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Pset #5 Solutions
3.11 Fall 2003
Solution Problem
To understand the shear forces and bending moments in a beam, we will look at a simple problem. We are looking at a simply supported 20 ft. beam with a load of 10,000 lb. acting downward right at the center of the beam. Due to symmetry the two support forces will be equal, with a value of 5000 lb. each. This is the static equilibrium condition for the whole beam.
Next let's examine a section of the beam. We will cut the beam a arbitrary distance (x) between 0 and 10 feet, and apply static equilibrium conditions to the left end section as shown in Diagram 2 below. We can do this since as the entire beam is in static equilibrium, then a section of the beam must also be in equilibrium.
In Diagram 2a, we have shown left section of the beam, x feet, long - where x
is an arbitrary distance greater than 0 ft. and less than 10 ft. Notice if we just
include the 5000 lb. external support force, the section of the beam is clearly
not in equilibrium. Neither the sum of forces (translational equilibrium), nor
the sum of torque (rotational equilibrium) will sum to zero - as required for
equilibrium. Therefore, since we know the beam section is in equilibrium ,
there must be some forces and/or torque not accounted for. In diagram 2b, we have shown the missing force and torque. The 10,000 lb.
load which we originally applied to the beam, and the support force cause
internal "shearing forces" and internal torque called "bending moments" to
develop. (We have symbolically shown these in Diagram 2c.) When we cut the beam, the internal shear force and bending moment at that point then become
an external force and moment (torque) acting on the section. We have shown
these in Diagram 2b, and labeled them V (shear force) and M (bending
moment).
Please note that M is a moment or torque - not a force. It does not appear in
the sum of forces equation when we apply static equilibrium to the section -
which will be our next step.
Equilibrium Conditions:
Sum of Forces in y-direction : + 5000 lb. - V = 0 , solving V = 5000 lb.
Sum of Toque about left end : -V * x + M = 0 , we next substitute the value of
V from the force equation into the torque equation: - 5000 lb. * x + M = 0 ,
then solving for M = 5000x (ft-lb.)
These are the equations for the shear force and bending moments for the
section of the beam from 0 to 10 feet. Notice that the internal shear force is a
constant value of 5000 lb. for the section, but that the value of the internal
torque (bending moment) varies from 0 ft-lb. at x = 0, to a value of 50,000 ft-
lb. at x = 10 ft.
[We really should not put exactly 0 ft., and 10 ft. into our equation for the bending moment. The reason is that at 0 and 10 ft., there are 'point loads/forces' acting. That is, we have our forces acting at point - and a point has zero area, so the stress (F/A) at these points would in theory be infinite. Of course, a stress can not be infinite, and we can not apply a force at a point - it is actually applied over some area (even if the area if small). However, in 'book' problems we normally apply forces at a point. To deal with this difficulty, we actually skip around these points. We cut our section at 0' < x < 6'. Still when we put values into our expressions we put in values such as x = 9.99999999 ft, and round it off (numerically) to 10 ft. This is, in effect, cheating a bit. We are putting in the value x = 10 ft., but only because the number we actually put in was rounded off to 10 ft. It all may sound confusing, but it works, and will become clear as we do several examples.]
First, however we will finish analyzing our simple beam. So far we have found
expressions for the shear force and bending moments (V1 = 5000 lb, M1 =
5000x ft-lb) for section 1 of the beam, between 0 and 10 ft. Now we will look
at the next section of the beam. We cut the beam at distance x (ft) from the
left end, where x is now greater than 10 ft. and less then 20 ft. and then look
at entire section to the left of where we cut the beam (See Diagram 3). Where
the beam was cut, we have an internal shear force and bending moment -
which now become external. These are shown in Diagram 3 as V2 and M2. (We
add the '2', to indicate we are looking at section two of the beam.)
MORE DIFFICULT PROBLEMS #2-
Solution Problem
Solution :
Part A :
STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions.
Sum Fx = Ax = 0 Sum Fy = -5,000 lbs - (1,000 lbs/ft)(8 ft) + Ay = 0 Sum TA = - (5,000 lbs)(12 ft) - (8,000 lbs)(12 ft) + Mext = 0 Solving for the unknowns: Ay = 13,000 lbs; Mext = 156,000 ft-lbs
Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. For this process we will ‘cut’ the beam into sections, and then use Statics - Sum of Forces to determine the Shear Force expressions, and Integration to determine the Bending Moment expressions in each section of the beam.
