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AEE Civil Engineering Core · Chapter 2

SFD/BMD, Deflection and Trusses

What to remember

  • Shear force and bending moment are related by dM/dx = V and dV/dx = −w; bending moment is maximum where shear force is zero.
  • Standard deflections: cantilever end load WL³/3EI; simply supported central load WL³/48EI; simply supported UDL 5wL⁴/384EI.
  • A plane pin-jointed truss is perfect (statically determinate) when m = 2j − 3. Analyse by method of joints or method of sections.

Types of beams, loads and sign convention

Supports: simple (roller) gives one vertical reaction; hinge gives two reaction components; fixed support gives two forces and a moment. Beam types: simply supported, cantilever, overhanging, fixed and continuous. Loads: point load, uniformly distributed load (UDL), uniformly varying load (UVL) and applied moment.

Sign convention: shear force is positive when it tends to rotate the segment clockwise (left up, right down). Bending moment is positive when it sags (tension at the bottom). A hogging moment is negative.

Static equilibrium: ΣV = 0, ΣH = 0, ΣM = 0. A plane structure is statically determinate if reactions r = 3 (for a beam), and the degree of static indeterminacy is r − 3 for a beam, and 3m + r − 3j for a rigid frame.

Relations and shapes of diagrams

  • dV/dx = −w (slope of the shear diagram equals the negative load intensity); dM/dx = V.
  • Under a point load SFD jumps by the load; under a UDL SFD is a straight sloping line and BMD is a parabola; with no load SFD is horizontal and BMD is a sloping line.
  • A couple causes a sudden jump in BMD with no change in SFD.
  • Point of contraflexure is where BM changes sign (BM = 0 with sign change).
Beam and loadMax shear forceMax bending moment
Cantilever, end load WWWL (at fixed end)
Cantilever, UDL wwLwL²/2
Simply supported, central WW/2WL/4
Simply supported, UDL wwL/2wL²/8 (centre)
Fixed beam, UDL wwL/2wL²/12 (support); wL²/24 (centre)
Fixed beam, central WW/2WL/8 (both support and centre)
  • Example: simply supported 6 m beam with UDL 10 kN/m: reactions 30 kN each, M max = 10 × 36/8 = 45 kN·m.
  • An overhanging beam has points of contraflexure; a simply supported beam under only downward loads has none.
  • A propped cantilever (fixed at one end, simple support at other) with UDL: reaction at prop = 3wL/8; fixed-end moment = wL²/8.

Fixed and continuous beams

Fixed-end beams are statically indeterminate to the second degree for vertical loads. Fixing end moments reduce the deflection (central deflection under a central load is one-fourth that of the simply supported beam). A sinking support changes the end moments.

Continuous beam methods:

  • Clapeyron's three-moment theorem: relates the hogging moments at three consecutive supports.
  • Moment distribution (Hardy Cross): stiffness of a member = 4EI/L when the far end is fixed and 3EI/L when it is hinged. Carry-over factor is ½ (zero for a hinged far end). Distribution factor = member stiffness/sum of stiffness at the joint.
  • Slope-deflection method uses joint rotations as unknowns.
  • Example: fixed-end moment for a UDL of 12 kN/m on a 5 m span = 12 × 25/12 = 25 kN·m.

Deflection of beams

Differential equation: EI d²y/dx² = M. Methods: double integration, Macaulay's method (for several loads), moment-area method (Mohr's theorems), conjugate beam method, strain energy and unit load method.

