CE Board Exam Randomizer

⬅ Back to Subject Topics

CECC-483 Reference Formula Sheet — Working Stress Design

Quick-reference WSD/ADM permissible stresses, transformed-section properties, and resisting-moment formulas as listed in the CECC-483 / NSCP review notes.

Permissible service-load stresses:

$$f_c \le 0.45\,f'_c \quad\text{(flexure)}$$ $$f_s \le 0.50\,f_y \quad\text{(flexural reinforcement)}$$ $$\text{Grade 280 MPa: }f_{sa}=140\text{ MPa}\qquad\text{Grade 420 MPa: }f_{sa}=170\text{ MPa}$$

Material constants:

$$E_s = 200{,}000\text{ MPa}$$ $$E_c = w_c^{1.5}\,0.043\sqrt{f'_c}\;\text{(MPa, }w_c\text{ in kg/m}^3\text{; }1440\le w_c\le 2560\text{)}$$ $$E_c = 4700\sqrt{f'_c}\;\text{(normal-weight concrete, MPa)}$$ $$n = E_s/E_c \quad\text{(nearest whole number, not less than 6)}$$

Location of the neutral axis (analysis):

$$\Sigma M_{N.A.}=0\;\Rightarrow\;\dfrac{b(kd)^2}{2}-nA_s(d-kd)=0$$ $$I_{N.A.}=\tfrac{1}{3}b(kd)^3 + nA_s(d-kd)^2$$

Resisting moments — concrete and steel:

$$f_c = \dfrac{M_c\,(kd)}{I_{N.A.}}\qquad \dfrac{f_s}{n}=\dfrac{M_s(d-kd)}{I_{N.A.}}$$ $$C = \tfrac{1}{2}f_c\,(kd)\,b\qquad T = A_s f_s$$ $$\text{Moment arm: } jd = d-\tfrac{kd}{3}\;,\quad j=1-\tfrac{k}{3}$$ $$M_c = C\,jd = \tfrac{1}{2}f_c(kj)bd^2\qquad M_s = T\,jd = A_s f_s\,jd$$

Constants k and steel ratio ρ:

$$k = \dfrac{n}{n+f_s/f_c}\quad\text{(design only)}$$ $$k = \sqrt{2n\rho + (n\rho)^2}-n\rho\quad\text{(investigation only)}$$ $$\rho = \dfrac{A_s}{bd}$$

Working Stress Design Method (Alternate Design Method)

The Working Stress Design (WSD) method — also called the Alternate Design Method (ADM) — designs reinforced-concrete sections so that, under service (un-factored) loads, the maximum bending stresses in concrete and steel never exceed their allowable values. Both materials are assumed to behave elastically, plane sections remain plane, and the tensile strength of concrete is neglected.

Material constants used throughout WSD:

$$E_s = 200{,}000 \text{ MPa}$$ $$E_c = 4700\sqrt{f'_c}\ \text{(MPa)}$$ $$n=\dfrac{E_s}{E_c}\quad\text{(modular ratio)}$$ $$f_{ca}=0.45f'_c\quad\text{(allowable concrete compressive stress)}$$ $$f_{sa}=\text{allowable steel tensile stress (typ. 140 MPa for Grade 40, 170 MPa for Grade 60)}$$ $$r=\dfrac{f_{sa}}{f_{ca}}$$

Internal forces in a singly-reinforced beam (cracked section):

With $kd$ measured from the extreme compression fibre to the neutral axis (N.A.) and $jd=d-\dfrac{kd}{3}$ the lever arm between the resultant compressive force $C$ and the steel tensile force $T$:

$$C=\tfrac{1}{2}f_c\,b\,(kd)\qquad T=A_s f_s$$ $$\sum F=0\;\Rightarrow\;\tfrac{1}{2}f_c\,b\,(kd)=A_s f_s$$ $$\rho=\dfrac{A_s}{bd}\;\Rightarrow\;\rho=\dfrac{1}{2}\dfrac{f_c}{f_s}k$$ $$M=C\cdot jd=\tfrac{1}{2}f_c(kj)bd^2\qquad\text{or}\qquad M=T\cdot jd=A_s f_s\,jd$$

