Answer:
Step-by-step 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
