Problem 2.1. Scarf joint – Solution Manual

Scarf joint of a timber element is given in the Figure. Determine the normal and shear stresses in the plane of the glued surface. Give the allowable tensile force, F based on the resistance of the glue, in the plane of the glued surface

a) if the tensile strength of the glue is: 12 N/mm², and the shear strength is: 5 N/mm²,

Stresses are defined in the plane of the glued surface

b) applying the von Mises yield criterion, the strength of the glue is 12 N/mm².

Solve Problem

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Problem a)

Allowable force, F [kN]=

Problem b)

Allowable force, F [kN]=

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Steps

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Question a)

Step 1. Normal stress arising from the unidirectional tensile force must be transformed to the coordinate system attached to the glued surface. Draw the original x-y and the rotated x’-y’ coordinate systems.

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Step 2. Calculate the angle of rotation and write the transformation matrix. Check angleHide angle

$β = \tan^{- 1} \frac{50}{250} = 0.197$ .

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$T_{σ} = [\begin{matrix} \cos^{2} β & \sin^{2} β & 2 \sin β \cos β \\ \sin^{2} β & \cos^{2} β & - 2 \sin β \cos β \\ - \sin β \cos β & \sin β \cos β & \cos^{2} β - \sin^{2} β \end{matrix}] = [\begin{matrix} 0.962 & 0.038 & 0.385 \\ 0.038 & 0.962 & - 0.385 \\ - 0.192 & 0.192 & 0.923 \end{matrix}]$

Eq.(2-9)

Step 3. Write the stress vector in x-y coordinate system in the function of the unknown tensile force.

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$\{\begin{matrix} σ_{x} \\ σ_{y} \\ τ_{x y} \end{matrix}\} = \{\begin{matrix} \frac{F}{A} \\ 0 \\ 0 \end{matrix}\} = \{\begin{matrix} \frac{F}{150 \times 50} \\ 0 \\ 0 \end{matrix}\}$

Step 4. Perform transformation of stresses

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\{\begin{matrix} σ ‘_{x} \\ σ ‘_{y} \\ τ ‘_{x y} \end{matrix}\} = T_{σ} σ = [\begin{matrix} \cos^{2} β & \sin^{2} β & 2 \sin β \cos β \\ \sin^{2} β & \cos^{2} β & - 2 \sin β \cos β \\ - \sin β \cos β & \sin β \cos β & \cos^{2} β - \sin^{2} β \end{matrix}] \{\begin{matrix} σ_{x} \\ σ_{y} \\ τ_{x y} \end{matrix}\} = [\begin{matrix} 0.96 & 0.04 & 0.38 \\ 0.04 & 0.96 & - 0.38 \\ - 0.19 & 0.19 & 0.92 \end{matrix}] \{\begin{matrix} \frac{F}{150 \times 50} \\ 0 \\ 0 \end{matrix}\} = \{\begin{matrix} 12.82 \\ 0.512 \\ - 2.564 \end{matrix}\} 10^{- 5} F

Step 5. Calculate the allowed forces from tension and from shear. Choose the relevant maximum.

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σ ‘_{x} = f \to 12.82 \times 10^{- 5} F = 12 \frac{N}{{mm}^{2}} \to F_{σ} = 93.6 kN

Check allowed force from shearHide force

τ ‘_{x y} = f_{τ} \to 2.56 \times 10^{- 5} F = 5 \frac{N}{{mm}^{2}} \to F_{τ} = 195 kN

Check maximum allowed forceHide result

The maximum allowed force is the minimum of the above two limits:

F_{m a x} = \min (F_{σ}, F_{τ}) = 93.6 kN

Question b)

Step 1. Determine principal stresses.

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The member is in pure tension stage. Before transformation we had a unidirectional stress, thus $σ_{1} = σ_{x} and σ_{2} = 0$ .

It can be checked also by introducing the stresses, into Eq.2-11.

See Mohr circle in Figure 2.7b

Step 2. Write Von Mises failure criterion and determine the maximum allowed force.

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Von Mises failure criterion gives the maximum allowed force

$\sqrt{σ_{1}^{2} + σ_{2}^{2} - σ_{1} σ_{2}} = f \to \sqrt{{(\frac{F}{A})}^{2} + 0 - 0} = f \to \frac{F}{150 \times 50} = 12 \frac{N}{{mm}^{2}} \to F = 90 kN$

Eq.2-24

Results

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Question a)

Normal stress arising from the unidirectional tensile force must be transformed to the coordinate system attached to the glued surface.

The transformation of the stresses is the following:

Eq.(2-9)

\{\begin{matrix} σ ‘_{x} \\ σ ‘_{y} \\ τ ‘_{x y} \end{matrix}\} = T_{σ} σ = [\begin{matrix} \cos^{2} β & \sin^{2} β & 2 \sin β \cos β \\ \sin^{2} β & \cos^{2} β & - 2 \sin β \cos β \\ - \sin β \cos β & \sin β \cos β & \cos^{2} β - \sin^{2} β \end{matrix}] \{\begin{matrix} σ_{x} \\ σ_{y} \\ τ_{x y} \end{matrix}\} = [\begin{matrix} 0.96 & 0.04 & 0.38 \\ 0.04 & 0.96 & - 0.38 \\ - 0.19 & 0.19 & 0.92 \end{matrix}] \{\begin{matrix} \frac{F}{A} \\ 0 \\ 0 \end{matrix}\} = \{\begin{matrix} 12.82 \\ 0.512 \\ - 2.564 \end{matrix}\} 10^{- 5} F

where the angle of transformation is

$β = \tan^{- 1} \frac{50}{250} = 0.197$ .

The allowed force from tension is

$σ ‘_{x} = f \to 12.82 \times 10^{- 5} F = 12 \frac{N}{{mm}^{2}} \to F_{σ} = 93.6 kN$ .

The allowed force from shear is

$τ ‘_{x y} = f_{τ} \to 2.564 \times 10^{- 5} F = 5 \frac{N}{{mm}^{2}} \to F_{τ} = 195 kN$ .

The maximum allowed force is the minimum of the above two limits:

$F_{m a x} = \min (F_{σ}, F_{τ}) = 93.6 k N$ .

Question b)

The member is in pure tension stage. Before transformation we had a unidirectional stress, thus

$σ_{1} = σ_{x} and σ_{2} = 0$ .

It can be checked also by introducing the stresses, into Eq.2-11.

See Mohr circle in Figure 2.7b

Von Mises failure criterion gives the maximum allowed force

$\sqrt{σ_{1}^{2} + σ_{2}^{2} - σ_{1} σ_{2}} = f \to \sqrt{{(\frac{F}{A})}^{2} + 0 - 0} = f \to \frac{F}{150 \times 50} = 12 \frac{N}{{mm}^{2}} \to F = 90 kN .$

Eq.2-24