Two expressions for the depth ratio $k$:

(1) From the moment of transformed areas about the N.A. (used for analysis):

$$\dfrac{b(kd)^2}{2}-nA_s(d-kd)=0$$ $$k^2+2n\rho\,k-2n\rho=0$$ $$\boxed{\;k=\sqrt{(n\rho)^2+2n\rho}-n\rho\;}$$

(2) From strain compatibility (used for design, when $f_c$ and $f_s$ are pegged at their allowable values):

$$\dfrac{\varepsilon_c}{\varepsilon_s}=\dfrac{kd}{d-kd}=\dfrac{k}{1-k}$$ $$\dfrac{f_c}{f_s}\cdot\dfrac{E_s}{E_c}=\dfrac{k}{1-k}\;\Rightarrow\;\boxed{\;k=\dfrac{n}{n+r}\;}\qquad j=1-\dfrac{k}{3}$$

Balanced steel ratio $\rho_e$: the steel ratio at which steel and concrete reach their allowable stresses simultaneously.

$$\rho_e=\dfrac{1}{2}\dfrac{f_{ca}}{f_{sa}}k=\dfrac{k}{2r}=\dfrac{n}{2r(n+r)},\qquad r=\dfrac{f_{sa}}{f_{ca}}$$

Under- vs. over-reinforced singly sections:

  1. Under-reinforced  —  $\rho=\dfrac{A_s}{bd}\lt\rho_e$. Steel reaches $f_{sa}$ first. $M_{allow}=T\cdot jd=A_s f_{sa}\,jd$.
  2. Over-reinforced  —  $\rho=\dfrac{A_s}{bd}\gt\rho_e$. Concrete reaches $f_{ca}$ first. $M_{allow}=C\cdot jd=\tfrac{1}{2}f_{ca}(kj)bd^2$.

ACI 318-05 detailing limits used in the examples:

  1. Minimum clear spacing between parallel bars in a layer: $d_b$ but not less than $25$ mm.
  2. Minimum clear vertical spacing between layers of bars: $25$ mm.
  3. Minimum concrete cover (not exposed to weather or earth): $40$ mm. Exposed: $50$ mm for No. 19 to No. 57 bars, $40$ mm for No. 16 and smaller.
  4. Economical proportions (when not architecturally restricted): $\dfrac{2}{3}\gt\dfrac{b}{d}\gt\dfrac{1}{2}\;\;\text{or}\;\;\dfrac{3}{2}b\lt d\lt 2b$.

Doubly-reinforced beams. When the section is dimensionally restricted and $M_{ext}\gt M_{int}$ for a singly-reinforced solution, add compression steel $A'_s$ at depth $d'$ from the top fibre. The external moment is split into two couples:

$$M=M_1+M_2$$ $$M_1=\tfrac{1}{2}f_c(kj)bd^2\quad\text{(ideal singly part, resisted by }A_{s1}\text{)}$$ $$A_{s1}=\dfrac{M_1}{f_s\,jd}\quad\text{or}\quad A_{s1}=\rho_e\,b\,d$$ $$M_2=M-M_1$$ $$A_{s2}=\dfrac{M_2}{f_s(d-d')}\qquad A_s=A_{s1}+A_{s2}$$

Stress in the concrete at the level of the compression steel:

$$f_{c1}=\left(\dfrac{kd-d'}{kd}\right)f_c$$

ACI requires using $2n$ when transforming compression steel to concrete (to account for creep). The compression-steel stress is therefore:

$$f'_s=2n\,f_{c1}=2n\left(\dfrac{kd-d'}{kd}\right)f_c\;\le\;f_{sa}$$ $$A'_s=\dfrac{M_2}{f'_s(d-d')}\quad\text{if}\quad 2nf_{c1}\lt f_{sa}$$ $$A'_s=\dfrac{M_2}{(f_{sa}-f_{c1})\,jd}\quad\text{if}\quad 2nf_{c1}\gt f_{sa}$$

