## Subject:

**Situation**

The bridge truss shown in the figure is to be subjected by uniform load of 10 kN/m and a point load of 30 kN, both are moving across the bottom chord

Calculate the following:

1. The maximum axial load on member *JK*.

A. 64.59 kN | C. -64.59 kN |

B. -63.51 kN | D. 63.51 kN |

2. The maximum axial load on member *BC*.

A. 47.63 kN | C. -47.63 kN |

B. -74.88 kN | D. 74.88 kN |

3. The maximum compression force and maximum tension force on member *CG*.

A. -48.11 kN and 16.36 kN |

B. Compression = 0; Tension = 16.36 kN |

C. -16.36 kN and 48.11 kN |

D. Compression = 48.11 kN; Tension = 0 |

**Answer Key**

Part 2: [ A ]

Part 3: [ C ]

**Solution**

**Member**

*JK**M*= 0

_{C}Location of Unit Load = Point

*C*

$U_{JK} = -\dfrac{ab}{Ld} = -\dfrac{8(8)}{16(8\sin 60^\circ)}$

$U_{JK} = -\sqrt{3}/3$

**Influence Diagram for Member**

*JK*

$F_{JK} = 10\left[ \frac{1}{2}(16)\left( -\sqrt{3}/3 \right) \right] + 30\left( -\sqrt{3}/3 \right)$

$F_{JK} = -63.51 ~ \text{kN}$ ← [ B ] *answer for part 1*

**Member BC**

*M*= 0

_{J}Location of Unit Load = Point

*B*

$U_{BC} = +\dfrac{ab}{Ld} = \dfrac{4(12)}{16(8\sin 60^\circ)}$

$U_{BC} = \sqrt{3}/4$

**Influence Diagram for Member**

*BC*

$F_{BC} = 10\left[ \frac{1}{2}(16)\left( \sqrt{3}/4 \right) \right] + 30\left( \sqrt{3}/4 \right)$

$F_{BC} = 47.63 ~ \text{kN}$ ← [ A ] *answer for part 2*

**Member CG**

*F*= 0

_{V}Location of Unit Load = Point *B* (Within the Section)

$U_{CGv} = -\dfrac{a}{L}$

$U_{CG}\sin 60^\circ = -\dfrac{4}{16}$

$U_{CG} = -\sqrt{3}/6$

Location of Unit Load = Point *C* (Outside the Section)

$U_{CGv} = +\dfrac{b}{L}$

$U_{CG}\sin 60^\circ = \dfrac{8}{16}$

$U_{CG} = \sqrt{3}/3$

**Influence Diagram for Member**

*CG*

$y = \sqrt{3}/3 - \left( -\sqrt{3}/6 \right) = \sqrt{3}/2$

$\dfrac{x - 8}{\sqrt{3}/3} = \dfrac{4}{\sqrt{3}/2}$ → $x = 32/3$

Maximum Compression Force:

$F_{CG} = 10\left[ \frac{1}{2}\left(16 - \frac{32}{3} \right)\left( -\sqrt{3}/6 \right) \right] + 30\left( -\sqrt{3}/6 \right)$

$F_{CG} = -16.36 ~ \text{kN}$

Maximum Tension Force:

$F_{CG} = 10\left[ \frac{1}{2}\left(\frac{32}{3} \right)\left( \sqrt{3}/3 \right) \right] + 30\left( \sqrt{3}/3 \right)$

$F_{BC} = 47.63 ~ \text{kN}$

Answer for part 3 = [ C ]