{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2.BMDiagram

# 2.BMDiagram - Simply Supported Beam under UDL Theory and...

This preview shows pages 1–3. Sign up to view the full content.

Bending Moment Diagrams & Shearing Bending Moment Diagrams & Shearing Force Diagrams for Beams Force Diagrams for Beams Theory and Design of Structures I Theory and Design of Structures I Simply Supported Beam under UDL Consider the simply supported beam AB carrying a uniformly distributed load (UDL) w. The reactions at A and B are found by considering the entire beam as a free body. A B w kN/m wl /2 kN wl /2 kN x l m Simply supported beam under UDL • To evaluate the shearing force and bending moment at a chosen section x from A, we consider the equilibrium of the portion of the beam to the left of this section, namely wx wl V x = 2 2 2 2 wx x wl M x = A B w kN/m wl /2 kN wl /2 kN x l m Simply supported beam under UDL A w wl /2 M x V x x • If we consider the equilibrium of that portion of the beam to the right of the section, we get B wl /2 l - x w M x V x Free body diagram () 2 wl x l w V x = 2 2 2 x l w x l wl M x = A B w kN/m wl /2 kN wl /2 kN x l m Simply supported beam under UDL Simply Supported Beam under UDL Simply Supported Beam under UDL • Since the entire beam is in equilibrium, we must obtain the same values for shear force and bending moment at a chosen section by using the equilibrium conditions for either portion of the beam •B y v a r y i n g x from 0 to l , the values of M x and V x can be obtained. The bending moment diagram and the shear force diagram can thus be constructed accordingly wl /2 kN wl /2 kN l m Simply supported beam under UDL A B w kN/m + Bending moment diagram wl 2 /8 + Shear force diagram - wl /2 - wl /2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
• Consider structure ABCD with a hinge at C as shown A 20 kN/m 3 m B D C 40 kN 40 kN/m 10 m 2 m Frame ABCD with hinge at C Structural Analysis of a Simple Frame Structural Analysis of a Simple Frame Reactions Reactions • Consider free body ABC. Taking moment about C gives • Taking moment about B, 0 6 240 12 40 10 = × × × B V kN 192 = B V 0 2 40 4 240 10 = × × + × C V kN 88 = C V 40 V C H C V B H B B C 240 A A 20 kN/m B D C 40 k N 40 k N / m 10 m 2 m Frame ABCD with hinge at C • Then consider free body CD. Taking moment about C gives • Consider the equilibrium of forces in the horizontal and vertical directions ( x -& y -directions).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

2.BMDiagram - Simply Supported Beam under UDL Theory and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online