Section 1: Cut the beam at x, where 0 < x < 8 ft. Analyze left hand section.
- FBD. (Shown in Diagram)
- All forces in x & y components (yes)
- Apply translational equilibrium conditions (forces only):
Sum Fx = 0 (no net external x- forces) Sum Fy = 13,000 lbs - V1 = 0 Solving: V1 = 13,000 lbs
- Integration
M1 = 13,000x + C
a)Boundary condition to find C1: at x=0 M=-156,000 ft-lbs (That is, for a cantilever beam, the value of the bending moment at the wall is equal to the negative of the external moment.) Apply BC: -156,000 = 13000(0) + C
Solving: C1 = -156, Therefore… M1 = [13,000x - 156,000] ft-lbs for 0 < x < 8 ft.
Section 2: Cut the beam at x, where 8 < x < 12 ft. Analyze left hand section.
- FBD. (Shown in Diagram)
- All forces in x & y components (yes)
- Apply translational equilibrium conditions (forces only):
Sum Fx = 0 (no net external x- forces) Sum Fy = 13,000 lbs - (1,000 lbs/ft)((x - 8)ft) - V2 = 0 Solving: V2 = [-1,000x + 21,000] lbs
Solving: C3 = -128,000 ft-lbs Therefore… M3 = [-500x^2 + 16,000x - 128,000] ft-lbs for 12 < x < 16
Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above, we can draw the shear force and bending moment diagrams for our loaded beam.
V1 = 13,000 lb, V2 = [-1,000x+21,000] lb, V3 = [-1,000x + 16,000] lb M1 =[13,000x-156,000] ft-lb, M2 = [-500x^2 +21,000x-188,000] ft-lb, M3 = [-500x^2 + 16,000x - 128,000] ft-lb
Solution Problem
Part A: STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any
needed angles and dimensions. STEP 2: Break any forces not already in x and y direction into their x and y components. STEP 3: Apply the equilibrium conditions.
Sum Fy = (-1,000 lbs/ft)(4 ft) - (1,500 lbs/ft)(4 ft) + By + Cy = 0 Sum TB = (Cy)(6 ft) + (1,000 lbs/ft)(4 ft)(2 ft) - (1, lbs/ft)(4 ft)(8 ft) = 0 Solving for the unknowns: Cy = 6,670 lbs; By = 3,330 lbs
Part B: Determine the Shear Forces and Bending Moments expressions for each section of the loaded beam. For this process we will ‘cut’ the beam into sections, and then use Statics - Sum of Forces to determine the Shear Force expressions, and Integration to determine the Bending Moment expressions in each section of the beam.
Section 1: Cut the beam at x, where 0 < x < 4 ft. Analyze left hand section.
- FBD. (Shown in Diagram)
- All forces in x & y components (yes)
- Apply translational equilibrium conditions (forces only):
Sum Fx = 0 (no net external x- forces) Sum Fy = -1,000 lbs/ft(x) - V1 = 0
Solving: V1 = -1,000x lbs
- Integration
M1 = -500 x^2 + C
Section 3: Cut the beam at x, where 10 < x < 14 ft. Analyze left hand section.
- FBD. (Shown in Diagram)
- All forces in x & y components (yes)
- Apply translational equilibrium conditions (forces only):
Sum Fx = 0 (no net external x- forces) Sum Fy = -1,000 lbs/ft(4 ft) + 3,330 lbs + 6,670 lbs - 1500lbs/ft(x-10)ft - V3 = 0
Solving: V3 = [-1,500x + 21,000] lbs
- Integration
M3 = -750x^2 + 21,000x + C
a)Boundary condition to find C3: at x=14 ft M=0 ft-lbs (end of beam, no external torque so M3=0) Apply BC: 0 = -750(14)^2 + 21,000(14) + C
Solving: C3 = -147,000 ft-lbs Therefore… M3 = [-750x^2 + 21,000x - 147,000] ft-lbs for 10 < x < 14
Part C: Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above, we can draw the shear force and bending moment diagrams for our loaded beam.