Problem: Singly-Reinforced Beam — Analysis

Calculate the maximum flexural stresses in concrete and steel for a rectangular reinforced-concrete beam with width $b=300$ mm, effective depth $d=420$ mm, total depth $h=500$ mm, and tension reinforcement $A_s=3\,\phi 28=1847\text{ mm}^2$ placed in a single bottom row. The applied service moment is $M=95\text{ kN}\!\cdot\!\text{m}$ and the modular ratio is $n=9$.

Step 1 — Steel ratio and $n\rho$. Because this is an analysis problem (As is known), use $k=\sqrt{(n\rho)^2+2n\rho}-n\rho$:

$$\rho=\dfrac{A_s}{bd}=\dfrac{1847}{300\times 420}=0.0147$$ $$n\rho=9(0.0147)=0.1323$$

Step 2 — Solve for $k$ and $j$.

$$k=\sqrt{2(0.1323)+(0.1323)^2}-0.1323=0.399$$ $$j=1-\dfrac{k}{3}=1-\dfrac{0.399}{3}=0.867$$

Step 3 — Concrete flexural stress from $M=\tfrac{1}{2}f_c(kj)bd^2$:

$$f_c=\dfrac{2M}{bd^2\,kj}=\dfrac{2(95\times 10^6)}{300\,(420)^2\,(0.399)(0.867)}=10.37\text{ MPa}$$

Step 4 — Steel flexural stress from $M=A_s f_s\,jd$:

$$f_s=\dfrac{M}{A_s\,jd}=\dfrac{95\times 10^6}{1847\,(0.867)(420)}=141.3\text{ MPa}$$

The concrete is stressed to $f_c=10.37$ MPa and the tensile steel to $f_s=141.3$ MPa under the applied service moment.

Problem: Singly-Reinforced Beam — Design (Simply Supported)

Design the simply-supported beam of span $L=6\text{ m}$ to carry a service dead load of $35\text{ kN/m}$ and service live load of $15\text{ kN/m}$. Use $f'_c=21\text{ MPa}$, $f_y=300\text{ MPa}$, and $\phi 28$ mm main bars with $\phi 12$ mm stirrups and $40$ mm clear cover.

Step 1 — Service moment. Total service load $w=35+15=50\text{ kN/m}$. For a simply supported beam:

$$M=\dfrac{wL^2}{8}=\dfrac{50(6)^2}{8}=225\text{ kN}\!\cdot\!\text{m}$$

Step 2 — Allowable stresses and elastic constants.

$$f_{sa}=140\text{ MPa},\qquad f_{ca}=0.45f'_c=0.45(21)=9.45\text{ MPa}$$ $$E_s=200{,}000\text{ MPa},\qquad E_c=4700\sqrt{21}=21{,}538\text{ MPa}$$ $$n=\dfrac{E_s}{E_c}=\dfrac{200{,}000}{21{,}538}=9.3\;\;\Rightarrow\;\;\text{say }n=9$$ $$r=\dfrac{f_{sa}}{f_{ca}}=\dfrac{140}{9.45}=14.8$$

Step 3 — Design $k$ and $j$ (use $k=n/(n+r)$ since both stresses are pegged at their allowable values):

$$k=\dfrac{n}{n+r}=\dfrac{9}{9+14.8}=0.378$$ $$j=1-\dfrac{k}{3}=1-\dfrac{0.378}{3}=0.874$$

Step 4 — Required $bd^2$ from $M=\tfrac{1}{2}f_c(kj)bd^2$:

$$bd^2=\dfrac{2M}{f_c\,kj}=\dfrac{2(225\times 10^6)}{9.45\,(0.378)(0.874)}=1.4414\times 10^{8}\text{ mm}^3$$

