A beam is shown in the figure below. (figure 1)

A beam is shown in the figure below. (Figure 1)

Draw the shear diagram for the beam.

4 of 1 0 kip S kip 2 kip/it 40 kip ft 4 ft 6 ft

Draw the moment diagram for the beam.

Part C

Determine the shear throughout the beam as a functions of x, where 0?x<6ft.

Part D

Determine the moment throughout the beam as a functions of x, where 0?x<6ft.

Part E

Determine the shear throughout the beam as a functions of x, where 6ft<x?10ft.

Part F

Determine the moment throughout the beam as a functions of x, where 6ft<x?10ft.

Answer

General guidance

Concepts and reason
Equilibrium condition:

A member or object is said to be in mechanical equilibrium when the summation of all the forces and moments acting on it is zero.

Force Moment:

The magnitude of the moment can be determined by the product of force and the perpendicular distance to the force from reference point.

Distributed load

A load on a member which is evenly distributed over the entire length of the member is distributed load. It is expressed in the units of weight per length.

Shear force:

A reaction induced in a beam by an external force which causes the plane layers of beam to slide in opposite direction to each other.

Bending moment:

A reaction induced in a beam by external forces or moments which cause the beam to bend.

Bending moment is defined as the reaction moment in a structural element due to application of external moment or force, which causes the element to bend.

Equilibrium force and moment conditions can be applied to the free body diagram of the loaded beam to determine the support reactions. Then consider the free body diagram of the sectioned beam, and apply equilibrium force equations at different locations to determine the shear forces. Similarly, apply the equilibrium moment condition at the section to determine the bending moment at different locations.

Fundamentals

Distributed loads acts on a beam or member as shown in Figure (1):

KL2
L/2
2L/3
RL W.
W.
vy
AVV
VVV
L
(a) Rectangular Load
(b) Triangular Load
Figure 1

The resultant force of a distributed load is the area of the loading diagram. The equivalent force for the rectangular loading in Figure 1(a) is,

R=WL

Here, the distributed load is and the length of the beam is L.

Write the equivalent force for triangular loading in Figure 1(b).

Write the formula for the moment due to force.

M = Fd

Here, the force is and the perpendicular distance is .

Shear force and bending moment:

Consider the beam as shown in Figure (2).

MB
NB
(9)
(9)
Figure 2

In the Figure 2(b), the force acting perpendicular to the cross section is normal force and the force acting parallel to the cross section is shear force . Also the couple moment at B is the bending moment at B.

The forces and moment at any location can be determined by considering equilibrium condition for either side of the sectioned beam.

Write the equilibrium force equation in y direction.

ΣΕ, = 0

Here, the total force acting in y direction is .

Write the equilibrium moment condition about any point.

ΣΜ = 0

Here, the summation of moment due to torsion is .

Force sign conventions:

Forces acting in right direction are considered positive and forces acting in left direction are considered negative. Similarly, upward vertical forces are considered positive and downward vertical forces are considered negative.

General sign conventions of moment:

Moment is considered as positive in counter clockwise direction and is considered negative in clockwise direction.

Shear Force:

At any section of the beam, the shear force acting internally is the summation of the forces acting perpendicular to the beam on either side of the beam section considered.

Sign conventions for the shear force calculations is shown in figure (3) below

RHS of the beam section
IF
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Figure 3
+F
LHS of the beam section

Bending moment:

At any section of the beam, the bending moment acting internally is the summation of the moments about the section of all the forces acting on either side of the beam section considered.

Sign conventions for the bending moment calculations is shown in figure (4) below

+M
RHS of the beam section
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Figure 4
LHS of the beam section
+M
-M

Step-by-step

Step 1 of 7

Draw the free body diagram for the loading on beam as in Figure (5).

10 kip
2 kip/ft
8 kip
V
V
V
V
40 kip-ft
—
В
6
ft
* 4 ft
Figure 5

Here, vertical reaction force at fixed support A is and the moment acting at A is .

Write the equilibrium force equation in vertical direction from the Figure (5).

F, = 0
RA-10 kip-(2 kip/ftx6 ft) -8 kip = 0)
R, = 30 kip

Write the equilibrium moment equation about A from the Figure (5).

ΣΜ, = 0
Μ,-10 kip(6 ft)-(2 kip/fx6 ft)(3 H)- 8 kip(10 ft) - 40 kip-ft = 0
Μ,=-216 kip - ft

The free body diagram for the loading on the beam is drawn as in Figure (5). Then equilibrium force and moment conditions are applied to determine the support reactions.

Step 2 of 7

(C)

Draw the free body diagram of the beam sectioned at x distance as in Figure (6) for the length .

2 kip/ft
MLA
nex
V
V
V
V
V
Figure 6

Here, shear force at x distance is and the bending moment at x distance is .

