Question

second column benn

+ 1 Links BC and DE are both made of steel (E = 29 x 106 psi and a = 7.2 x 10- /deg F) and are 0.5 in wide and 0.125 in thick as shown in Figure 1. Table 1 details the values of P, 0, Lbc and Lpe. Determine: a) The stresses acting in each link when a load P is applied at an angle to the rigid member AF [9 marks), b) The stresses acting each link when a load P is applied at an angle 0 together with a temperature decrease by 212 deg F [6 marks). 4 in B С 2 in. DO in. F Figure 1 Table 1:P, , Lacand pe values. Student Last Name P (lbf) (deg) Acosta to Barounian 350 45 Benn to Chattopadhyay 450 40 Chaudhry to Francia 550 35 Ganeshamoorthy to Hashi 650 30 He to Kristipati 750 25 Madhu to Morales 300 45 Moreno to Paul-Bamidele 400 40 Qadeer to Saeed 500 35 Sanchez to Tahal 600 30 Tiamiyu to Zhang 700 25 Lec (in) LDE (in) 2.5 3.0 2.5 3.5 2.5 4.0 2.5 4.5 2.5 5.0 3.5 3 3.5 3.5 4 3.5 4.5 3.5 3.5

Answer 1

A)

The veritical component of the force will not do anything as it will be directly countered by a reaction from the bottom. However, the horizontal component will actually cause the stresses.

Horizontal component of force =

Pcos = 450 x cos(40) = 344.719 lb

FBD:

344.719 lb(f) 4 2.5 R(BC) B C2 3.5 E R(DE) D 2 O

Since the point "O" is hinged,

ΣΜ, = 0

344.719 x 8 + RDE X 2 = RBC X 4

Also,

FH = 0

$-344.719+R_{BC}-R_{DE}=0$

Solving these two simultaneously gives us,

R(DE) = 689.438 lb(f)

R(BC) = 1034.157 lb(f)

Stresses in BC and DE,

OBC RBC Area

where

• R(BC) = 1034.157 lbf
• Area = 0.5 x 0.125 = 0.0625 in2

OBC RBC Area

$\mathbf{\sigma _{BC}}=\frac{1034.157}{0.0625}=\mathbf{16546.512\:psi=16.546\:ksi=-16.546\:ksi}$ [Compressive stress]

$\mathbf{\sigma _{DE}}=\frac{R_{DE}}{Area}=\frac{689.438}{0.0625}=\mathbf{11031.008\:psi=11.031\:ksi}$ [Tensile stress]

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B)

In this case,

- RBC - (aTE) OBC Area

where

• R(BC) = 1034.157 lbf
• Area = 0.0625 in2
• $\alpha$ = 7.2e-6 1/F
• T = Temperature change = -212 F
• E = 29e6 psi

- RBC - (aTE) OBC Area

$\mathbf{\sigma _{BC}}=\frac{-1034.157}{0.0625}-\left (7.2e-6\times -212\times 29e6\right )=\mathbf{+27719.088\:psi}$ [Tensile]

Similarly,

where

• R(DE) = 689.438 lbf
• Area = 0.0625 in2
• $\alpha$ = 7.2e-6 1/F
• T = Temperature change = -212 F
• E = 29e6 psi

$\sigma _{DE}=\frac{R_{DE}}{Area}-\left (\alpha TE\right )$

$\mathbf{\sigma _{DE}}=\frac{689.438}{0.0625}-\left (7.2e-6\times -212\times 29e6\right )=\mathbf{+55296.508\:psi}$[Tensile]

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