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**Q.1 A portal frame shown in figure (not drawn to scale) has a hinge support at joint P and a roller support at joint R. A point load of 50 kN is acting at joint R in the horizontal direction. The flexural rigidity. El, of each member is 10 ^{6} kNm^{2}. Under the applied load, the horizontal displacement (in mm, round off to 1 decimal place) of joint R would be______. [2019 : 2 Marks, Set-I]**

Solution:

(a) 7.5

(b) 3.0

(c) 48.0

(d) 0.1

Ans. (A)

Assume 'R sinks by Î”.

Sway analysis:

End moment distribution,

Sway force

Alternate method:

The rotation at Q (in rad, up to two decimal places) is ________ . [2018 : 2 Marks, Set-II]

The vertical reaction at support Q is

(a) 0.0 kN

(b) 2.5 kN

(c) 7.5 kN

(d) 10.0 kN

Ans.

Bending moment about hinge point A = 0

(consider the right hand side of A)

The rotation (in degrees, up to one decimal place) at the rigid joint Q would b e _________ . [2017 : 2 Marks, Set-II)

= -700 kNm (ACW)

EI = 2.5 x 10

= 20 x 10

= 20000 kNm

As the two members QR and QS are identical, moment will be equally divided i.e., 350 kNm each.

Fixed end moment:

Slope deflection equation:

Joint equilibrium equation at joint Q,

The value of support reaction (in kN) at B should be equal to ________. [2017 : 2 Marks, Set-I]

Fixed end moment:

By super position,

The bending moment in the beam at the join â€˜Q â€™ is [2016 : 2 Marks, Set-II]

(a) zero

(b)

(c)

(d)

Ans.

As there is no horizontal force,

Hence,

H

âˆ´ BM at Q = 0

By symmetry we can consider two parts of beam separately with half the load.

Moment at joint B = 10 kNm

Carry over moment at joint A due to 10 kNm moment at propped end = 10/2= 5kN-m

(a) 170

(b) 172

(c) 176

(d) 178

Ans.

As the frame and loading is symmetrical, Î¸

Fixed end moment:

Distribution factor:

**Q.10 All members in the rigid-jointed frame shown are prismatic and have the same flexural stiffness EI. Find the magnitude of BM at Q (in kN-m) due to given loading. [2013 : 2 Marks]**

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