Beam and loadMaximum slopeMaximum deflection
Cantilever, end load WWL²/2EIWL³/3EI
Cantilever, UDL wwL³/6EIwL⁴/8EI
Simply supported, central WWL²/16EIWL³/48EI
Simply supported, UDL wwL³/24EI5wL⁴/384EI
Fixed beam, central W0 at endsWL³/192EI
Fixed beam, UDL w0 at endswL⁴/384EI
  • Example 1: cantilever L = 3 m, W = 10 kN, EI = 9000 kN·m²: δ = 10 × 27/(3 × 9000) = 10 mm.
  • Example 2: simply supported L = 4 m, UDL 12 kN/m, EI = 8000 kN·m²: δ = 5 × 12 × 256/(384 × 8000) = 5 mm.
  • Deflection is proportional to L³ for point loads and L⁴ for UDL, and inversely proportional to EI. Doubling depth of a rectangular beam increases I by 8 times.
  • Maximum deflection occurs where slope is zero. Mohr's first theorem: change of slope between two points = area of M/EI diagram. Second theorem: deviation of a tangent = moment of that area about the point.
  • Castigliano's theorem: deflection = partial derivative of strain energy with respect to the load. Maxwell's reciprocal theorem: deflection at A due to a unit load at B equals deflection at B due to unit load at A.

Influence lines and arches

Influence line diagram (ILD) shows the effect at a section for a unit load moving over the structure. Moving load rules: for a UDL longer than the span, maximum SF or BM occurs when the load covers the whole relevant portion. Absolute maximum bending moment for a simply supported beam under a moving single load is at midspan. Under a train of loads, it occurs under a load such that the centre of the span bisects the distance between that load and the resultant.

Three-hinged arch: statically determinate, since the hinge gives the extra equation (moment at the crown hinge is zero). Horizontal thrust for a UDL on a parabolic arch H = wL²/8h. Example: w = 10 kN/m, L = 20 m, rise h = 5 m: H = 10 × 400/40 = 100 kN. Bending moment at a section = beam moment − H × y, and it is zero everywhere for a parabolic arch under full UDL.

Trusses

A plane truss is made of straight members with pin joints, loaded at joints only, so each member carries only axial force.

ConditionResult
m = 2j − 3Perfect (determinate) truss
m < 2j − 3Deficient (unstable)
m > 2j − 3Redundant (indeterminate)

With reactions r > 3, the condition becomes m + r = 2j. Example: j = 6 gives m = 9 for a perfect truss. A simple truss is built by adding two members and one joint at a time to a triangle.

Methods of analysis:

  • Method of joints: apply ΣV = 0 and ΣH = 0 at a joint with at most two unknowns. Start from a support. Tension is taken as positive (member pulls on the joint).
  • Method of sections (Ritter): cut through three members and take moments about the point where two cut members meet. It is best for finding forces in a few members.
  • Graphical method: Maxwell diagram and Bow's notation.

Zero-force members: at an unloaded joint with two non-collinear members, both are zero; at an unloaded joint with three members where two are collinear, the third is zero. A joint with two members and a load acting along one of them leaves the other member at zero.

For a truss with a diagonal under a horizontal load, tension and compression diagonals are used in counter bracing. Deflection of a truss joint is found by the unit load method: δ = Σ(P u L/AE).

Worked examples

  • 1. A simply supported beam of span 8 m carries a point load of 40 kN at 2 m from the left support. Left reaction = 40 × 6/8 = 30 kN, right reaction = 10 kN. Maximum bending moment is under the load = 30 × 2 = 60 kN·m. Shear force changes sign at the load.
  • 2. A cantilever of 2 m carries 10 kN at the free end and a UDL of 5 kN/m over the whole length. Fixed-end moment = 10 × 2 + 5 × 4/2 = 30 kN·m.
  • 3. Stiffness and distribution: at a joint a member with its far end fixed (stiffness 4EI/L = 8 units) meets another member with its far end hinged (stiffness 3EI/L = 4 units). Distribution factors = 8/12 and 4/12.

Exam traps

  • Maximum bending moment is at the section of zero shear, not always at midspan.
  • Cantilever UDL moment is wL²/2, not wL²/8.
  • Fixed beam UDL has wL²/12 at the supports and wL²/24 at the centre; the hogging and sagging moments differ.
  • The factor 5/384 belongs to the simply supported UDL case, 1/48 to the central load.
  • Shear force is positive when a left-hand section is pushed up, not down.
  • The condition m = 2j − 3 is for pin-jointed plane trusses; for space trusses it is m = 3j − 6.
  • Stiffness is 4EI/L (far end fixed) and 3EI/L (far end hinged).
  • Distribution factors at a joint add up to 1.