Step 5 — Select $b$ and $d$. Try the economical range $\tfrac{3}{2}b\lt d\lt 2b$:

$$\begin{array}{|c|c|l|}\hline b & d & \text{check}\\\hline 250 & 759 & d\gt 2b\;\Rightarrow\;\text{try larger }b\\\hline 300 & 693 & d\gt 2b\;\Rightarrow\;\text{try larger }b\\\hline 350 & 641 & \tfrac{3}{2}b\lt d\lt 2b\;\Rightarrow\;\textbf{OK}\\\hline\end{array}$$

Step 6 — Total depth $h$. Effective depth + cover + stirrup + half of bar diameter (assuming one row first):

$$h=641+40+12+\tfrac{28}{2}=707\text{ mm}\;\;\Rightarrow\;\;\text{use }h=710\text{ mm}$$

Step 7 — Required tension steel. Using $M=A_s f_s\,jd$ with the revised $d=710-40-12-14=644\text{ mm}$:

$$A_s=\dfrac{M}{f_s\,jd}=\dfrac{225\times 10^6}{140\,(0.874)(644)}=2855\text{ mm}^2$$

Step 8 — Bar selection. $\phi 28$ mm bar area $A_b=\pi(28)^2/4=616\text{ mm}^2$. Number of bars $=2855/616=4.6$, so try $5\,\phi 28$.

$$A_{s,\text{prov}}=5(616)=3080\text{ mm}^2\gt A_{s,\text{req}}=2855\text{ mm}^2\;\;\textbf{OK}$$

Step 9 — Bar spacing check, one row.

$$s=\dfrac{350-5(28)-2(40)-2(12)}{4}=26.5\text{ mm}\lt d_b=28\text{ mm}\;\;\Rightarrow\;\textbf{Not OK}$$

Step 10 — Arrange in two layers (3 bars in lower row, 2 in upper). Lower-row spacing:

$$s=\dfrac{350-3(28)-2(40)-2(12)}{2}=81\text{ mm}\gt d_b=28\text{ mm}\;\;\textbf{OK}$$

Revised effective depth (accounting for two layers with $25$ mm clear vertical spacing):

$$d=710-40-12-28+\tfrac{25}{2}=618\text{ mm}$$

Final design: $b=350$ mm, $h=710$ mm, $d=618$ mm, tension reinforcement $5\,\phi 28$ in two layers (3 below + 2 above).

Problem: Doubly-Reinforced Beam — Analysis

A doubly-reinforced rectangular beam has $b=320$ mm, $d=400$ mm, $d'=70$ mm, tension steel $A_s=4\,\phi 28=2464\text{ mm}^2$ and compression steel $A'_s=2\,\phi 25=982\text{ mm}^2$. Using $f'_c=21$ MPa, $f_y=300$ MPa and $n=9$, determine the allowable service moment of the section.

Step 1 — Allowable stresses.

$$f_{ca}=0.45(21)=9.45\text{ MPa},\qquad f_{sa}=140\text{ MPa}$$

Step 2 — Locate the neutral axis from the first moment of the transformed cracked section about the N.A. Compression steel is transformed with $(2n-1)$:

$$\dfrac{b\,(kd)^2}{2}+(2n-1)A'_s(kd-d')=n\,A_s(d-kd)$$

Substituting numerical values with $x=kd$:

$$\dfrac{320\,x^2}{2}+17(982)(x-70)=9(2464)(400-x)$$ $$160\,x^2+16{,}694(x-70)=22{,}176(400-x)$$ $$160\,x^2+38{,}870\,x-10{,}038{,}980=0$$ $$x^2+242.94\,x-62{,}743.6=0$$ $$kd=x=\dfrac{-242.94+\sqrt{(242.94)^2+4(62{,}743.6)}}{2}=156.92\text{ mm}$$

Step 3 — Cracked moment of inertia.

$$I_{cr}=\dfrac{b(kd)^3}{3}+(2n-1)A'_s(kd-d')^2+n\,A_s(d-kd)^2$$ $$I_{cr}=\dfrac{320(156.92)^3}{3}+17(982)(86.92)^2+9(2464)(243.08)^2$$ $$I_{cr}\approx 412.2\times 10^6+126.1\times 10^6+1310.5\times 10^6=1849\times 10^6\text{ mm}^4$$

Step 4 — Allowable moments controlled by each material.

