Question

Using the method of joints, determine FAB, FBC, and FBD, the magnitude of the force in each of the members connected to joint B. Assume for your calculations that each member is in tension, and include in your response the sign of each force that you obtain by applying this assumption. Express your answers numerically in pounds to three significant figures separated by commas.

Answer 1

• FBC = 417.9 lb

• FBD = 146.371 lb

• FAB = 192 lb

Explanation:

Given

P₁ = 192 lb

P₂ = 399 lb

As shown, the truss is loaded by the forces P₁ and P₂.

For the truss, we apply equilibrium as follows

∑Fₓ = 0 (→ +)

⇒ Aₓ - P₁ = 0

⇒ Aₓ = P₁ (→)

∑MA = 0 (+ counterclockwise)

- P₂*a + Cy*2a - P₁*a = 0

⇒ Cy = (P₁ + P₂)/2 (↑)

∑Fy = 0 (↑+)

⇒ Ay - P₂ + Cy = 0

⇒ Ay = P₂ - Cy

⇒ Ay = P₂ - ((P₁ + P₂)/2)

⇒ Ay = (P₂ - P₁)/2 (↑)

Using the method of joints, for the node C we have

∑Fy = 0 (↑+)

Cy - FBCy = 0

where FBCy = FBC*Sin ∅ ⇒ FBCy = FBC*Sin 45°

⇒ FBCy = (√2/2)*FBC

then

Cy - ((√2/2)*FBC) = 0

⇒ FBC = √2*Cy

⇒ FBC = √2*((P₁ + P₂)/2)

⇒ FBC = (√2/2)*(P₁ + P₂) (Compression)

For the join B we have

∑Fy = 0 (↑+)

- P₂ + FBCy + FBDy = 0

Knowing that

FBCy = FBC*Sin 45° = (√2/2)*(P₁ + P₂)*(√2/2)

⇒ FBCy = (P₁ + P₂)/2

and

FBDy = FBD*Sin ∅ ⇒ FBDy = FBD*Sin 45°

⇒ FBDy = (√2/2)*FBD

we have

- P₂ + ((P₁ + P₂)/2) + (√2/2)*FBD = 0

⇒ FBD = (√2/2)*(P₂ - P₁) (Compression)

Finally, we apply

∑Fx = 0 (→+)

- FBCx + FBDx + FAB = 0

⇒ FAB = FBCx - FBDx (i)

If

FBCx = FBC*Cos ∅ ⇒ FBCx = FBC*Cos 45°

⇒ FBCx = ((√2/2)*(P₁ + P₂))*(√2/2)

⇒ FBCx = (P₁ + P₂)/2

and

FBDx = FBD*Cos ∅ ⇒ FBDx = FBD*Cos 45°

⇒ FBDx = ((√2/2)*(P₂ - P₁))*(√2/2)

⇒ FBDx = (P₂ - P₁)/2

we use eq (i)

⇒ FAB = ((P₁ + P₂)/2) - ((P₂ - P₁)/2)

⇒ FAB = P₁ (Compression)

Knowing the magnitudes of P₁ = 192 lb and P₂ = 399 lb we have

FBC = (√2/2)*(192 lb + 399 lb)

⇒ FBC = 417.9 lb

FBD = (√2/2)*(399 lb - 192 lb)

⇒ FBD = 146.371 lb

FAB = 192 lb