Step 1: Define coordinate system and load function
Let \(x\) measure distance from left support, \(0 \leq x \leq 6\,m\). The load intensity is constant: \(w(x) = 2\, \text{kN/m}\).

Step 2: Integrate load to get shear force
\[ V(x) = - \int w(x) \, dx + C_1 = - \int 2 \, dx + C_1 = -2x + C_1 \]

Step 3: Integrate shear force to get bending moment
\[ M(x) = \int V(x) \, dx + C_2 = \int (-2x + C_1) \, dx + C_2 = -x^2 + C_1 x + C_2 \]

Step 4: Apply boundary conditions (supports)
For simply supported beam: - At \(x=0\), moment \(M=0\) - At \(x=L=6\,m\), moment \(M=0\) Also, shear force at supports can be used, but moment zero is sufficient here.

Apply \(M(0)=0\):\quad \[ 0 = -0 + 0 + C_2 \Rightarrow C_2 = 0 \]

Apply \(M(6)=0\):\quad \[ 0 = -(6)^2 + C_1 (6) + 0 \Rightarrow 0 = -36 + 6C_1 \Rightarrow C_1 = 6 \]

Final expressions:
\[ V(x) = -2x + 6 \quad \text{(kN)} \] \[ M(x) = -x^2 + 6x \quad \text{(kN·m)} \]

Step 5: Sketch diagrams

Interpretation: Shear force starts at +6 kN at left support, decreases linearly to -6 kN at right support. Bending moment forms a parabola, zero at ends, maximum at center \(x=3\ m\).

Step 1: Define load function \(w(x)\)
The load is zero everywhere except at \(x=4\,m\) where the point load acts. Mathematically, a point load can be represented by a Dirac delta function \(\delta(x - 4)\) scaled by magnitude \(P\):

\[ w(x) = P \cdot \delta(x - 4) \]

For integration, we treat the beam in two parts: \(0 \leq x < 4\) and \(4 < x \leq 8\).

Step 2: Integrate load to find shear force
In regions without load, shear force is constant; at the point load, shear force experiences a jump of magnitude \(P\).

Let \(V(x)\) be the shear force function:

• For \(0 \leq x < 4\), no load yet, so \(w(x) = 0\), \(V(x) = C_1\).
• At \(x=4\), shear force drops by \(P\), so for \(4 < x \leq 8\), \(V(x) = C_1 - P\).

Step 3: Integrate shear force for bending moment
For \(0 \leq x < 4\): \[ M(x) = \int V(x) dx + C_2 = C_1 x + C_2 \]

For \(4 < x \leq 8\): \[ M(x) = \int (C_1 - P) dx + C_3 = (C_1 - P)x + C_3 \]

Step 4: Apply boundary and continuity conditions

• At \(x=0\), \(M=0 \Rightarrow C_2 = 0\).
• At \(x=8\), \(M=0 \Rightarrow (C_1 - P)(8) + C_3 = 0\).
• Moment is continuous at \(x=4\):\quad \(M(4^{-}) = M(4^{+})\): \[ C_1 (4) + 0 = (C_1 - P)(4) + C_3 \Rightarrow 4C_1 = 4C_1 - 4P + C_3 \Rightarrow C_3 = 4P \]
• Shear force sum equals reactions at supports. By symmetry: \[ R_A = R_B = \frac{P}{2} = 5\, kN \] So, \[ V(0) = R_A = C_1 = 5 \]
• Using boundary \(M(8)=0\): \[ (5 - 10)(8) + 4 \times 10 = (-5)(8) + 40 = -40 + 40 = 0 \] Satisfied.

Final forms:

For \(0 \leq x < 4\):
\[ V(x) = 5\, kN \] \[ M(x) = 5x\, \text{kN·m} \] For \(4 < x \leq 8\):
\[ V(x) = 5 - 10 = -5\, kN \] \[ M(x) = -5x + 40\, \text{kN·m} \]

Step 5: Sketch diagrams
Shear force is a step function: +5 kN on left half, -5 kN on right half.
Bending moment increases linearly to max at mid-span, then decreases linearly to zero.

Step 1: Define load function
The combined load function is: \[ w(x) = 1.5 + 5 \cdot \delta(x - 6) \] where \(\delta\) is the Dirac delta for the point load at 6 m.

Step 2: Integrate load for shear force
The load has two parts: continuous uniform load and a point load.

Over the beam segments, shear force changes as follows:

• For \(0 \leq x < 6\), \[ V(x) = -\int 1.5\, dx + C_1 = -1.5x + C_1 \]
• At \(x=6\), shear drops by 5 kN due to point load.
• For \(6 < x \leq 10\), \[ V(x) = -1.5x + C_1 - 5 \]

Step 3: Integrate shear force for bending moment
For \(0 \leq x < 6\): \[ M(x) = \int (-1.5x + C_1) dx + C_2 = -0.75x^2 + C_1 x + C_2 \] For \(6 < x \leq 10\): \[ M(x) = \int (-1.5x + C_1 - 5) dx + C_3 = -0.75x^2 + (C_1 - 5) x + C_3 \]