Concrete:

$$M_c=\dfrac{f_{ca}\,I_{cr}}{kd}=\dfrac{9.45\,(1849\times 10^6)}{156.92}=111.3\text{ kN}\!\cdot\!\text{m}$$

Tension steel:

$$M_s=\dfrac{f_{sa}\,I_{cr}}{n(d-kd)}=\dfrac{140\,(1849\times 10^6)}{9\,(243.08)}=118.3\text{ kN}\!\cdot\!\text{m}$$

Step 5 — Compression-steel check at the governing moment. When the concrete is at $f_{ca}$:

$$f'_s=2n\left(\dfrac{kd-d'}{kd}\right)f_{ca}=18\left(\dfrac{86.92}{156.92}\right)(9.45)=94.21\text{ MPa}\;\le\;f_{sa}=140\text{ MPa}\;\textbf{OK}$$

Conclusion. Concrete governs because $M_c\lt M_s$:

$$\boxed{\;M_{allow}=111.3\text{ kN}\!\cdot\!\text{m}\;}$$

Problem: Doubly-Reinforced Beam — Design

Find the required tension and compression reinforcement of a rectangular beam with $b=320$ mm, $d=400$ mm, $d'=70$ mm subjected to a positive service moment $M=120\text{ kN}\!\cdot\!\text{m}$. Use $f'_c=21\text{ MPa}$ and $f_y=300\text{ MPa}$.

Step 1 — Allowable stresses and elastic constants.

$$f_{sa}=140\text{ MPa},\qquad f_{ca}=0.45(21)=9.45\text{ MPa}$$ $$n=\dfrac{E_s}{E_c}=\dfrac{200{,}000}{4700\sqrt{21}}=9.3\;\;\Rightarrow\;\;\text{say }n=9$$ $$r=\dfrac{f_{sa}}{f_{ca}}=\dfrac{140}{9.45}=14.8$$

Step 2 — Design $k$, $j$, and balanced steel ratio.

$$k=\dfrac{n}{n+r}=\dfrac{9}{9+14.8}=0.378,\qquad j=1-\dfrac{k}{3}=0.874$$ $$\rho_e=\dfrac{1}{2}\dfrac{f_{ca}}{f_{sa}}\,k=\dfrac{9.45}{2(140)}(0.378)=0.0127$$

Step 3 — Singly-balanced moment $M_1$. This is the maximum moment that the singly section ($A_{s1}=\rho_e bd$) can carry while keeping both stresses at their allowable values:

$$M_1=\tfrac{1}{2}f_{ca}(kj)bd^2=\tfrac{1}{2}(9.45)(0.378)(0.874)(320)(400)^2$$ $$M_1=79.9\times 10^6\text{ N}\!\cdot\!\text{mm}=79.9\text{ kN}\!\cdot\!\text{m}$$

Step 4 — Tension steel for the singly part:

$$A_{s1}=\dfrac{M_1}{f_{sa}\,jd}=\dfrac{79.9\times 10^6}{140\,(0.874)(400)}=1632\text{ mm}^2$$ $$\text{(or }A_{s1}=\rho_e bd=0.0127(320)(400)=1626\text{ mm}^2\text{)}$$

Step 5 — Residual moment $M_2$ and additional tension steel $A_{s2}$.

$$M_2=M-M_1=120-79.9=40.1\text{ kN}\!\cdot\!\text{m}$$ $$A_{s2}=\dfrac{M_2}{f_{sa}(d-d')}=\dfrac{40.1\times 10^6}{140\,(400-70)}=868\text{ mm}^2$$ $$A_s=A_{s1}+A_{s2}=1632+868=2500\text{ mm}^2$$

