Force method: frame with 120 k load

Reading of the figure: beam a-b is 24 ft (12 ft + 12 ft) with stiffness $2EI$, carrying 120 kip at d (mid-span); column b-c is 18 ft high with stiffness $3EI$; a and c are hinged. Units: kip and ft.

$r = 4$, so $D_s = 1$. Take the horizontal reaction $H_a$ at a as the redundant (a becomes a roller).

        120 k
   a o----d----b
   ^           |
   H_a         | 18 ft
               |
               o c
   |<- 12 ->|<- 12 ->|

Primary structure (a on a roller)

Reactions: $V_a = V_c = 60$ kip.

($M_0 = 60\times12 = 720$ at d. A unit horizontal force at a causes a vertical reaction pair of $18/24 = 0.75$, so $m_d = -0.75\times12 = -9$ and $m_b = -18$ (hogging).)

Coefficients (unit-load method; beam stiffness $2EI$, column stiffness $3EI$)

$$\begin{aligned} f_{11} &= \frac{1}{2EI}\left[\frac{12}{3}(81) + \frac{12}{3}(81 + 162 + 324)\right] + \frac{1}{3EI}\left[\frac{18}{3}(324)\right] = \frac{162 + 1134 + 648}{EI} = \frac{1944}{EI} \\ \Delta_{10} &= \frac{1}{2EI}\left[\frac{12}{3}(720)(-9) + \frac{12}{6}\{2(720)(-9) + 720(-18)\}\right] = \frac{-12960 - 25920}{EI} = \frac{-38880}{EI} \end{aligned}$$

Compatibility (a does not move horizontally)

$$f_{11}H_a + \Delta_{10} = 0 \ \Rightarrow\ H_a = \frac{38880}{1944} = 20.00\ \text{kip}$$

Reactions

• a: $H_a = 20.00$ kip ($\rightarrow$), $V_a = 45.00$ kip ($\uparrow$)
• c: $H_c = 20.00$ kip ($\leftarrow$), $V_c = 75.00$ kip ($\uparrow$)

(The unit force on the primary structure gives a vertical reaction at a of $0.75$ per unit $H$, so $V_a = 60 - 0.75\times20 = 45$ kip and $V_c = 60 + 15 = 75$ kip.)

Bending moment $M = M_0 + H_a m$ (kip-ft)

Shear force (kip)

Axial force (kip)

Beam: $-20.0$ (compression); column: $-75.0$ (compression).

Answer: $H_a = H_c = 20.00$ kip; $V_a = 45.00$ kip, $V_c = 75.00$ kip; $M_d = 540.0$ kip-ft (sagging), $M_b = -360.0$ kip-ft (hogging).