# Solution (Digital Format) --- ## Frame Analysis Given: - \( E = 70 \, \text{GPa} \) - \( I = 129 \times 10^6 \, \text{mm}^4 \) - Horizontal member lengths: \( AB = 5\,\text{m} \), \( BC = 4\,\text{m} \), \( CD = 3\,\text{m} \) - Vertical load at \( D \): \( 200 \, \text{kN} \) (rightward) - Horizontal load at \( A \): \( 200 \, \text{kN} \) (leftward) - Moment at \( A \): \( 400 \, \text{kN}\cdot\text{m} \) (counterclockwise) --- ### Step 1: Equilibrium Consider the frame in equilibrium: \[ \sum F_x = \] - \(\rightarrow\) positive - \(\leftarrow\) negative \[ H_A = 200\, \text{kN} \, \leftarrow \] \[ \sum M_A = \] - \(\circlearrowright\) positive - \(\circlearrowleft\) negative \[ 200 \times 2 - M_A = \] \[ M_A = 400 \, \text{kN}\cdot\text{m} \] --- ## (a) Rotation at Point D - Apply a unit moment (\(1\,\text{kN}\cdot\text{m}\)) at \( D \) for moment-area method. --- ### Section Analysis For each section, the moment equations are: - **Section AB:** \( 200x - 400 \) - **Section BC:** \( 200 \times 3 = 600 \) - **Section CD:** \( 200x \) The value of \( m \) (from the moment due to unit load at \( D \)) for each section is \(-1\). - For \( AB \): \( EI = 2EI \) - For \( BC \): \( EI = EI \) - For \( CD \): \( EI = EI \) --- ### Rotation Calculation \[ \theta_D = \int_^5 \frac{(200x-400)(-1)}{2EI} dx + \int_^4 \frac{(600)(-1)}{EI} dx + \int_^3 \frac{(200x)(-1)}{EI} dx \] --- ### Material Properties \[ E = 70\,\text{GPa} \] \[ E = 70 \times 10^6\, \text{kN}/\text{m}^2 \] \[ I = 129 \times 10^6\, \text{mm}^4 = 1.29 \times 10^{-3} \, \text{m}^4 \] --- ### Substitution & Simplification \[ \theta_D = \frac{-250 - 240 - 900}{EI} \] \[ \theta_D = \frac{-355}{70 \times 10^6 \times 1.29 \times 10^{-3}} \] \[ \theta_D = -.039 \, (\text{units}) \] --- **The negative sign will change the assumed direction.** \[ \boxed{\theta_D = .039 \, (\text{units}) \, \curvearrowright} \] --- - Positive: Clockwise rotation - Negative: Counter-clockwise rotation This provides the required rotation at point \( D \) for the given frame and loading conditions.

# Solution (Digital Format) --- ## Frame Analysis **Given:** - \( E = 70 \, \text{GPa} \) - \( I = 129 \times 10^6 \, \text{mm}^4 \) - Horizontal member lengths: - \( AB = 5\,\text{m} \) - \( BC = 4\,\text{m} \) - \( CD = 3\,\text{m} \) - Vertical load at \( D \): \( 200 \, \text{kN} \) (rightward) - Horizontal load at \( A \): \( 200 \, \text{kN} \) (leftward) - Moment at \( A \): \( 400 \, \text{kN}\cdot\text{m} \) (counterclockwise) --- ## Step 1: Equilibrium Consider the frame in equilibrium: \[ \sum F_x = 0 \] - Positive: \(\rightarrow\) - Negative: \(\leftarrow\) \[ H_A = 200\, \text{kN} \, \leftarrow \] \[ \sum M_A = 0 \] - Positive: \(\circlearrowright\) - Negative: \(\circlearrowleft\) \[ 200 \times 2 - M_A = 0 \] \[ M_A = 400 \, \text{kN}\cdot\text{m} \] --- ## (a) Rotation at Point D - Apply a unit moment (\(1\,\text{kN}\cdot\text{m}\)) at \( D \) for moment-area method. --- ### Section Analysis For each section, the moment equations are: - **Section AB:** \( 200x - 400 \) - **Section BC:** \( 600 \) - **Section CD:** \( 