V1 = -1,000x lb, V2 = -667 lb, V3 = -1,500x+21,000 lb M1 = -500x^2 ft-lb, M2 = -667x-5,330 ft-lb, M3 = - 750x^2 +21,000x-147,000 ft-lb
Solution Problem
Overall Equilibrium
We start by drawing a free-body diagram
(Fig. 2) of the beam and determining the
support reactions. Summing moments about
the left end of the beam
Fig. 2
MA = 7RC - 2 [ 4 × 10 ]
(1a)
gives
RC = 45 kN (1b)
Then, summing forces in the vertical direction
F =RA +RC - 4 × 10 - 16 - 19 = 0 (2a)
gives
RA = 30 kN (2b)
Drawing the Shear Force Diagram
Distributed Load
The downward distributed load causes
the shear force graph to slope downward
(in the direction of the load). Since the
distributed load is constant, the slope of
the shear force graph is constant (dV/dx
=w = constant).
The total change in the shear force
graph between pointsA andB is 40 kN
(equal to the area under the distributed
load between pointsA andB) from +30 kN
to -10 kN.
We also need to know where the shear
force becomes zero. We know that the
full 4 m of the distributed load causes a change in the shear force of 40 kN.
So how much of the distributed load will it take to cause a change of 30 kN
(from +30 kN to 0 kN)? Since the distributed load is uniform, the area
(change in shear force) is just 10 ×b = 30, which givesb = 3 m. That is, the
shear force graph becomes zero at x = 3 m (3 m from the beginning of the
uniform distributed load).
Fig. 4
Concentrated Force (^) Fig. 5
The 16-kN concentrated force atB
causes the shear force graph to jump
down (in the direction of the force) by 16
kN (the magnitude of the force) from -
kN to -26 kN.
No Loads
Fig. 6
Since there are no loads between
pointsB andC, the shear force graph is
constant (the slope dV/dx =w = 0) at -
kN.
Fig. 7
Concentrated Force
The 45-kN concentrated force (support reaction) atC causes the shear
force graph to jump up (in the direction of the force) by 45 kN (the
magnitude of the force) from -26 kN to +19 kN.
Concentrated Force
The 19-kN concentrated force atD causes the shear force graph to jump
down (in the direction of the force) by 19 kN (the magnitude of the force)
from +19 kN to 0 kN.
Drawing the Bending Moment Diagram
Since there are no concentrated moments acting on this beam, the
bending moment diagram (graph) will be continuous (no jumps) and it will
start and end at zero.
Decreasing Shear Force
The bending moment graph starts out at
zero and with a large positive slope (since
the shear force starts out with a large
positive value and dM/dx =V ). As the shear
force decreases, so does the slope of the
bending moment graph. Atx = 3 m the shear
force becomes zero and the bending moment
is at a local maximum (dM/dx =V = 0 ) For
values of x greater than 3 m (3 <x < 4 m) the
shear force is negative and the bending
moment decreases (dM/dx =V < 0).
The shear force graph is linear (1st^ order
function ofx ), so the bending moment graph
is a parabola (2nd^ order function ofx ).
The change in the bending moment
betweenx = 0 m andx = 3 m is equal to the area under the shear graph
between those two points. The area of the triangle is
Fig. 10
M = (1/2)(30 × 3) = 45 kN·m
So the value of the bending moment atx = 3 m isM = 0 + 45 = 45 kN·m. The
change in the bending moment betweenx = 3 andx = 4 m is also equal to the
area under the shear graph
M = (1/2)(-10 × 1) = -5 kN·m
So the value of the bending moment atx = 4
m isM = 45 - 5 = 40 kN·m.
Constant Shear Force
Although the bending moment graph is
continuous atx = 4 m, the jump in the shear
force atx = 4 m causes the slope of the
bending moment to change suddenly from
dM/dx =V = -10 kN·m/m to dM/dx = -
kN·m/m.
Since the shear force graph is constant
betweenx = 4 m andx = 7 m, the bending
moment graph has a constant slope between
x = 4 m andx = 7 m (dM/dx =V = -
kN·m/m). That is, the bending moment graph
is a straight line.
The change in the bending moment
betweenx = 4 m andx = 7 m is equal to the
area under the shear graph between those two points. The area of the
rectangle is just M = (-26 × 3) = -78 kN·m. So the value of the bending
moment atx = 7 m isM = 40 - 78 = -38 kN·m.
Fig. 11
