac{0.85f_c'b\beta_1d}{P-A_sf_y}<\frac{\epsilon_y}{-\epsilon_c}-1$

Scenario 2 – Compression Controlled

Scenario criteria:
$\epsilon_s < \epsilon_y$

$\Sigma F = P - F_C - F_T = 0$

$F_C=P-F_T$

$F_T=A_s\epsilon_sE_s$

$A_c = -\frac{F_C}{0.85f_c'} = a*b$

$a = -\frac{F_C}{0.85f_c'b} = \beta_1 c$

$c = -\frac{F_C}{0.85f_c'b\beta_1}$

$d_m = d - \frac{a}{2}$

$M_n = F_Td_m$

$\epsilon_s = \epsilon_c\left(\frac{d}{c}-1\right)$

Use some algebra to find $\epsilon_s$ :

$\epsilon_s = 0.003\left(\frac{0.85f_c'b\beta_1d}{F_C}+1\right)$

$\epsilon_s = 0.003\left(\frac{0.85f_c'b\beta_1d}{P-A_s\epsilon_sE_s}+1\right)$

$\left(\epsilon_s\frac{1}{0.003}-1\right)(P-A_s\epsilon_sE_s) = 0.85f_c'b\beta_1d$

$\epsilon_s^2\left(\frac{A_sE_s}{0.003}\right)+\epsilon_s\left(-A_sE_s-\frac{P}{0.003}\right)+\left(P+0.85f_c'b\beta_1d\right) = 0$

Solve for $\epsilon_s$ using quadratic formula then plug in to find $M_n$

$M_n = A_s\epsilon_sE_s\left(d + \frac{P-A_s\epsilon_sE_s}{2*0.85f_c'b\beta_1}\right)$

Scenario 3 – No Compression

Scenario criteria:
$F_C = 0$
$\epsilon_s >= 0.005$

$\Sigma F =P-F_{T1}-F_{T2}=0$

$\Sigma M_{h/2} =M_n+F_{T1}d_{h1}-F_{T2}d_{h2}=0$

Maximum moment occurs at

$F_{T2}=A_{s2}f_y$

$F_{T1} = P-F_{T2}$

$M_n=F_{T2}d_{h2}-F_{T1}d_{h1}$

Putting the pieces together gives:

$M_n=A_{s2}f_yd_{h2}+(A_{s2}f_y-P)d_{h1}$

Tension Limit

Given moment

$P=\left(d-\frac{M}{A_sf_y}\right)(2*0.85f_cb)-A_sf_y$

This equation works as long as $F_C<0$ , which occurs at:

$P=A_sf_y$

$M=\left(d-\frac{2A_sf_y}{2*0.85f_c'b}\right)A_sf_y$

For moments below this threshold, the maximum tension will put the member into pure tension flexure (no compression)

$\Sigma F =P-F_{T1}-F_{T2}=0$

$\Sigma M_{h/2} =M_n-F_{T1}d_{h1}+F_{T2}d_{h2}=0$

Unless the applied moment is very small, maximum tension occurs at

$F_{T1}=A_{s1}f_y$

$F_{T2}d_{h2}=A_{s1}f_yd_{h1}-M_n$

$F_{T2}=\frac{A_{s1}f_yd_{h1}-M_n}{d_{h2}}$

$P=F_{T1}+F_{T2}=A_{s1}f_y+\frac{A_{s1}f_yd_{h1}-M_n}{d_{h2}}$

Where the second term has a maximum value of $A_{s2}f_y$

Compression Limit

Given moment

Scenario 1: Compression Controlled

Scenario criteria:
$\epsilon_s < 0.005$

$F_T=A_s\epsilon_sE_s$

With a bunch of algebra:

$\epsilon_s^2(2dA_sE_s)+\epsilon_s(\epsilon_cdA_sE_s-2M)-2\epsilon_cM=0$

Solve for $\epsilon_s$ using quadratic formula the plug into the following equation to find P

$P=\frac{A_s\epsilon_sE_s(\epsilon_c+\epsilon_s)-0.85f_c'bd\beta_1\epsilon_c}{\epsilon_c+\epsilon_s}$

Scenario 2: Tension Controlled

Scenario criteria:
$\epsilon_s >= 0.005$