Solutions Q1
Q. 1 Attempt any Four questions
a) State and explain Castiglianos First theorem and Principal of Superposition. [05 M]
Ans :
Castiglianos First Theorem :-
In any beam or truss subjected to any load system, the deflection at any point r is given by the partial differential coefficient of the total strain energy stored with respect to a force Pr, acting at the point r in the direction in which the deflection is desired.
y_r=\frac{\partial W_i}{\partial P_r}= Partial differential coefficient of the total strain energy stored with respect to Pr
Principal of Superposition :-
The principle of superposition states that when there are numbers of loads are acting together on an elastic material, the resultant strain will be the sum of individual strains caused by each load acting separately.
Q. 1 b) Explain the two moment area theorem with necessary diagrams. [05 M]
Ans :
As we know, the moment area method applies to prismatic and non-prismatic beams. It is suitable when the point of zero slopes and zero deflection is known.
Thus, it is unsuitable when internal hinges are present because the internal hinge slope suddenly changes. This method is based on the two theorems described below.
Theorems of moment area method:
Theorem 1 :
Change in slope from any point x to B is equal to the Area of the M/EI diagram between x and B.
Mathematically,
θxB = θB – θx = Area of diagram between x and B
If the point of zero slopes is known (A), then the change in slope from A to B equals the Area of the M/EI diagram between A and B.
θAB = θB – θA = Area of M/EI diagram between A and B
But θA = 0, So
θB – 0 = -(1/2) L (PL/EI)
⇒ θB = – (PL2/2EI)
Theorem 2 :
The deflection of any point B, w.r.t tangent at any point x (δB/x) is equal to the moment of Area of M/EI diagram between x and B about B.
δB/x = moment of Area of M/EI diagram between x and B about B
δB/x = Ax¯
By the above diagram,
Total deflection of B = δB1+δB2+δB3
BB1 = ΔB = δx + θx LxB + Ax¯
Q 1 c) For a Three Hinged parabolic arch of span L and rise h carries a UDL of intensity ‘w’ per unit run over the whole span. Show that horizontal thrust at each support of arch is wL2/8h and bending moment at any section of the arch is zero. [05 M]
Ans :
(i) Reactions :
| \sum M_A=0 (↻+ve) | \sum V=0 (↑+ve) |
| w \times l \times \frac{l}{2}-V_a\times l=0 | VA + VB – w = 0 |
| V_A+\frac{wl}{2} = 0 | |
| V_B=\frac{wl}{2} | V_A=\frac{wl}{2} |
∑ H = 0
HA – HB = 0
HA = HB
∑ MC = 0 (↻+ve) (considering only AC part)
V_A\times \frac{l}{2}-H_A\times h-w\times \frac{l}{2}\times \frac{l}{4}=0 \frac{wl}{2}\times \frac{l}{2}-H_A\times h-w\times \frac{l}{2}\times \frac{1}{4} = 0 H_A\times h=\frac{wl}{2}\times \frac{l}{2}-w\times \frac{l}{2}\times \frac{l}{4}=\frac{wl^2}{4}-\frac{wl^2}{8}=\frac{wl^2}{8} H_A= \frac{wl^2}{8h}OR
∑ MC = 0 (↻+ve) (considering only BC part)
-V_B\times \frac{l}{2}-H_B\times h+w\times \frac{l}{2}\times \frac{l}{4}=0 H_B= \frac{wl^2}{8h}Hence Proved.
Bending moment at any section :
The equation to the arch with the end A as a origin is
y= \frac{4h}{l^2} x(l-x)
Bending moment at any section X having coordinates (x. y) with respect to A as origin is given by
M_x=\frac{wl}{2}\times x -\frac{wx^2}{2}-H_Ay=\frac{wl}{2}\times x -\frac{wx^2}{2}-\frac{wl^2}{8h}\times \frac{4h}{l^2}x(l-x)=0Q 1 d) Explain the application of virtual work method for finding deflection in Trusses. [05 M]
Ans :
The real system will consist of the truss with all the real applied external loads. For that system, we will define an unknown deflection of a specific joint in a specific direction (this is the deflection that we want to find). This is our real external deformation. From the real system, we must also calculate all internal deformations of the truss elements. We can find these by first finding the forces in the truss elements (using method of joints) and then applying mechanics to calculate all the axial deformations of each.
The virtual system will of course be the same as the real system but without the real loads. It will only have a single external unit load (the virtual external force) which will be placed on the joint that we want to find the deflection for. This force must also be in the same direction as the deflection that we want to find. This virtual external force will create known virtual internal forces which may be found using an analysis of the truss (with the virtual external force only).
Together, these will result in the following virtual work balance for trusses :
w v,e = w v,i – – – – – (1)
1(\Delta_r)=\sum_{j=1}^{n}\left(p_{vj}\delta_{rj} \right) – – – – – (2)
where 1 is the virtual external unit load, Δr is the external real deflection, pvj is the virtual internal axial force in each truss member in the virtual system, and δrj is the real internal deformation of each truss element in the real system.
For truss member deformations caused by axial force, the real internal deformation of an individual member is :
\delta_{rj}=\frac{p_{rj}L}{AE} – – – – – (3)
where prj is the real internal force in that member, L is the length of the bar, E is the Young’s Modulus of the bar material, and A is the cross-sectional area of the bar.
Q 1 e) Explain the function of each components of a suspension bridge consisting of suspension cable and three hinged stiffening girder. [05 M]
Ans :
A suspension bridge is referred to a type bridge supported by cables. This type of bridge has been with mankind since ancient times. Today’s large and magnificent suspension bridges were made possible through the establishment of structural analysis methods, material developments, construction methods, and computer technology developments. Suspension bridges are one of the most beautiful special bridges, and are considered one of the types of bridges many structural engineers dream to design.
A suspension bridge is composed of the following members :
– Girder
– Main cable
– Pylon
– Suspender
– Anchorage
– Saddle
Stiffening Girder :
Stiffening girders of suspension bridges are longitudinal structures that support or distribute vehicle loads. Furthermore, since the stiffening girders are supported by cables, aerodynamic stability is required. In the past, I-girders were mostly used for stiffening girders, but later they were developed into truss structures due to aerodynamic stability problems, and most recently been developed into box-shape cross-sections.