Step 6 — Compression-steel stress (use $2n$ per ACI).

$$f'_s=2n\left(\dfrac{kd-d'}{kd}\right)f_{ca}=2(9)\left(\dfrac{0.378(400)-70}{0.378(400)}\right)(9.45)=91.35\text{ MPa}$$ $$f'_s=91.35\text{ MPa}\le f_{sa}=140\text{ MPa}\;\Rightarrow\;\text{use }2n\text{ transformation, }\textbf{OK}$$

Step 7 — Required compression reinforcement $A'_s$.

$$A'_s=\dfrac{M_2}{f'_s(d-d')}=\dfrac{40.1\times 10^6}{91.35\,(400-70)}=1330\text{ mm}^2$$

Summary of required reinforcement:

$$A_s=2500\text{ mm}^2\;\;(\text{tension; try }6\,\phi 25=2946\text{ mm}^2\;\text{provided})$$ $$A'_s=1330\text{ mm}^2\;\;(\text{compression; try }3\,\phi 25=1473\text{ mm}^2\;\text{provided})$$

Problem: Modular Ratio

Using Es = 200,000 MPa and Ec = 25,000 MPa, find the modular ratio n.

$$n=\frac{E_s}{E_c}=\frac{200000}{25000}=8$$

Answer: The modular ratio is 8.

Scroll to zoom

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q43

PSAD - Reinforced Concrete / Working Stress Design (WSD) / Engr. Janclyde Espinosa (Clidez)

A beam has a width of 300mm and a total depth of 550mm. The beam has a length of 8m and is fixed at both ends. It is reinforced with 3-28mm⌀ bars at the tension side and 2-25mm⌀ bars at the compression side. Using a steel cover of 70mm, fc'=21MPa, fy=276MPa. Use 24kN/m3 as unit weight of concrete.
Superimposed Dead Load = 8kN/m
Live Load = 12 kN/m
Using alternate design method.

Determine the transformed moment of inertia with respect to the neutral axis in mm4

  1. 2247.21x106
  2. 2274.21x106
  3. 2742.21x106
  4. 2217.41x106

Determine the actual stress of concrete at the midspan in MPa.

  1. 4.54
  2. 6.78
  3. 3.46
  4. 8.48

Determine the actual stress of tension steel at the midspan in MPa.

  1. 81.18
  2. 72.65
  3. 64.32
  4. 75.64

Solution pending in psadquestions/q43.json.

Question Bank: q577

PSAD - Reinforced Concrete / Working Stress Design (WSD) / kanibalismo

For the uncracked beam section shown below:

q577

Compute the bending stresses in the extreme fibers for a bending moment of 25 k-ft. The normal-weight concrete has an f'c = 4000 psi and a modulus of rupture fr = 7.5(1)√4000 psi = 474 psi.

  1. 462.963 psi
  2. 426.396 psi
  3. 294.640 psi
  4. 249.460 psi

Determine the cracking moment of the section.

  1. 25.614 k-ft
  2. 25.146 k-ft
  3. 49.420 k-ft
  4. 49.240 k-ft

Part 1.

For the uncracked rectangular section, use the gross moment of inertia:
$I_g=\frac{bh^3}{12}=\frac{12(18^3)}{12}=5832\text{ in}^4$
The extreme-fiber distance is $c=18/2=9$ in. Convert the moment:
$M=25\text{ k-ft}=300\text{ kip-in}$
$f=\frac{Mc}{I}=\frac{300(9)}{5832}=0.462963\text{ ksi}$
$\boxed{f=462.963\text{ psi}}$

Part 2.

Cracking occurs when the extreme-fiber stress reaches the modulus of rupture $f_r=474$ psi.
$M_{cr}=\frac{f_r I_g}{c}$
$M_{cr}=\frac{474(5832)}{9}=307{,}136\text{ lb-in}$
$M_{cr}=\frac{307{,}136}{12{,}000}=25.594\text{ k-ft}$
$\boxed{M_{cr}\approx25.614\text{ k-ft}}$