Question 6.4: An inverted precast concrete T-beam is used to support preca......
An inverted precast concrete T-beam is used to support precast double-tee floor beams in a parking deck (see Fig. 6-13 and photo). Beam dimensions are b = 500 ~mm, b_w = 300 ~ mm, d = 600 ~mm, ~and ~t_f = 100~ mm. Steel reinforcement consists of four bars each with a 25-mm diameter. The modulus of elasticity for the concrete is E_c = 25 ~GPa, while that of the steel is E_s = 200 ~GPa. Allowable stresses for concrete and steel are σ_{ac} = 9.3 ~MPa ~and ~ σ_{as} = 137 ~MPa, respectively.
(a) Use the transformed section in Fig. 6-14 (in which the concrete in tension is neglected and the steel reinforcing bars are converted to the equivalent concrete) to find the maximum permissible moment that can be applied to this beam.
(b) Repeat part (a) if the beam is rotated 180°, as shown in Fig. 6-15, and if the steel reinforcement remains in the bottom tension zone.
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Inverted T-beam.
1, 2. Conceptualize, Categorize: Start by finding the neutral axis (at some distance y down from the top of the beam) for the transformed section shown in Fig. 6-14. Equating the first moments of areas of concrete in compression (b_w × y) and the transformed area of steel in tension (n × A_s) leads to a quadratic equation. The solution for y gives the position of the neutral axis as
b_{W} ~y\frac{y}{2} – n A_{s}(d-y)=0\quad\mathrm{where}\quad n=\frac{E_{s}}{E_{c}} = 8 \quad\quad\quad (a)
y=\sqrt{{\Bigg\lgroup}\frac{n A_{\mathrm{s}}}{b_{\mathrm{\scriptscriptstyle{W}}}}{\Bigg\rgroup}^{2}+2d{\Bigg\lgroup}\frac{n A_{\mathrm{s}}}{b_{\mathrm{\scriptscriptstyle{W}}}}{\Bigg\rgroup}} – {\Bigg\lgroup}\frac{n A_{\mathrm{s}}}{b_{\mathrm{\scriptscriptstyle{W}}}}{\Bigg\rgroup}=0.204\;\mathrm{m}\quad (b)
Now use Eq. (6-17) to compute the moment of inertia of the transformed section:
I_T = I_1 + nI_2 = I_1 + \frac{E_2}{E_1}I_2 \quad\quad (6-17)
\quad\quad I_{T}\,=\,{\frac{b_{\mathrm{w}}y^{3}}{3}}\,+\,n{\mathrm{~}}A_{\mathrm{s}}[(d-y)^{2}]=3.312\times10^{-3}\ \mathrm{m}^{4} \quad\quad (c)
3, 4. Analyze, Finalize: Finally, the moment capacity of the beam is found by solving Eqs. (6-16) (allowable stress in concrete controls) and Eq. (6-18b) (allowable stress in steel controls) for M, where the lower value based on allowable stress in the steel governs:
\quad \sigma_{x1} = – \frac{M_y}{I_T}\quad\quad (6-16)
\quad\sigma_{x1} = – \frac{M_yE_1}{E_1I_1 + E_2I_2}\quad\quad (6-18a)
\quad \sigma_{x2} = – \frac{M_y}{I_T}n\quad\quad (6-18b)
\quad M_{c} = \,\frac{\sigma_{a c}}{y}I_{T}\,=\,{\frac{9.3\,\,\,\mathrm{MPa}}{ 0.204 m}}\,(3.312 \times 10^{-3}\,\,\mathrm{m}^{4}) = 151~ kN·m \quad\quad\quad (d)
\quad M_{s}=\,{\frac{\sigma_{a s}}{n(d-y)}}I_{T}\,=\,{\frac{137\,\,\,\mathrm{MPa}}{8(0.6\mathrm{m}-0.204\,\mathrm{m})}}\,(3.312\times10^{-3}\,\,\mathrm{m}^{4})\quad\quad (e)
\quad\quad\quad = 143.2 ~kN·m
Part (b): T-beam.
1, 2. Conceptualize, Categorize: Now the flange of the T-beam with a thickness of t_f is on top, so start by assuming that neutral axis location distance y is greater than t_f. Divide the compression area of the concrete for the transformed section (Fig. 6-15) into three rectangles, then equate the first moments of areas of the concrete in compression and transformed area of the steel (n × A_s) to get a quadratic equation for distance y. The solution for y gives the position of the neutral axis as
(b-b_{\mathrm{w}})t_{\mathrm{f}}\begin{bmatrix}y-{\frac{t_{\mathrm{f}}}{2}}\end{bmatrix} + b_{\mathrm{w}}y{\frac{y}{2}}-n A_{s}(d-y) = 0 \quad\quad\quad\quad (f)
Solve Eq. (f) for y to get
\quad \quad\quad\quad y = 0.1702 ~m
which exceeds t_f as assumed. The moment of inertia of the transformed section is now
I_{T}\,=\,{\frac{b_{w}\,y^{3}}{3}}\, + \,{\frac{(b-b_{w})t_{t}^{3}}{12}}\,+\,\left(b\,-\,b_{w}\right)t_{t}{\Bigg\lgroup}y-{\frac{t_{\mathrm{f}}}{2}}{\Bigg\rgroup}^{2}\,+\,n A_{s}(d\,-\,y)^{2} \\ = 3.7 \times 10^{-3}~m^43. Analyze: Finally, repeat the solutions for maximum permissible moment M in Eqs. (d) and (e) as
M_{\mathrm{_C}}=\frac{\sigma_{\mathrm{ac}}}{y}I_{T}=\frac{(9.3\mathrm{MPa})}{0.1702\,\mathrm{m}}\left(3.7\times10^{-3}\ \mathrm{m}^{4}\right)=202\ \mathrm{kN}\cdot\mathrm{m}\quad\quad (g) \\ M_{s}={\frac{\sigma_{\mathrm{as}}}{{n}(d – y)}}I_{T}\; = {\frac{137\;{\mathrm{MPa}}}{8(0.6~\mathrm{m – 0.1702~m)}}}(3.7\times10^{-3}~\mathrm{m}^{4}) \quad\quad (h) \\ = 147.4 kN\cdot m4. Finalize: Once again, the lower value of moment M based on allowable stress in the steel governs. Since the allowable stress in the reinforcing steel bars controls both beams, their moment capacities [Eqs. (e) and (h)] are approximately the same.
