Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

seat work 1. a beam supported on three reactions as shown consists of t…

Question

seat work

  1. a beam supported on three reactions as shown consists of two segment joined at a frictionless hinge at which the bending moment is zero.

Explanation:

Analyze the beam segments and equilibrium of segment AC

$$ LATEXBLOCK0 $$

Analyze segment DE and solve for reactions

$$ LATEXBLOCK1 $$

Determine maximum shear and maximum bending moment

\[
\begin{aligned}
&\text{Shear force } V(x) \text{ along the beam (from } A \text{ to } E\text{):}\\
&\quad V(0^+) = R_1 = 58\text{ kN}\\
&\quad V(2^-) = 58 - 20 \times 2 = 18\text{ kN}\\
&\quad V(2^+) = 18 - 40 = -22\text{ kN}\\
&\quad V(4^-) = -22\text{ kN}\\
&\quad V(4^+) = -22 + R_2 = -22 - 10 = -32\text{ kN}\\
&\quad V(5) = -32\text{ kN}\\
&\quad V(6) = -32\text{ kN}\\
&\quad V(10^-) = -32 - 20 \times 4 = -112\text{ kN}\\
&\quad V(10^+) = -112 + R_3 = -112 + 48 = -64\text{ kN} \implies V_{\max} = 112\text{ kN}\\
&\text{Bending moment } M(x) \text{ along the beam:}\\
&\quad M(0) = 0\\
&\quad M(2) = 58(2) - 20 \times 2 \times 1 = 76\text{ kN}\cdot\text{m}\\
&\quad M(4) = 58(4) - 40(2) - 40(3) = 232 - 80 - 120 = 32\text{ kN}\cdot\text{m}\\
&\quad M(5) = 0\text{ (at hinge } D\text{)}\\
&\quad M(6) = -32(1) = -32\text{ kN}\cdot\text{m}\\
&\quad M(10) = 0\\
&\quad \text{Maximum positive moment: } M_{\max}^+ = 76\text{ kN}\cdot\text{m}\\
&\quad \text{Maximum negative moment: } M_{\max}^- = -32\text{ kN}\cdot\text{m} \implies M_{\max} = 76\text{ kN}\cdot\text{m}
\end{aligned…

Answer:

  • Reaction Forces:
  • \(R_1 = 58\text{ kN}\) (upward)
  • \(R_2 = -10\text{ kN}\) (downward)
  • \(R_3 = 48\text{ kN}\) (upward)
  • Maximum Shear Force:
  • \(V_{\max} = 112\text{ kN}\) (at the left of support \(R_3\))
  • Maximum Bending Moment:
  • \(M_{\max} = 76\text{ kN}\cdot\text{m}\) (at \(2\text{ m}\) from the left end \(A\))