Let's solve the problem step by step. ### **Step 1: Beam Description and Supports** - **Span lengths:** \( CA = 3\,m \), \( AD = 3\,m \), \( DB = 6\,m \) - **Supports:** - \( C \): Free end - \( A \): Hinge (pin) - \( B \): Roller (right support) - **Point D** is at 6 m from A, or 12 m from C. --- ## **Step 2: Influence Line for Reaction at A (\(R_A\))** This is a statically determinate beam (hinge at A, roller at B). Let's denote \( x \) as the position of a unit load, measured from C (left end). **Reaction at A due to a unit load at distance \( x \) from C:** Taking moments about B: \[ \text{Let total span} = 3 + 3 + 6 = 12\,m \] \[ R_A \times 12 = (12 - x) \times 1 \] \[ R_A = \frac{12 - x}{12} \] - At \( C \) (\( x = 0 \)): \( R_A = 1 \) - At \( B \) (\( x = 12 \)): \( R_A = 0 \) The influence line is a straight line from 1 at C to 0 at B. --- ## **Step 3: Influence Line for Shear Just Right of A (\(V_{A^+}\))** Shear just to the right of A is the upward reaction at A minus any downward load just right of A. - When the unit load is to the **left of A** (\(0 \leq x < 6\)), \(V_{A^+} = R_A\). - When the unit load is **right of A** (\(x > 6\)), \(V_{A^+} = R_A - 1\). So: - For \( 0 \leq x < 6 \), \( V_{A^+} = \frac{12 - x}{12} \) - For \( 6 < x \leq 12 \), \( V_{A^+} = \frac{12 - x}{12} - 1 = \frac{-x}{12} \) At \( x = 6 \): - Just left: \( V_{A^+} = \frac{12 - 6}{12} = 0.5 \) - Just right: \( V_{A^+} = 0.5 - 1 = -0.5 \) So, there is a jump of -1 at \( x = 6 \). --- ## **Step 4: Influence Line for Moment at A (\(M_A\))** Moment at A due to unit load at \( x \): - For \( x < 6 \): \( M_A = 0 \) (since left of hinge) - For \( x \geq 6 \): \( M_A = R_A \cdot 0 - (x - 6) \cdot 1 = -(x - 6) \) But hinge at A does not resist moment, so \( M_A = 0 \) everywhere. - So, **Influence line for moment at A is zero everywhere**. --- ## **Step 5: Influence Line for Moment at D (\(M_D\))** Moment at D (6 m to the right of A): - For a unit load at \( x \): If \( x < 9 \) (left of D): - \( M_D = R_A \cdot (9 - 6) \) - \( R_A = \frac{12 - x}{12} \) - So, \( M_D = \frac{12 - x}{12} \cdot 3 \) If \( x > 9 \) (right of D): - \( M_D = R_A \cdot 3 - (x - 9) \cdot 1 \) - \( M_D = \frac{12 - x}{12} \cdot 3 - (x - 9) \) So the influence line for \( M_D \): \[ M_D = \begin{cases} \frac{12-x}{4} & \text{for } 6 < x < 9 \\ \frac{12-x}{4} - (x-9) & \text{for } 9 < x < 12 \\ \end{cases} \] --- ## **Step 6: Maximum Values for Moving Loads** ### (1) **Concentrated load \(P = 6\,kN\)** - The maximum effect occurs where the influence line is maximum. #### (a) **Reaction at A** - Max value: \( R_A \) max at \( x=0 \): \( 1 \cdot 6 = 6\,kN \) #### (b) **Shear just right of A** - Max positive value: Just left of A (\( x=6^- \)): \( 0.5 \cdot 6 = 3 \) kN - Max negative value: Just right of A (\( x=6^+ \)): \( -0.5 \cdot 6 = -3 \) kN #### (c) **Moment at A** - Always zero for hinge. #### (d) **Moment at D** - Max value is at \( x \) where the influence line is maximum. - For \( 6 < x < 9 \): \( M_D = \frac{12-x}{4} \) Maximum at \( x=6 \): \( M_D = \frac{12-6}{4} = 1.5 \) - \( M_{D,\text{max}} = 1.5 \cdot 6 = 9\,kNm \) ### (2) **Uniform load \(w = 3\,kN/m\)** The maximum value is the area under the positive portion of the influence line, times \( w \). #### (a) **Reaction at A** - Area under \( R_A = \frac{12-x}{12} \) from 0 to 12: - Area = \( \frac{1}{2} \times 12 \times 1 = 6 \) - Max reaction: \( R_{A,\text{max}} = 6 \cdot 3 = 18\,kN \) #### (b) **Shear just right of A** - Area under the influence line from 0 to 12. - From \( 0 \) to \( 6 \): triangle, height 0.5, base 6: \( \frac{1}{2} \cdot 6 \cdot 0.5 = 1.5 \) - From \( 6 \) to \( 12 \): triangle, height -0.5, base 6: \( \frac{1}{2} \cdot 6 \cdot (-0.5) = -1.5 \) - Net area = 1.5 - 1.5 = 0 But for **maximum positive shear**, consider only the left triangle: - Max positive shear: \( 1.5 \cdot 3 = 4.5\,kN \) - Max negative shear: \( -1.5 \cdot 3 = -4.5\,kN \) #### (c) **Moment at A** - Always zero. #### (d) **Moment at D** - Area under positive part of influence line for \( M_D \): - From \( x=6 \) to \( x=9 \): \( M_D = \frac{12-x}{4} \) - This is a triangle from \( 1.5 \) at \( x=6 \) to 0 at \( x=9 \), base = 3 - Area = \( \frac{1}{2} \cdot 3 \cdot 1.5 = 2.25 \) - \( M_{D,\text{max}} = 2.25 \cdot 3 = 6.75\,kNm \) --- ## **Step 7: Summary Table** | Quantity | Max (6 kN load) | Max (3 kN/m load) | |--------------------|-----------------|-------------------| | Reaction at A | 6 kN | 18 kN | | Shear right of A | ±3 kN | ±4.5 kN | | Moment at A | 0 | 0 | | Moment at D | 9 kNm | 6.75 kNm | --- ### **Influence Line Sketches** 1. **Reaction at A:** Straight line from 1 at C to 0 at B. 2. **Shear just right of A:** From 1 at C dropping linearly to 0.5 at A, then jumping to -0.5, then linearly to 0 at B. 3. **Moment at A:** Zero everywhere. 4. **Moment at D:** Zero up to D, then drops linearly from 1.5 at D to 0 at B. --- ### **Final Answers** **(a) Reaction at A:** - Max value: \(6\,kN\) (point load), \(18\,kN\) (UDL) **(b) Shear just right of A:** - Max positive: \(3\,kN\) (point load), \(4.5\,kN\) (UDL) - Max negative: \(-3\,kN\) (point load), \(-4.5\,kN\) (UDL) **(c) Moment at A:** - Always zero. **(d) Moment at D:** - Max value: \(9\,kNm\) (point load), \(6.75\,kNm\) (UDL) If you need the **influence line diagrams** sketched, let me know!