Draw the Shear Force (V) and Bending Moment (M) diagrams of statically indeterminate beamshown in figure using “Force Method". The (roller) support at “B" settles 35 mm. The moment of inertiais given by (I) for regions “AB", “BC" and “CD"; however it is equal to (21) for the region “DE".(“B" is the roller and “E" is the fixed type of support). [The flexural rigidity: El=40000 KNM²]60 kN10 kN/m►XВ(I)(1)(I)(21)1.5m

Answered: Draw the Shear Force (V) and Bending…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Draw the Shear Force (V) and Bending Moment (M) diagrams of statically indeterminate beam
shown in figure using “Force Method". The (roller) support at “B" settles 35 mm. The moment of inertia
is given by (I) for regions “AB", “BC" and “CD"; however it is equal to (21) for the region “DE".
(“B" is the roller and “E" is the fixed type of support). [The flexural rigidity: El=40000 KNM²]
60 kN
10 kN/m
►X
В
(I)
(1)
(I)
(21)
1.5m

Expert Solution

Step 1: Diagram and Solution

Solution:

As per the diagram

As support sinks by 35 mm

$δ_{1} + 35 m m = δ_{2} C a l c u l a t i o n f o r δ_{1} C o n s i d e r a x - x s e x t i o n a n d a s s u m e a p o i n t l o a d p a t B M o m e n t a t s e c t i o n x - x, M_{x} = p (x) + 60 (x - 1) + \frac{10 {(x - 2)}^{2}}{2} \frac{d M_{x}}{d p} = x δ = \int_{}^{} \frac{d M_{x}}{d p} . \frac{M_{x}}{E I} = \int_{0}^{2} \frac{(p x + 60 (x - 1) x)}{E I} + \int_{2}^{4} \frac{10 {(x - 2)}^{2}}{2 E (E I)} = \int_{0}^{2} \frac{60 x^{2} - 60 x}{E I} + \int_{2}^{4} \frac{10 (x^{2} + 4 - 4 x)}{4 E I} = \frac{1}{E I} {[60 (\frac{x^{3}}{3}) - 60 (\frac{x^{2}}{2})]}_{0}^{2} + \frac{10}{4 E I} {[\frac{x^{3}}{3} + 4 x - \frac{4 x^{2}}{2}]}_{2}^{4} = \frac{40}{E I} + \frac{10}{4 E I} [\frac{56}{3} + 8 - 24] = \frac{40}{E I} + \frac{80}{12 E I} = \frac{46.66}{40000} = 1.16 \times 10^{- 3} m = 1.16 m m$ \

Step 2: Shear force values and bending moment values

$δ_{2} = \frac{R_{B} l^{3}}{3 E I} = \frac{R_{B} {(5)}^{3}}{3 E I} = \frac{125 R_{B}}{3 E I} δ_{1} + 35 = (\frac{R_{B} (125)}{3 E I}) \frac{(1.16 + 35) \times 3 (40000)}{125} = R_{B} R_{B} = 34.71 k N$

Shear Force Values

$A t, A = 0 B = 34.71 k N l e f t o f C = 34.71 k N R i g h t o f C = - 25.29 k N D = - 25.29 k N E = - 55.29 k N B e n d i n g m o m e n t V a l u e s A = 0 B = 0 C = 34.71 k N . m D = 34.71 (2) - 60 (1) = 9.42 k N E = 34.71 (5) - 60 (4) - 10 \times 3 \times \frac{3}{2} = 111.45 k N . m$