Write the expression for the shear force as a function of x for the length .

V =R-(2 kip/ft)x

Substitute 30 kip for .

V = 30-(2 kip/ft)
= 30- 2x

Calculate the shear forces from Figure (6).

V
= 30 kip

V-61- = 30 kip-2 kip/ftx6 ft
= 18 kip

Part C

The expression for the shear force as a function of x for the length is V = 30 - 2x.


The free body diagram of the beam sectioned at x distance as in Figure (6) for the length is considered. Then equilibrium force equation in vertical direction is used to determine the expression for the shear force as a function of .

Step 3 of 7

(D)

Write the equilibrium moment equation about A from the Figure (6).

ΣΜ = 0
M - R, (x)+ (2 kip/n)(x n) ( )Μ=0

Substitute 216 kip-ft for , and 30 kip for .

216 kip-ft + (30 kip)(x)+(2kip/4)(xA)(**)+MP=0
M
= -x +30x-216

Perform the calculations for the bending moment from Figure (6).

M-0 =-(0)² +30(0)-216
=-216 kip-ft

MX-6=-(6 ft) + 30(6 ft)-216
=-72 kip-ft

Part D

The expression for the bending moment as a function of x for the length is M. =-x +30x -216.


The free body diagram of the beam sectioned at x distance as in Figure (6) for the length is considered. Then equilibrium moment equation is used to determine the expression for the bending moment as a function of .

Step 4 of 7

(E)

Draw the free body diagram of the beam sectioned at x distance as in Figure (7) for the length .

10 kip
2 kip/ft
6
ft
Figure 7

Write the expression for the shear force as a function of x for the length from Figure (7).

V = R4 -(2 kip/ft)(6 ft)-10 kip

Substitute 30 kip for .

V = 30 kip -(2 kip/ft)(6 ft)-10 kip
V = 8 kip

Part E

The expression for the shear force as a function of x for the length is V = 8.


The free body diagram of the beam sectioned at x distance as in Figure (7) for the length is considered. Then equilibrium force equation in vertical direction is used to determine the expression for the shear force as a function of .

Step 5 of 7

(F)

Write the equilibrium moment equation about the section from Figure (7) for the length .

ΣΜ = 0
Μ,-R, (x)+(2 kip/n)(on)6.1):10 kip(x-6)t + Μ= 0

Substitute 216 kip-ft for , and 30 kip for .

216 kip. f1 – 30 kip(x)+(2 kip/f)(64)(x-61 ) +10 kip(x - 6) ft+M, = 0
216-30x+12(x-3)+10(x-6)+M, = 0
M. =(8x – 120) kip.ft

Calculate the bending moment acting on the beam as follows:

MX-6R = 8(6)-120
=-72 kip.ft

M-1on = 8(10)-120
= -40 kip-ft

Part F

The expression for the bending moment as a function of x for the length is M. =(8x –120) kip-ft.


The free body diagram of the beam sectioned at x distance as in Figure (7) for the length is considered. Then equilibrium moment equation is used to determine the expression for the bending moment as a function of .

Step 6 of 7

(A)

Tabulate the calculated values of shear force as in Table.

Location on beam (ft)
0
Shear force (kip)
30
18
6-
6+
10

Draw the shear force diagram for the beam as in Figure (8).

V(kip)
30H
→x(ft)
6 ft
Aft
Figure 8

Part A

The shear force diagram for the loading on the beam is as follows:

V(kip)
► x(ft)
6 ft
*
4 ft


The calculated values of the shear force are tabulated. Then the shear force diagram is drawn as in Figure (8). The shear force for the uniformly distributed length of beam varies linearly.

Step 7 of 7

(B)

Tabulate the calculated values of the bending moment.

Location on the beam (ft)
Bending moment (kip-ft)
-216
0
-72
6
10
-40

Draw the bending moment diagram for the loading on the beam as in Figure (9).

Figure 9
4 ft
ft
6
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
x(ft)
-
M (kip-ft) 4

Part B

The bending moment diagram for the loading on the beam is as follows:

M (kip-ft)
x(ft)
6 ft
4 ft


The calculated values of the bending moment are tabulated. Then the bending moment diagram is drawn as in Figure (9).

Answer

Part C

The expression for the shear force as a function of x for the length is V = 30 - 2x.

Part D

The expression for the bending moment as a function of x for the length is M. =-x +30x -216.

Part E

The expression for the shear force as a function of x for the length is V = 8.

Part F

The expression for the bending moment as a function of x for the length is M. =(8x –120) kip-ft.

Part A

The shear force diagram for the loading on the beam is as follows:

V(kip)
► x(ft)
6 ft
*
4 ft

Part B

The bending moment diagram for the loading on the beam is as follows:

M (kip-ft)
x(ft)
6 ft
4 ft

Hottest videos

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts