Home   Tell-A-Friend    Discussion Board   Contact Us

civil engineer

Treasure of Civil Engineering Resources

Course materials, Books, Computer software, Quiz,  Conferences, Journals, Research theses, Jobs, Products, Services etc.

          Deflection of pin-jointed planeTruss 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 mm2 and same modulus of elasticity 200GPa.

pin-jointed plane truss

Figure 7-5(a)


According to unit load method the deflection of a joint of truss is given by the following formula

     δE = (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).

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

Featured Links

Moment Distribution CalculatorNew

Easy to use calculator for solving  Indeterminate beams with different load

Fixed Beam Calculator

Shear force and bending moment calculations for different loading cases of Fixed beam

Moment of Inertia calculatorNew

For different sections including I-section and T-section.

Deflection Calculator

Easy to use calculator for different loads on beams

Problem SolverNew

A collection of illustrated solved examples for civil engineers.

RC Beam CalculatorNew

Calculate the strength of reinforced concrete beams

Bending Moment calculatorNew

Calculate Bending moments for simply supported beams

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 FDE =0 and FDC =0;

Similarly the equilibrium conditions at joint F gives; 

   FFE =0   and   FFA =0.

Consider the equilibrium of joint B along y-axis we get FBE =0.

Now consider the equilibrium of joint A.

Σ Fy = 0;

FAE sin45 + Ay - FAF = 0              (i)

FAE sin45 + 0.5 - 0 =0

Therefore, FAE = -0.5/sin45 = -0.707 kN

Σ Fx = 0;  

FAE cos45 + FAB = 0                  (ii)

Therefore, FAB = -FAE 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, FCE = -0.707 kN, FCB = 0.5 kN

All the calculated values are entered in Table 7-5

       Σ NnL =129.9 kN2 m

External virtual work = Internal virtual work

1 kN. δE = (Σ NnL)/AE

 δE = 129.9/[(250x10-6)(200x106)]

  vertical deflection at E =0.0026 m = 2.6 mm      (Ans)

For other problems please visit our Problem SolverNew

Reinforced concreteNew

Analysis and design of reinforced concrete structures


A collection of quiz in different areas of civil engineering

CE HorizonNew

Online Civil Engineering Journal and Magazine

Profile of Civil EngineersNew

Get to know about distinguished civil engineers

Table 7-5 Result of Member Forces Calculations
Member N (kN) n (kN) L (m) NnL (kN2 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

If you don't find the required information please suggest us We are updating the website regularly.

Join the mailing list to get informed about new products or links


Applied Mechanics 

Structural Analysis 

Design of Structures

Construction Materials

Engineering Graphics

Disaster Management




Land Surveying


Environmental Engineering

Irrigation Engineering

Offshore Engineering

Construction Management 

Quantity Surveying

Construction Disputes

Construction Technology

Construction Equipments

Research Papers

Journals & Magazines

Construction Companies


Professional Societies

Computer Software 


Photo Album



Join us

Job Search

Book Store


Colleges & Universities

Learning Support

Last updated on Monday February 04, 2013