# Strength of Temporary Earth Retaining Wall Made from Wooden Planks

**Situation**

A temporary earth retaining wall consists of wooden plank driven vertically into the ground. The wall is designed to resist 2.4 m height of soil.

Cross-sectional dimensions of the plank = 300 mm wide × 75 mm thick

Allowable bending stress of the plank = 10.4 MPa

Allowable shear stress of the plank = 0.8 MPa

Unit weight of retained soil = 17.3 kN/m

^{3}

Active earth pressure coefficient = 1/3

1. Calculate the maximum flexural stress.

A. 12.7 MPa | C. 8.6 MPa |

B. 14.2 MPa | D. 10.1 MPa |

2. Calculate the maximum shear stress.

A. 1.11 MPa | C. 0.99 MPa |

B. 0.33 MPa | D. 0.77 MPa |

3. Calculate the minimum thickness of the plank to prevent failure.

A. 90 mm | C. 110 mm |

B. 80 mm | D. 100 mm |

**Answer Key**

Part 2: [ B ]

Part 3: [ A ]

**Solution**

$F_a = \left( \frac{1}{2}K_a \gamma H^2 \right)(0.3)$

$F_a = \left[ \frac{1}{2}(\frac{1}{3})(17.3)(2.4^2) \right](0.3)$

$F_a = 4.9824 ~ \text{kN}$

$V_{max} = R = F_a$

$V_{max} = 4.9824 ~ \text{kN}$

$M_{max} = M = 0.8F_a$

$M_{max} = 0.8(4.9824)$

$M_{max} = 3.985\,92 ~ \text{kN}\cdot\text{m}$

Maximum Bending Stress

$f_b = \dfrac{6(3.985\,92)(1000^2)}{300(75^2)}$

$f_b = 14.17 ~ \text{MPa}$ ← [ B ] *answer for part 1*

Maximum Shear Stress

$f_v = \dfrac{3(4.9824)(1000)}{2(300)(75)}$

$f_v = 0.332 ~ \text{MPa}$ ← [ B ] *answer for part 2*

Minimum thickness of plank to prevent failure

(

*f*= 14.17 MPa) > (

_{b}*F*= 10.4 MPa) ← failed in bending

_{b}From part (2)

(*f _{v}* = 0.332 MPa) < (

*F*= 0.8 MPa) ← passed in shear

_{v}To prevent failure, increase the thickness of the plank according to bending

$F_b = \dfrac{6M}{bd^2}$

$10.4 = \dfrac{6(3.985\,92)(1000^2)}{300d^2}$

$d = 87.55 ~ \text{mm}$

Use d = 90 mm ← [ A ] *answer for part 3*