One-liners

  • 1. dM/dx = V and dV/dx = −w.
  • 2. SS beam with UDL: M max = wL²/8.
  • 3. SS beam with central point load: M max = WL/4.
  • 4. Cantilever end load: deflection WL³/3EI.
  • 5. SS beam with UDL: deflection 5wL⁴/384EI.
  • 6. Fixed beam central deflection under UDL is wL⁴/384EI.
  • 7. Perfect plane truss: m = 2j − 3.
  • 8. Space truss: m = 3j − 6.
  • 9. Carry-over factor in moment distribution is 1/2.
  • 10. Stiffness of a member with a fixed far end: 4EI/L.
  • 11. Three-hinged arch thrust under UDL: wL²/8h.
  • 12. Maxwell's theorem: reciprocal deflections are equal.

Practice questions

  1. A simply supported beam of span 6 m carries a UDL of 10 kN/m over the full span. The maximum bending moment is

    1. 30 kN-m
    2. 90 kN-m
    3. 60 kN-m
    4. 45 kN-m
    Answer

    D. 45 kN-m

    M = wL^2/8 = 10 x 36/8 = 45 kN-m.

  2. A simply supported beam of span 8 m carries a point load of 40 kN at 2 m from the left support. The left reaction is

    1. 10 kN
    2. 30 kN
    3. 40 kN
    4. 20 kN
    Answer

    B. 30 kN

    R_left = 40 x 6/8 = 30 kN.

  3. A cantilever of 2 m carries 10 kN at its free end and a UDL of 5 kN/m over the whole length. The fixed-end moment is

    1. 40 kN-m
    2. 20 kN-m
    3. 30 kN-m
    4. 25 kN-m
    Answer

    C. 30 kN-m

    M = 10 x 2 + 5 x 2^2/2 = 20 + 10 = 30 kN-m.

  4. The bending moment is maximum at the section where

    1. the shear force is zero (changes sign)
    2. the deflection is zero
    3. the slope of the BMD is maximum
    4. the load is maximum
    Answer

    A. the shear force is zero (changes sign)

    dM/dx = V, so M is stationary when V = 0.

  5. The relation between load intensity w and shear force V is

    1. dM/dx = w
    2. dV/dx = w x
    3. dV/dx = M
    4. dV/dx = -w
    Answer

    D. dV/dx = -w

    The slope of the SFD equals the negative load intensity.

  6. Under a UDL, the shear force diagram and bending moment diagram are respectively

    1. parabolic and linear
    2. linear and parabolic
    3. linear and cubic
    4. constant and linear
    Answer

    B. linear and parabolic

    Integrating a constant load gives a linear V and a parabolic M.

  7. A simply supported beam of 6 m carries a central point load of 20 kN. The maximum bending moment is

    1. 15 kN-m
    2. 60 kN-m
    3. 30 kN-m
    4. 120 kN-m
    Answer

    C. 30 kN-m

    M = WL/4 = 20 x 6/4 = 30 kN-m.

  8. A point of contraflexure is a point where

    1. the bending moment changes sign
    2. shear force is maximum
    3. slope is maximum
    4. deflection is maximum
    Answer

    A. the bending moment changes sign

    At this point BM is zero and the curvature changes direction.

  9. The fixed-end moment of a 5 m span fixed beam with UDL 12 kN/m is

    1. 12.5 kN-m
    2. 25 kN-m
    3. 30 kN-m
    4. 50 kN-m
    Answer

    B. 25 kN-m

    M = wL^2/12 = 12 x 25/12 = 25 kN-m.

