# Truss With Tension-Only Diagonals

**Situation**

Diagonals *BG*, *CF*, *CH*, and *DG* of the truss shown can resist tension only.

If *W* = 3 kN and *P* = 0, find the following:

1. the force in member *CF*.

A. 4.76 kN | C. 4.67 kN |

B. 4.32 kN | D. 4.23 kN |

2. the force in member *BF*.

A. 3.2 kN | C. 3.4 kN |

B. 3.3 kN | D. 3.5 kN |

3. the force in member *DH*.

A. 2.8 kN | A. 2.5 kN |

B. 2.8 kN | D. 2.7 kN |

**Answer Key**

Part 2: [ B ]

Part 3: [ D ]

**Solution**

**Reaction at**

*B*

$\Sigma M_E = 0$

$7.5R_B = 3(1.5) + 4.5(3) + 2.25(3)$

$R_B = 3.3 ~ \text{kN}$

**Section Through M-M**

$\Sigma F_V = 0$

$CF_V = R_B$

$CF \sin 45^\circ = 3.3$

$CF = 3.3\sqrt{2} ~ \text{kN tension}$

$CF = 4.67 ~ \text{kN tension}$ ← [ C ] *answer for part 1*

**At Joint B**

$\Sigma F_V = 0$

$BF = R_B$

$BF = 3.3 ~ \text{kN compression}$ ← [ B ] *answer for part 2*

**Section Through N-N**

$\Sigma F_V = 0$

$DG_V + 3 = R_B$

$DG(4/5) + 3 = 3.3$

$DG = 0.375 ~ \text{kN tension}$

**From Joint D**

$\Sigma F_V = 0$

$DH + (4/5)DG = 3$

$DH + (4/5)(0.375) = 3$

$DH = 2.7 ~ \text{kN tension}$ ← [ D ] *answer for part 3*