Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a10 mm diameter, determine the maximum value of the average normal stress in the link BD if θ = 15 degrees.

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 Step 1

Assumption

• Horizontal shear acting on the inner column is equal to the double of horizontal shear on outer columns.
• Shear is acting at midpoint of each member.

Shear forces assumed on each member

IJ=R

JK=S

KL=T

IE=P

JF=2P

KG=2P

LH=P

EF=U

FG=V

GH=W

EA=Q

BF=2Q

CG=2Q

DH=Q

Step 2

Horizontal shear

For storey first

$$\begin{gathered} \mathrm{P }+ \mathrm{2 }\mathrm{P }+ \mathrm{2 }\mathrm{P }+ \mathrm{P }= \mathrm{20 } \\ \mathrm{6 }\mathrm{P }= \mathrm{20 } \\ \mathrm{P }= \mathrm{3 }. \mathrm{33 }\mathrm{k }N \end{gathered}$$

For bottom storey

$$\begin{gathered} \mathrm{Q }+ \mathrm{2 }\mathrm{Q }+ \mathrm{2 }\mathrm{Q }+ \mathrm{Q }= \mathrm{40 }+ \mathrm{20 } \\ \mathrm{6 }\mathrm{Q }= \mathrm{60 } \\ \mathrm{Q }= \mathrm{10 }\mathrm{k }N \end{gathered}$$

Taking joint I

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{I }}= \mathrm{0 } \\ \mathrm{R }\times \mathrm{4 }- \mathrm{P }\times \mathrm{2 }+ \mathrm{50 }\times \mathrm{0 }= \mathrm{0 } \\ \mathrm{R }= \frac{\mathrm{3 }. \mathrm{33 }\times \mathrm{2 }}{\mathrm{4 }} \\ \mathrm{R }= \mathrm{1 }. \mathrm{665 }\mathrm{k }N \end{gathered}$$

Taking joint J

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{I }}= \mathrm{0 } \\ - \mathrm{S }\times \mathrm{3 }+ \mathrm{2 }\mathrm{P }\times \mathrm{2 }- \mathrm{R }\times \mathrm{4 }= \mathrm{0 } \\ - \mathrm{S }\times \mathrm{4 }+ \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }\times \mathrm{2 }- \mathrm{1 }. \mathrm{665 }\times \mathrm{4 }= \mathrm{0 } \\ \mathrm{S }= \mathrm{2 }. \mathrm{22 }\mathrm{k }N \end{gathered}$$

Taking joint K

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{I }}= \mathrm{0 } \\ - \mathrm{S }\times \mathrm{3 }+ \mathrm{2 }\mathrm{P }\times \mathrm{2 }- \mathrm{T }\times \mathrm{4 }= \mathrm{0 } \\ - \mathrm{2 }. \mathrm{22 }\times \mathrm{3 }+ \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }\times \mathrm{2 }- \mathrm{T }\times \mathrm{4 }= \mathrm{0 } \\ \mathrm{T }= \mathrm{1 }. \mathrm{665 }\mathrm{k }N \end{gathered}$$

For bottom storey

Taking joint E

$$\begin{gathered} \underset{}{\overset{}{\sum }}\mathrm{M }= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{Q }+ \mathrm{2 }\times \mathrm{P }- \mathrm{4 }\times \mathrm{U }= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{10 }+ \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }- \mathrm{4 }\times \mathrm{U }= \mathrm{0 } \\ \mathrm{U }= \mathrm{9 }. \mathrm{165 }\mathrm{k }N \end{gathered}$$

Taking joint F

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{F }}= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{2 }\mathrm{Q }+ \mathrm{2 }\times \mathrm{2 }\mathrm{P }- \mathrm{4 }\times \mathrm{U }- \mathrm{3 }\times \mathrm{V }= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{2 }\times \mathrm{10 }+ \mathrm{2 }\times \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }- \mathrm{4 }\times \mathrm{9 }. \mathrm{165 }- \mathrm{3 }\times \mathrm{V }= \mathrm{0 } \\ \mathrm{V }= \mathrm{12 }. \mathrm{22 }\mathrm{k }N \end{gathered}$$

Taking joint G

$$\begin{gathered} \underset{}{\overset{}{\sum }}{\mathrm{M }}_{\mathrm{F }}= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{2 }\mathrm{Q }+ \mathrm{2 }\times \mathrm{2 }\mathrm{P }- \mathrm{4 }\times \mathrm{W }- \mathrm{3 }\times \mathrm{V }= \mathrm{0 } \\ \mathrm{3 }\times \mathrm{2 }\times \mathrm{10 }+ \mathrm{2 }\times \mathrm{2 }\times \mathrm{3 }. \mathrm{33 }- \mathrm{4 }\times \mathrm{W }- \mathrm{3 }\times \mathrm{12 }. \mathrm{22 }= \mathrm{0 } \\ \mathrm{W }= \mathrm{9 }. \mathrm{165 }\mathrm{k }N \end{gathered}$$

Shear forces

P= 3.33 kN

Q= 10 kN

R= 1.665 kN

S= 2.22 kN

T= 1.665 kN

U= 9.165 kN

V= 12.22 kN

W= 9.165 kN

Step 3

Axial force calculations

In column AEI

Axial force at AEI=10.33 kN

In column BFJ

Axial force at BFJ=-2.5 kN (Upward direction)

In column CGK

Axial force at CGK= 2.5 kN

In column DHL

Axial force at DHL=8 kN