Concept explainers
Calculate the horizontal deflection and vertical deflection at joint B.
Answer to Problem 1P
The horizontal deflection at joint B is 0.567 in.←_.
The vertical deflection at joint B is 0.126 in.↑_.
Explanation of Solution
Given information:
The truss is given in the Figure.
The value of E is 29,000 ksi and the value of A is 5.25 in.2
Procedure to find the deflection of truss by virtual work method is shown below.
- For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.
- For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces (Fv) in all the member of the truss.
- Finally use the desired deflection equation.
Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.
- For summation of forces along x-direction is equal to zero (∑Fx=0), consider the forces acting towards right side as positive (→+) and the forces acting towards left side as negative (←−).
- For summation of forces along y-direction is equal to zero (∑Fy=0), consider the upward force as positive (↑+) and the downward force as negative (↓−).
- For summation of moment about a point is equal to zero (∑Mat a point=0), consider the clockwise moment as negative and the counter clockwise moment as positive.
Method of joints:
The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates Tension (T).
Condition for zero force member:
- 1. If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.
- 2. If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.
Calculation:
Consider the real system.
Find the member axial force (F) for the real system using method of joints:
Let Ay be the vertical reaction at the roller support A.
Let Cx and Cy be the horizontal and vertical reactions at the hinged support C.
Sketch the free body diagram of the truss as shown in Figure 1.
Find the reactions at the supports using equilibrium equations:
Summation of forces along x-direction is equal to 0.
+→∑Fx=0Cx−75=0Cx=75 k→
Summation of moments about C is equal to 0.
∑MC=0−Ay(12)+75(16)=0Ay=100 k↑
Summation of forces along y-direction is equal to 0.
+↑∑Fy=0Ay+Cy=0Cy=−Ay
Substitute 100 k for Ay.
Cy=−100 k=100 k↓
Sketch the free body diagram of joint A as shown in Figure 2.
Apply equilibrium equation to the free body diagram of joint A:
Summation of forces along y-direction is equal to 0.
+↑∑Fy=0100+FABsin53.13°=0FAB=−125 k
Summation of forces along x-direction is equal to 0.
+→∑Fx=0FAC+FABcos53.13°=0
Substitute −125 k for FAB.
FAC+(−125)cos53.13°=0FAC=75 k
Sketch the free body diagram of joint C as shown in Figure 3.
Apply equilibrium equation to the free body diagram of joint C:
Summation of forces along y-direction is equal to 0.
+↑∑Fy=0−100+FBC=0FBC=100 k
Sketch the resultant diagram of the real forces as shown in Figure 4.
Consider the virtual system:
For horizontal deflection apply l k at joint B in horizontal direction.
Find the member axial force (Fv1 ) due to virtual load using method of joints:
Sketch the free body diagram of the truss as shown in Figure 5.
Refer Figure 1 and Figure 5.
The real system is similar to the virtual system except the load of 75k acting in the horizontal direction. So divide the real force (F) by 75 to obtain the virtual force (Fv1).
The virtual force in the member AB is (Fv1)AB=FAB75.
Substitute −125 k for FAB.
(Fv1)AB=−12575=−53 k
The virtual force in the member AC is (Fv1)AC=FAC75.
Substitute 75 k for FAC.
(Fv1)AC=75 k75=1 k
The virtual force in the member BC is (Fv1)BC=FBC75.
Substitute 100 k for FAC.
(Fv1)BC=100 k75=43 k
Sketch the resultant diagram of the virtual system (Fv1) as shown in Figure 6.
Consider the virtual system:
For vertical deflection apply l k at joint B in vertical direction.
Find the member axial force (Fv2) due to virtual load using method of joints:
Sketch the free body diagram of the truss as shown in Figure 7.
Find the reactions at the supports using equilibrium equations:
Summation of forces along x-direction is equal to 0.
+→∑Fx=0Cx=0
Summation of moments about C is equal to 0.
∑MC=0Ay=0
Summation of forces along y-direction is equal to 0.
+↑∑Fy=0Ay+Cy−1=0
Substitute 0 for Ay.
0+Cy−1=0Cy=1 k↑
The reaction at A is zero. So the member AB and AC satisfies the zero force member condition.
Therefore, (Fv2)AB=0 and (Fv2)BC=0.
Sketch the free body diagram of joint C as shown in Figure 8.
Apply equilibrium equation to the free body diagram of joint C:
Summation of forces along y-direction is equal to 0.
+↑∑Fy=01+FBC=0(Fv2)BC=−1 k
Sketch the resultant diagram of the virtual system (Fv2) as shown in Figure 9.
The expression to find the deflection 1(Δ) is shown below:
1(Δ)=∑Fv(FLAE)
Here, L is the length of the member, A is the area of the member, and E is the young’s modulus of the member.
For E and A is constant, the expression becomes,
(1 k)Δ=1AE∑FvFL (1)
Length of the member AC is,
LAC=12 ft=12 ft(12 in.1 ft)=144 in.
Similarly convert the length of the remaining member and show in column 2 of Table 1.
Find the product of Fv1FL for the member AC:
Substitute 1 k for Fv1 , 75 k for F, and 144 in. for L.
Fv1FL=(1)(75)(144)=10,800 k2-in.
Similarly calculate for the remaining member and show in column 5 of Table 1.
Find the product of Fv2FL for the member AC:
Substitute 0 for Fv2 , 75 k for F, and 144 in. for L.
Fv1FL=(0)(75)(144)=0
Similarly calculate for the remaining member and show in column 7 of Table 1.
Frame a table for the above calculated values.
Summarize the calculated values as shown in Table 1.
| Member |
L (in.) | F(k) | Fv1(k) | Fv1FL(k2-in.) | Fv2(k) | Fv2FL(k2-in.) |
| AC | 144 | 75 | 1 | 10,800 | 0 | 0 |
| AB | 240 | −125 | −53 | 50,000 | 0 | 0 |
| BC | 192 | 100 | 43 | 25,600 | −1 | −19,200 |
| ∑ | 86,400 | −19,200 |
Find the horizontal deflection at joint B (ΔBH):
Substitute 86,400 k2-in. for ∑Fv1FL, 5.25 in.2 for A, and 29,000 ksi for E in Equation (1).
(1 k)ΔBH=86,400 k2-in.(5.25 in.2)(29,000 ksi)(1 k)ΔBH=0.567 k-in.ΔBH=0.567 in.
Therefore, the horizontal deflection at joint B is 0.567 in.←_.
Find the vertical deflection at joint B (ΔBV):
Substitute −19,200 k2-in. for ∑Fv2FL, 5.25 in.2 for A, and 29,000 ksi for E in Equation (1).
(1 k)ΔBV=−19,200 k2-in.(5.25 in.2)(29,000 ksi)(1 k)ΔBV=−0.126 k-in.ΔBV=0.126 in.↑
Therefore, the vertical deflection at joint B is 0.126 in.↑_.
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