HW11+Solutions - HOMEWORK SOLUTIONS Week 11: Due Wednesday...

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HOMEWORK SOLUTIONS Week 11: Due Wednesday 23 November 1. The grain of the wooden block in Figure 1 is inclined at 75 o to the horizontal. If the block is loaded by the stresses shown, find the tensile stress perpendicular to the grain and the shear stress parallel to the grain. 1.2 0.8 1.5 1.2 0.8 0.8 0.8 1.5 all stress components are in MPa Figure 1 Solution The stress components illustrated correspond to σ x = - 1 . 5 ; σ y = 1 . 2 ; τ xy = - 0 . 8 and the inclination of the grain corresponds to a clockwise rotation of the x -plane through 15 o (i.e. a counterclockwise rotation of - 15 o ). The stress components on this rotated plane are given by the equations σ x = σ x cos 2 θ + σ y sin 2 θ + 2 τ xy sin θ cos θ τ x y = τ xy (cos 2 θ - sin 2 θ ) + ( σ y - σ x ) sin θ cos θ 1
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with θ = - 15 o . We obtain σ x = - 1 . 5 cos 2 ( - 15) + 1 . 2 sin 2 ( - 15) - 2 × 0 . 8 sin( - 15) cos( - 15) = - 0 . 919 MPa τ x y = - 0 . 8(cos 2 ( - 15) - sin 2 ( - 15)) + 2 . 7 sin( - 15) cos( - 15) = - 1 . 368 MPa . Notice that the σ x is negative, showing that the normal stress across the grain is compressive. 2
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2. Figure 2 shows a glued lap joint between two pieces of wood of height 3.5 inches and thickness (into the paper) 1.7 inches. The glue fails in shear at a stress τ = 800 psi and in tension at σ = 400 psi. Assuming these two failure modes are independent of each other, what is the maximum angle θ for which the joint would sustain a tensile force F = 1000 lb with a safety factor of 2.5? F
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HW11+Solutions - HOMEWORK SOLUTIONS Week 11: Due Wednesday...

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