CE Board Exam Randomizer

Question 1

Serviceability of Reinforced Concrete Beams

Serviceability ensures that a beam, slab, or column behaves acceptably under service (un-factored) loads: cracks remain narrow, deflections remain small, and stresses remain in the elastic range. Strength design (USD) prevents collapse; serviceability design keeps the structure usable. The NSCP 2015 / ACI 318 provisions below govern stress checks, cracking, and short- and long-term deflections.

Permissible service-load stresses (WSD):

$$f_c \le 0.45\,f'_c \quad \text{(concrete in flexure)}$$ $$f_s \le 140\text{ MPa} \quad \text{for Grade 280 reinforcement}$$ $$f_s \le 170\text{ MPa} \quad \text{for Grade 420 reinforcement}$$

Assumptions for service-load stress investigation (straight-line theory):

Strains vary linearly from the neutral axis (deep beams with $h/L \ge 2/5$ continuous or $4/5$ simple require a non-linear strain distribution).
The stress–strain relationship of concrete is linear within permissible service load stresses.
Concrete resists no tension.
Modular ratio $n = E_s / E_c$ is taken as the nearest whole number (but not less than 6). For deflection computations in lightweight concrete, $n$ is the same as for normal-weight concrete of the same strength.
For doubly-reinforced flexural members, an effective modular ratio of $2E_s/E_c$ is used to transform compression reinforcement; $f'_s$ shall not exceed $f_{sa}$.

Answer

Step 1 — Modulus of rupture:

$$f_r = 0.62\,\lambda\sqrt{f'_c} = 0.62(1.0)\sqrt{34} = 3.616\text{ MPa}$$

Step 2 — Gross section properties:

$$I_g = \dfrac{bh^3}{12} = \dfrac{300(600)^3}{12} = 5.4\times 10^9\text{ mm}^4$$ $$y_t = \dfrac{h}{2} = 300\text{ mm}$$

Step 3 — Cracking moment:

$$M_{cr} = \dfrac{f_r\,I_g}{y_t} = \dfrac{3.616(5.4\times 10^9)}{300} = 65.1\times 10^6\text{ N}\!\cdot\!\text{mm} = 65.1\text{ kN}\!\cdot\!\text{m}$$

Answer

Step 1 — Time-dependent factor for 5 years: $\xi = 2.0$.

Step 2 — Multiplier:

$$\lambda_\Delta = \dfrac{\xi}{1+50\rho'} = \dfrac{2.0}{1+50(0.012)} = \dfrac{2.0}{1.6} = 1.25$$

The additional long-term deflection equals $1.25\,\delta_{\text{sustained, immediate}}$. Total long-term deflection (sustained + live) becomes $\delta_{\text{immediate, all}} + 1.25\,\delta_{\text{sustained, immediate}}$.

Question 2

Modulus of Elasticity of Concrete

For values of unit weight $w_c$ between 1,440 and 2,560 kg/m³:

$$E_c = w_c^{1.5}\,0.043\sqrt{f'_c}\quad\text{(MPa)}$$

For normal-weight concrete:

$$E_c = 4700\sqrt{f'_c}\quad\text{(MPa)}$$ $$E_s = 200{,}000\text{ MPa}$$ $$n = \dfrac{E_s}{E_c}$$

Question 3

Modulus of Rupture of Concrete, f_r

The modulus of rupture (the tensile-flexural strength of plain concrete) governs the cracking moment:

$$f_r = 0.62\,\lambda\sqrt{f'_c}\quad\text{(MPa)}$$

Lightweight concrete modification factor $\lambda$:

Concrete	λ
All-lightweight	0.75
Lightweight, fine blend	0.75 to 0.85
Sand-lightweight	0.85
Sand-lightweight, coarse blend	0.85 to 1.00
Normal-weight	1.00

If the average splitting tensile strength $f_{ct}$ is measured by ASTM C330M, then:

$$\lambda = \dfrac{f_{ct}}{0.56\sqrt{f_{cm}}}\;\le\;1.0$$

where $f_{cm}$ = measured average compressive strength of concrete (MPa). The concrete mixture tested shall be representative of the mixture used in the work.

Question 4

Cracking Moment, M_cr

The bending moment that first opens a flexural crack is computed from the gross transformed section:

$$M_{cr} = \dfrac{f_r\,I_g}{y_t}$$

where

$I_g$ = moment of inertia of the gross concrete section about the centroidal axis, neglecting reinforcement.
$y_t$ = distance from the centroidal axis of the gross section to the extreme tension fibre.

