**Deflection of Truss by unit load method**

**Problem 7-5**

**A pin-jointed truss is shown in
figure 7-5(a). Determine the vertical displacement of joint E
by using **
**
unit load method****.
All the members have cross-sectional area of 250 mm ^{2} and same modulus
of elasticity 200GPa.**

**Figure 7-5(a)**

**Solution:**

According to unit load method (also known as virtual work method) the deflection of a joint of truss is given by the following formula

δ_{E} = (Σ
NnL)/AE

Here we have to find the member forces two times. First we will calculate member forces "N" due to the real loading and then "n" due to unit virtual load applied at the point of required deflection (in this case joint E). Tensile forces are considered as positive and compressive forces as negative. L is the length of the member, A is area of cross-section of the member and E is modulus of elasticity of the material of the truss member.

__ Step 1__: The member forces
"N" due to
real load are calculated in
Example 3-1 .So, we take those values.

**Figure 7-5(b)**

__ Step 2:__ Calculation of member
forces "n" due to unit virtual load applied at E as shown in
figure 7-5(b).

As the unit load is applied at center of the truss, the support reactions at A and C will be 0.5 kN each.

Considering the equilibrium of joint D,
we get F_{DE} =0 and F_{DC}
=0;

Similarly the equilibrium conditions at joint F gives;

F_{FE} =0 and F_{FA}
=0.

Consider the equilibrium of joint B
along y-axis we get F_{BE} =0.

Now consider the equilibrium of joint A.

Σ F_{y} = 0;

F_{AE} sin45 + A_{y}
- F_{AF} = 0
(i)

F_{AE}
sin45 + 0.5 - 0 =0

Therefore, F_{AE}
= -0.5/sin45 = -0.707 kN

Σ F_{x} = 0;

F_{AE}
cos45 + F_{AB} = 0
(ii)

Therefore, F_{AB} =
-F_{AE}
cos45 =0.5 kN

As the loading is symmetric, the equilibrium of joint C will also yield the forces similar to the member forces at A

Therefore, F_{CE} =
-0.707 kN, F_{CB} = 0.5 kN

All the calculated values are entered in Table 7-5

(In case of mobile device please scroll horizontally to view the full width of table)

Table 7-5 Result of Member
Forces Calculations |
||||

Member |
N (kN) |
n (kN) |
L (m) |
NnL
(kN^{2} m) |

AB | 22.5 | 0.5 | 2 | 22.5 |

AF | -25 | 0 | 2 | 0 |

AE | -31.82 | -0.707 | 2.83 | 63.67 |

BC | 22.5 | 0.5 | 2 | 22.5 |

BE | 20 | 0 | 2 | 0 |

CD | 0 | 0 | 2 | 0 |

CE | -10.61 | -0.707 | 2.83 | 21.23 |

DE | -15 | 0 | 2 | 0 |

EF | 0 | 0 | 2 | 0 |

**Σ
NnL =129.9 kN**^{2}** m**

External virtual work = Internal virtual work

1 kN. δ_{E} = (Σ
NnL)/AE

δ_{E}
= 129.9/[(250x10^{-6})(200x10^{6})]

**Vertical displacement
of Joint E =0.0026
m = 2.6 mm (Ans)**

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