## Subject:

**Situation**

Flexible cables *BE* and *CD* are used to brace the truss shown below.

1. Determine the load *W* to cause a compression force of 8.9 kN to member *BD*.

A. 7.80 kN | C. 26.70 kN |

B. 35.64 kN | D. 13.35 kN |

2. Which cable is in tension and what is the tensile reaction?

A. BE = 12.58 kN |
C. BE = 6.29 kN |

B. CD = 6.29 kN |
D. CD = 12.58 kN |

3. If *W* = 20 kN, what will be the tensile reaction of member *CE*?

A. 6.67 kN | C. 0 |

B. 13.33 kN | D. 10 kN |

**Answer Key**

Part 1: [ D ]

Part 2: [ C ]

Part 3: [ A ]

Part 2: [ C ]

Part 3: [ A ]

**Solution**

$\Sigma M_F = 0$

$12R_A = 4W$

$R_A = W/3$

**Part 1 ( F_{BD} = 8.9 kN):**

$8(\frac{1}{3}W) = 4F_{BD}$

$\frac{8}{3}W = 4(8.9)$

$W = 13.35 ~ \text{kN}$ Answer: [ D ]

**Part 2:**

$F_{BE}\sin 45^\circ = \frac{1}{3}W$

$F_{BE}\sin 45^\circ = \frac{1}{3}(13.35)$

$F_{BE} = 6.29 ~ \text{kN}$ Answer: [ C ]

**Part 3 ( W = 20 kN):**

$4F_{CE} = 4(\frac{1}{3}W)$

$4F_{CE} = \frac{4}{3}(20)$

$F_{CE} = 6.67 ~ \text{kN}$ Answer: [ A ]