200x \) The value of \( m \) (from the moment due to unit load at \( D \)) for each section is \(-1\): - For \( AB \): \( EI = 2EI \) - For \( BC \): \( EI = EI \) - For \( CD \): \( EI = EI \) --- ### Rotation Calculation \[ \theta_D = \int_0^5 \frac{(200x-400)(-1)}{2EI} \, dx + \int_0^4 \frac{(600)(-1)}{EI} \, dx + \int_0^3 \frac{(200x)(-1)}{EI} \, dx \] --- ### Material Properties \[ E = 70\,\text{GPa} = 70 \times 10^6\, \text{kN}/\text{m}^2 \] \[ I = 129 \times 10^6\, \text{mm}^4 = 1.29 \times 10^{-3} \, \text{m}^4 \] --- ### Substitution & Simplification \[ \theta_D = \frac{-250 - 240 - 900}{EI} \] \[ \theta_D = \frac{-355}{70 \times 10^6 \times 1.29 \times 10^{-3}} \] \[ \theta_D = -0.039 \, (\text{units}) \] --- **The negative sign indicates a change in assumed direction.** \[ \boxed{\theta_D = 0.039 \, (\text{units}) \, \curvearrowright} \] --- - Positive: Clockwise rotation - Negative: Counter-clockwise rotation This provides the required rotation at point \( D \) for the given frame and loading conditions.

# Solution (Digital Format) --- ## Frame Analysis **Given:** - \( E = 70 \, \text{GPa} \) - \( I = 129 \times 10^6 \, \text{mm}^4 \) - Horizontal member lengths: - \( AB = 5\,\text{m} \) - \( BC = 4\,\text{m} \) - \( CD = 3\,\text{m} \) - Vertical load at \( D \): \( 200 \, \text{kN} \) (rightward) - Horizontal load at \( A \): \( 200 \, \text{kN} \) (leftward) - Moment at \( A \): \( 400 \, \text{kN}\cdot\text{m} \) (counterclockwise) --- ## Step 1: Equilibrium Consider the frame in equilibrium: \[ \sum F_x = 0 \] - Positive: \(\rightarrow\) - Negative: \(\leftarrow\) \[ H_A = 200\, \text{kN} \, \leftarrow \] \[ \sum M_A = 0 \] - Positive: \(\circlearrowright\) - Negative: \(\circlearrowleft\) \[ 200 \times 2 - M_A = 0 \] \[ M_A = 400 \, \text{kN}\cdot\text{m} \] --- ## (a) Rotation at Point D - Apply a unit moment (\(1\,\text{kN}\cdot\text{m}\)) at \( D \) for moment-area method. --- ### Section Analysis For each section, the moment equations are: - **Section AB:** \( 200x - 400 \) - **Section BC:** \( 600 \) - **Section CD:** \( 200x \) The value of \( m \) (from the moment due to unit load at \( D \)) for each section is \(-1\): - For \( AB \): \( EI = 2EI \) - For \( BC \): \( EI = EI \) - For \( CD \): \( EI = EI \) --- ### Rotation Calculation \[ \theta_D = \int_0^5 \frac{(200x-400)(-1)}{2EI} \, dx + \int_0^4 \frac{(600)(-1)}{EI} \, dx + \int_0^3 \frac{(200x)(-1)}{EI} \, dx \] --- ### Material Properties \[ E = 70\,\text{GPa} = 70 \times 10^6\, \text{kN}/\text{m}^2 \] \[ I = 129 \times 10^6\, \text{mm}^4 = 1.29 \times 10^{-3} \, \text{m}^4 \] --- ### Substitution & Simplification \[ \theta_D = \frac{-250 - 240 - 900}{EI} \] \[ \theta_D = \frac{-355}{70 \times 10^6 \times 1.29 \times 10^{-3}} \] \[ \theta_D = -0.039 \, (\text{units}) \] --- **The negative sign indicates a change in assumed direction.** \[ \boxed{\theta_D = 0.039 \, (\text{units}) \, \curvearrowright} \] --- - Positive: Clockwise rotation - Negative: Counter-clockwise rotation This provides the required rotation at point \( D \) for the given frame and loading conditions.