Question 5

Calculation of Immediate Deflections

For non-prestressed members, the effective moment of inertia $I_e$ is calculated by the following formula unless obtained by a more comprehensive analysis. $I_e$ shall not be greater than $I_g$:

$$I_{cr}\;\le\;I_e = \left(\dfrac{M_{cr}}{M_a}\right)^{3}\!I_g + \left[1-\left(\dfrac{M_{cr}}{M_a}\right)^{3}\right]\!I_{cr}\;\le\;I_g$$

where:

$I_{cr}$ = moment of inertia of the cracked section transformed to concrete.
$I_e$ = effective moment of inertia for the computation of deflection.
$I_g$ = moment of inertia of the gross concrete section about the centroidal axis, neglecting reinforcement.
$M_{cr}$ = cracking moment $= f_r I_g / y_t$.
$M_a$ = maximum moment in the member at the stage at which deflection is computed.

Typical instantaneous deflection formulas for elastic members (use $I_e$ in place of $I$):

$$\delta_{\text{midspan, simple, UDL}} = \dfrac{5wL^4}{384 E_c I_e}$$ $$\delta_{\text{midspan, simple, P at midspan}} = \dfrac{P L^3}{48 E_c I_e}$$ $$\delta_{\text{cantilever, UDL}} = \dfrac{wL^4}{8 E_c I_e}$$ $$\delta_{\text{cantilever, P at tip}} = \dfrac{P L^3}{3 E_c I_e}$$

Question 6

Calculation of Time-Dependent Deflections

Unless determined by a more comprehensive analysis, the additional long-term deflection resulting from creep and shrinkage is taken as the immediate deflection caused by the sustained load multiplied by the factor $\lambda_\Delta$:

$$\lambda_\Delta = \dfrac{\xi}{1+50\rho'}$$

where:

$\lambda_\Delta$ = multiplier for additional deflection due to long-term effects.
$\xi$ = time-dependent factor for sustained load.
$\rho' = A'_s/(bd)$ — evaluated at midspan for simple/continuous beams and at the support for cantilevers.

Time-dependent factor $\xi$ for sustained loads:

Sustained Load Duration	Time-Dependent Factor, ξ
3 months	1.0
6 months	1.20
12 months (1 year)	1.40
60 or more months (5 years and up)	2.0

The total long-term deflection is then:

$$\delta_{\text{total}} = \delta_{\text{immediate, all loads}} + \lambda_\Delta\,\delta_{\text{immediate, sustained loads}}$$

Question 7

Minimum Thickness (Deflections Not Computed) — NSCP Table 409-1

For beams and one-way slabs not supporting or attached to partitions likely to be damaged by large deflections, deflections need not be computed if the member thickness satisfies:

Member	Simply Supported	One-end Continuous	Both-end Continuous	Cantilever
Solid one-way slabs	L/20	L/24	L/28	L/10
Beams or ribbed one-way slabs	L/16	L/18.5	L/21	L/8

Modification factors for the tabulated values:

For structural lightweight concrete with unit weight $w_c$ between 1500 and 2000 kg/m³, multiply by $(1.65 - 0.0003w_c)$ but not less than 1.09.
For $f_y$ other than 420 MPa, multiply by $(0.4 + f_y/700)$.

Question 8

Factored Load Combinations (Reference)

Although the factored combinations belong to strength design, they are listed here for completeness when checking serviceability against the un-factored counterparts:

$$(1)\;\;U=1.4D$$ $$(2)\;\;U=1.2D+1.6L+0.5(L_r\text{ or }S\text{ or }R)$$ $$(3)\;\;U=1.2D+1.6(L_r\text{ or }S\text{ or }R)+(0.5L\text{ or }0.8W)$$ $$(4)\;\;U=1.2D+1.6W+0.5L+0.5(L_r\text{ or }S\text{ or }R)$$ $$(5)\;\;U=1.2D+1.0E+0.5L+0.2S$$ $$(6)\;\;U=0.9D+(1.6W\text{ or }1.0E)$$

Question 9

Example: Cracking Moment of a Rectangular Beam

A plain concrete rectangular beam has $b=300\text{ mm}$, $h=600\text{ mm}$, $f'_c=34\text{ MPa}$, normal-weight concrete. Compute the cracking moment $M_{cr}$.

Question 10

Example: Long-Term Deflection Multiplier

A simply supported beam has a sustained load duration of 5 years and contains compression reinforcement with $\rho' = A'_s/(bd) = 0.012$. Compute the multiplier $\lambda_\Delta$ for additional long-term deflection.