STEM Problem Detail

Problem Statement

Prompt

Please give me step wise solution for Problem 4.21

Solution Rendering...

**Degree of static indeterminacy:**

$$
D_s = R_e - 3
$$

$$
R_e = 2 + 1 + 1 = 4
$$

$$
D_s = 4 - 3 = 1
$$

**Let the reaction at A be redundant:**

$$
R_A = R
$$

---

**Equilibrium equations:**

$$
\sum F_x = 0
$$

$$
H_A = 0
$$

$$
\sum F_y = 0
$$

$$
R + R_B + R_C = 30 \times 11
$$

$$
R_B + R_C = 330 - R \quad (i)
$$

**Taking moments about C:**

$$
R \times 11 + R_B \times 6 - 30 \times 11 \times 5.5 = 0
$$

$$
6R_B = 1815 - 11R
$$

$$
R_B = \frac{1815 - 11R}{6}
$$

From (i),

$$
R_C = 330 - R - R_B
$$

$$
R_C = \frac{165 + 5R}{6}
$$

---

**Total strain energy of the system:**

$$
U = U_{AB} + U_{BC}
$$

$$
U = \int_0^5 \frac{M^2}{2EI} dx + \int_0^6 \frac{M^2}{2EI} dx
$$

**Bending moment expressions:**

For portion $$AB$$,

$$
M = Rx - \frac{wx^2}{2}
$$

For portion $$BC$$,

$$
M = R_C x - \frac{wx^2}{2}
$$

---

**Condition for redundant reaction:**

$$
\frac{\partial U}{\partial R} = 0
$$

$$
\frac{\partial U}{\partial R}
= \int_0^5 \frac{2(Rx - \frac{wx^2}{2})x}{2EI} dx
+ \int_0^6 \frac{2(\frac{165+5R}{6}x - \frac{wx^2}{2})\frac{x}{6}}{2EI} dx
$$

$$
\int_0^5 \frac{Rx^2 - \frac{wx^3}{2}}{EI} dx
+ \int_0^6 \frac{(\frac{165+5R}{6}x - \frac{wx^2}{2})x}{EI} dx = 0
$$

---

**Evaluating the integrals:**

$$
12\left[\frac{R x^3}{3} - \frac{w x^4}{8}\right]_0^5
+ 5\left[\frac{165+5R}{6}\frac{x^3}{3} - \frac{w x^4}{8}\right]_0^6 = 0
$$

$$
12\left(\frac{125R}{3} - 2343.75\right)
+ 5\left(\frac{165+5R}{6} \times 72 - 4860\right) = 0
$$

$$
500R - 28125 + 9900 + 300R - 24300 = 0
$$

$$
800R = 42525
$$

$$
R = 53.15 \,\text{kN}
$$

---

**Reactions:**

$$
R_A = 53.15 \,\text{kN}
$$

$$
R_B = \frac{1815 - 11 \times 53.15}{6}
$$

$$
R_B = 205.05 \,\text{kN}
$$

$$
R_C = \frac{165 + 5 \times 53.15}{6}
$$

$$
R_C = 71.80 \,\text{kN}
$$

**Check:**

$$
\sum F_y = 53.15 + 205.05 + 71.80 - 330 = 0
$$

---

**Shear force diagram**

**Portion AB** 
$$(0 < x < 5)$$:

$$
S_x = R_A - wx
$$

$$
S_x = 53.15 - 30x
$$

At $$x = 0$$,

$$
S_A = 53.15 \,\text{kN}
$$

At $$x = 5$$,

$$
S_B^- = 53.15 - 150 = -96.85 \,\text{kN}
$$

**Maximum bending moment in AB:**

$$
M = Rx - \frac{wx^2}{2}
$$

$$
\frac{dM}{dx} = R - wx = 0
$$

$$
x = \frac{R}{w} = \frac{53.15}{30} = 1.77 \,\text{m}
$$

$$
M_{\max} = 53.15(1.77) - \frac{30(1.77)^2}{2}
$$

$$
M_{\max} = 47.08 \,\text{kN·m}
$$

---

**Portion BC**
$$(0 < x < 6)$$:

$$
S_x = -R_C + wx
$$

$$
S_x = -71.80 + 30x
$$

At $$x = 0$$,

$$
S_C = -71.80 \,\text{kN}
$$

At $$x = 6$$,

$$
S_B^+ = 108.2 \,\text{kN}
$$

**Bending moment:**

$$
M = R_C x - \frac{wx^2}{2}
$$

$$
\frac{dM}{dx} = R_C - wx = 0
$$

$$
x = \frac{71.80}{30} = 2.39 \,\text{m}
$$

$$
M_{\max} = 71.80(2.39) - 15(2.39)^2
$$

$$
M_{\max} = 85.92 \,\text{kN·m}
$$

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Problem #58