[9급 서울시 응용역학개론] 18년 1회차 기출문제 정답·해설 페이지

0.01m

0.015m

0.025m

$A_x = 0,\ A_y = 0.5P,\ B_y = 0.5P$

$A_x = 0,\ A_y = -0.25P,\ B_y = 1.75P$

$A_x = P,\ A_y = 0.5P,\ B_y = 1.5P$

15

16

17

$f_B = \dfrac{P}{bh} + \dfrac{P\,e_x}{I_y}x - \dfrac{P\,e_y}{I_x}y$

$f_C = \dfrac{P}{bh} - \dfrac{P\,e_x}{I_y}x + \dfrac{P\,e_y}{I_x}y$

$f_D = \dfrac{P}{bh} + \dfrac{P\,e_x}{I_y}x - \dfrac{P\,e_y}{I_x}y$

11kN

13kN

15kN

$\theta_c = \dfrac{5ML}{3EI}(\curvearrowright),\quad y_c = \dfrac{2ML^2}{3EI}(\downarrow)$

$\theta_c = \dfrac{2ML}{3EI}(\curvearrowright),\quad y_c = \dfrac{5ML^2}{3EI}(\downarrow)$

$\theta_c = \dfrac{5ML}{6EI}(\curvearrowright),\quad y_c = \dfrac{4ML^2}{3EI}(\downarrow)$

4kN·m

8kN·m

16kN·m

$wL^2/24$

$wL^2/6$

$wL^2/12$

A점은 비례한도(proportional limit)이다.

C점은 항복점(yield strength)이다.

D점은 한계응력(ultimate stress)이다.

$-\dfrac{wL^2}{6}$

$-\dfrac{wL^2}{2}$

$-\dfrac{wL^2}{3}$

$A_y = 0.25P,\ M_A = -PL,\ C_y = 0.5P$

$A_y = 0.5P,\ M_A = -PL,\ C_y = 0.5P$

$A_y = -0.25P,\ M_A = PL,\ C_y = 0.25P$

$\dfrac{bh}{2\sqrt{3}}$

$\dfrac{b}{\sqrt{6}}$

$\dfrac{h}{2\sqrt{3}}$

50kN/m

200kN/m

300kN/m

$R_A$=1.3$WL$

$R_B$=0.9$WL$

$R_A$=0.9$WL$

$M_{AO}$=11.4kN·m(↷), $M_{BO}$=8.5kN·m(↷)

$M_{AO}$=8.5kN·m(↷), $M_{BO}$=11.4kN·m(↷)

$M_{AO}$=4.2kN·m(↷), $M_{BO}$=5.7kN·m(↷)

$\dfrac{2}{3}wl^2$

$\dfrac{1}{12}wl^2$

$\dfrac{7}{72}wl^2$

중심면

중앙면

중간면

2.0m2

2.4m2

2.6m2

4MPa

16MPa

32MPa

20N/m

10N/m

5N/m