A laminated wood beam is built up by gluing together three 2- in.-by-4-in. members to form a solid beam 4-in.-by-6-in. in cross section, as shown. The allowable shear stress in the glued joints is 60 psi. The beam is a 3-ft-long cantilever with a load P applied at its free end. Neglecting the weight of the beam:
(a) Calculate the maximum allowable load P that can be applied at the free end.
(b) Calculate the maximum bending stress in the beam.
The correct answer and explanation is:
To solve this problem, we’ll determine the maximum allowable load PP based on the shear stress at the glued joints and then calculate the maximum bending stress.
(a) Maximum Allowable Load PP Based on Shear Stress
Key Dimensions and Properties:
- Cross-sectional dimensions of the beam:
- Overall beam dimensions: 4 in.×6 in.4 \, \text{in.} \times 6 \, \text{in.}
- Laminations are joined along the 4 in.4 \, \text{in.} direction.
- Length of the beam: L=3 ft=36 in.L = 3 \, \text{ft} = 36 \, \text{in.}
- Allowable shear stress in glued joints: τallow=60 psi\tau_{\text{allow}} = 60 \, \text{psi}
Shear Force at the Joint:
The critical shear force will act at the glued joints. The maximum horizontal shear force VV in the beam occurs at the fixed support due to the applied load PP. For a cantilever beam: V=PV = P
First Moment of Area QQ:
The glued joints split the beam into two horizontal laminations:
- For the middle lamination: 2 in.×4 in.2 \, \text{in.} \times 4 \, \text{in.}
- For the top lamination: 2 in.×4 in.2 \, \text{in.} \times 4 \, \text{in.}
The first moment of area QQ for the section above the joint is: Q=Area above joint×distance to centroid of area above jointQ = \text{Area above joint} \times \text{distance to centroid of area above joint} Q=(2 in.×4 in.)×(4 in.+2 in.2)Q = (2 \, \text{in.} \times 4 \, \text{in.}) \times \left( \frac{4 \, \text{in.} + 2 \, \text{in.}}{2} \right) Q=8 in.2×3 in.=24 in.3Q = 8 \, \text{in.}^2 \times 3 \, \text{in.} = 24 \, \text{in.}^3
Moment of Inertia II:
The moment of inertia of the rectangular beam ( 4 in.×6 in.4 \, \text{in.} \times 6 \, \text{in.}) about its neutral axis is: I=112bh3=112(4)(63)I = \frac{1}{12} b h^3 = \frac{1}{12} (4) (6^3) I=112⋅4⋅216=72 in.4I = \frac{1}{12} \cdot 4 \cdot 216 = 72 \, \text{in.}^4
Shear Stress Formula:
The horizontal shear stress in the glued joint is given by: τ=VQIb\tau = \frac{V Q}{I b}
Where:
- b=4 in.b = 4 \, \text{in.} (width of the glued joint).
Setting τ=τallow\tau = \tau_{\text{allow}}: 60=P(24)72(4)60 = \frac{P (24)}{72 (4)} P=60⋅72⋅424=720 lbP = \frac{60 \cdot 72 \cdot 4}{24} = 720 \, \text{lb}
(b) Maximum Bending Stress
Bending Stress Formula:
The bending stress in the beam is given by: σ=McI\sigma = \frac{M c}{I}
Where:
- M=PLM = P L (maximum moment at the fixed support),
- c=3 in.c = 3 \, \text{in.} (distance from the neutral axis to the outermost fiber),
- I=72 in.4I = 72 \, \text{in.}^4.
Substitute P=720 lbP = 720 \, \text{lb} and L=36 in.L = 36 \, \text{in.}: M=PL=720⋅36=25,920 lb-in.M = P L = 720 \cdot 36 = 25,920 \, \text{lb-in.} σ=McI=25,920⋅372\sigma = \frac{M c}{I} = \frac{25,920 \cdot 3}{72} σ=1,080 psi\sigma = 1,080 \, \text{psi}
Final Answers:
(a) The maximum allowable load PP is: P=720 lbP = 720 \, \text{lb}
(b) The maximum bending stress in the beam is: σ=1,080 psi\sigma = 1,080 \, \text{psi}