Calculate the horizontal displacement of the support ‘B’ for the
Calculate the horizontal displacement of the support ‘B’ for the truss shown in the figure below. All members have 4000 mm2 area and modulus of elasticity 2 × 105 MPa.
A. 10 mm
B. 0.15 mm
C. 0.2 mm
D. 5 mm
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Explanation:
Member forces due to 40 kN load.
RA + RB = 0
HA = 40 kN
∑ MA = 0 ⇒ RB × 3 = 40 × 4
\( \Rightarrow {R_B} = \frac{{160}}{3}kN\)
\(\Rightarrow {R_A} = \frac{{ - 160}}{3}kN\)
Considering joint A,
DAB = FAB = 40 kN (Tensile)
\({F_{AC}} = \frac{{160}}{3}kN\) (Tensile)
Considering joint B,
FBC = RB sin θ
\( = \frac{{160}}{3} \times \frac{4}{5} = \frac{{128}}{3}\left( {Comp.} \right)\)
Member forces due to unit load
KAB = 1 kN (Tensile)
KAC = KBC = 0
|
No. |
Member |
Length |
P |
K |
\(\frac{{PKL}}{{AE}}\) |
|
1 |
AB |
3 |
40 |
1 |
\(\frac{{40 \times 1 \times 3 \times {{10}^3}}}{{2 \times {{10}^5} \times 4000}}\) |
|
2 |
BC |
5 |
\(\frac{{ - 128}}{3}\) |
0 |
0 |
|
3 |
AC |
4 |
\(\frac{{160}}{3}\) |
0 |
0 |
|
|
|
|
|
|
\(\sum \frac{{PKL}}{{AE}} = 1.5 \times {10^{ - 4}}m\) = 0.15 mm |
(ΔH) of B = 0.15 mm