Step 4: Apply boundary and continuity conditions

• At \(x=0\), \(M=0 \Rightarrow C_2=0\).
• At \(x=10\), \(M=0\): \[ 0 = -0.75 (10)^2 + (C_1 - 5)(10) + C_3 = -75 + 10C_1 - 50 + C_3 = 10C_1 + C_3 -125 \] \[ \Rightarrow C_3 = 125 - 10C_1 \]
• Moment continuity at \(x=6\): \[ M(6^-) = M(6^+) \Rightarrow -0.75 (6)^2 + 6C_1 + 0 = -0.75(6)^2 + 6(C_1 - 5) + C_3 \] \[ 27C_1 = 36 + C_3 \Rightarrow 6C_1 = 6C_1 -30 + C_3 \Rightarrow 0 = -30 + C_3 \Rightarrow C_3 = 30 \]

Substitute \(C_3 = 30\) into previous equation: \[ 30 = 125 - 10 C_1 \Rightarrow 10 C_1 = 95 \Rightarrow C_1 = 9.5 \]

Step 5: Expressions:

For \(0 \leq x < 6\): \[ V(x) = -1.5 x + 9.5 \] \[ M(x) = -0.75 x^2 + 9.5 x \] For \(6 < x \leq 10\): \[ V(x) = -1.5 x + 9.5 - 5 = -1.5 x + 4.5 \] \[ M(x) = -0.75 x^2 + 4.5 x + 30 \]

Step 6: Support reactions (for verification):
Total load \(= 1.5 \times 10 + 5 = 20\, kN\). Using equilibrium: \[ R_A = ? \] Taking moments about \(A\): \[ R_B \times 10 = 1.5 \times 10 \times 5 + 5 \times 6 = 75 + 30 = 105 \] \[ R_B = 10.5\, kN, \quad R_A = 20 - 10.5 = 9.5\, kN \] Matches the \(C_1\) found, confirming solution consistency.

Step 1: Load function
\[ w(x) = 3\, \text{kN/m}, \quad 0 \leq x \leq 4 m \] Take \(x\) from the fixed end toward the free end.

Step 2: Integrate load for shear force
\[ V(x) = - \int w(x) dx + C_1 = -3 x + C_1 \]

Step 3: Integrate shear force for bending moment
\[ M(x) = \int V(x) dx + C_2 = \int (-3 x + C_1) dx + C_2 = -1.5 x^2 + C_1 x + C_2 \]

Step 4: Apply boundary conditions
At free end \(x = L = 4\, m\), both shear force and bending moment are zero because there is no support:

• Shear force at free end: \[ V(4) = -3 \times 4 + C_1 = 0 \Rightarrow C_1 = 12 \]
• Bending moment at free end: \[ M(4) = -1.5(4)^2 + 12 \times 4 + C_2 = -24 + 48 + C_2 = 24 + C_2 = 0 \Rightarrow C_2 = -24 \]

Step 5: Final expressions:

Interpretation:
At fixed end \(x=0\), shear force: \[ V(0) = 12\, kN \] and bending moment: \[ M(0) = -24\, kN \cdot m \] (Negative sign indicates direction).

Step 6: Sketch diagrams

Step 1: Load function
Since the load varies linearly from 0 to 4 kN/m over 8 m, the load intensity is: \[ w(x) = k x \] where \(k\) is the slope. At \(x=8\), \[ w(8) = 4 = k \times 8 \Rightarrow k = \frac{4}{8} = 0.5\, \text{kN/m}^2 \] So, \[ w(x) = 0.5 x \]

Step 2: Integrate load for shear force
\[ V(x) = - \int w(x) dx + C_1 = - \int 0.5 x \, dx + C_1 = -0.25 x^2 + C_1 \]

Step 3: Integrate shear force for bending moment
\[ M(x) = \int V(x) dx + C_2 = \int (-0.25 x^2 + C_1 ) dx + C_2 = -\frac{0.25}{3} x^3 + C_1 x + C_2 = -\frac{1}{12} x^3 + C_1 x + C_2 \]

Step 4: Apply boundary conditions
Simply supported beam: \[ M(0) = 0, \quad M(8) = 0 \] At \(x=0\): \[ M(0) = 0 + 0 + C_2 = 0 \Rightarrow C_2 = 0 \] At \(x=8\): \[ 0 = -\frac{1}{12} (8)^3 + 8 C_1 + 0 \Rightarrow 0 = -\frac{512}{12} + 8 C_1 \Rightarrow 8 C_1 = \frac{512}{12} = 42.67 \] \[ C_1 = \frac{42.67}{8} = 5.33 \]

Step 5: Final expressions \[ V(x) = -0.25 x^2 + 5.33 \] \[ M(x) = -\frac{1}{12} x^3 + 5.33 x \]

Step 6: Support reactions (verify)
Total load \(= \text{Area of triangular load curve} = \frac{1}{2} \times 8 \times 4 = 16\, kN\). Taking moments about left support: \[ R_B \times 8 = \text{Moment of load about A} = \frac{1}{2} \times 8 \times 4 \times \frac{2}{3} \times 8 = 16 \times \frac{16}{3} = 85.33 \, kN\cdot m \] \[ R_B = \frac{85.33}{8} = 10.67\, kN, \quad R_A = 16 - 10.67 = 5.33\, kN \] Shear at \(x=0\) is \(V(0) = 5.33\, kN\), matching \(R_A\), confirming our constants.

Step 7: Diagram sketch