  10. A fixed beam of span 4 m has a central load 40 kN. The fixed-end moment is

    1. 30 kN-m
    2. 10 kN-m
    3. 40 kN-m
    4. 20 kN-m
    Answer

    D. 20 kN-m

    M = WL/8 = 40 x 4/8 = 20 kN-m.

  11. The central deflection of a fixed beam with a central point load compared with that of a simply supported beam is

    1. one eighth
    2. one half
    3. one fourth
    4. same
    Answer

    C. one fourth

    WL^3/192EI is 1/4 of WL^3/48EI.

  12. A cantilever of 3 m with a 10 kN end load and EI = 9000 kN-m2 has a tip deflection of

    1. 30 mm
    2. 3.33 mm
    3. 10 mm
    4. 90 mm
    Answer

    C. 10 mm

    delta = WL^3/3EI = 10 x 27/(3 x 9000) = 0.01 m.

  13. A simply supported beam of 4 m with a UDL of 12 kN/m and EI = 8000 kN-m2 has a central deflection of

    1. 5 mm
    2. 2.5 mm
    3. 15 mm
    4. 10 mm
    Answer

    A. 5 mm

    delta = 5wL^4/384EI = 5 x 12 x 256/(384 x 8000) = 0.005 m.

  14. The maximum slope of a simply supported beam with a central point load W is

    1. WL^2/2EI
    2. WL^2/48EI
    3. WL^2/8EI
    4. WL^2/16EI
    Answer

    D. WL^2/16EI

    Slope at the support is WL^2/16EI.

  15. The factor 5/384 appears in the maximum deflection of

    1. a simply supported beam with a central load
    2. a simply supported beam with a UDL
    3. a fixed beam with a UDL
    4. a cantilever with a UDL
    Answer

    B. a simply supported beam with a UDL

    delta = 5wL^4/384EI.

  16. If the span of a simply supported beam under a central point load is doubled, the maximum deflection becomes

    1. 4 times
    2. 8 times
    3. 16 times
    4. 2 times
    Answer

    B. 8 times

    delta is proportional to L^3.

  17. The depth of a rectangular beam is doubled keeping width and span the same. The deflection under a given load becomes

    1. one fourth
    2. half
    3. one eighth
    4. one sixteenth
    Answer

    C. one eighth

    I is proportional to d^3, so delta falls to 1/8.

  18. Mohr's first moment-area theorem states that the change in slope between two points is equal to

    1. the moment of the M/EI diagram
    2. the shear force between them
    3. the deflection at the points
    4. the area of the M/EI diagram between them
    Answer

    D. the area of the M/EI diagram between them

    Change of slope = area under M/EI.

  19. Castigliano's theorem gives the deflection as

    1. the partial derivative of strain energy with respect to the load
    2. the area of the BMD
    3. the integral of the SFD
    4. the product of load and span
    Answer

    A. the partial derivative of strain energy with respect to the load

    delta = dU/dP.

  20. Maxwell's reciprocal theorem states that

    1. reaction at A equals reaction at B
    2. moment at A equals moment at B
    3. deflection at A due to unit load at B equals deflection at B due to unit load at A
    4. slope equals deflection
    Answer

    C. deflection at A due to unit load at B equals deflection at B due to unit load at A

    This follows from the symmetry of the flexibility matrix.

  21. The number of members in a perfect plane truss with 6 joints is

    1. 8
    2. 9
    3. 12
    4. 10
    Answer

    B. 9

    m = 2j - 3 = 9.

  22. A plane truss with 8 joints and 14 members (3 reactions) is

    1. redundant by 1
    2. perfect
    3. deficient by 1
    4. redundant by 3
    Answer

    A. redundant by 1

    2j - 3 = 13 members are needed; m = 14 gives one redundant.

  23. In a space truss the condition for a perfect truss is

    1. m = 3j - 3
    2. m = 2j - 6
    3. m = 2j - 3
    4. m = 3j - 6
    Answer

    D. m = 3j - 6

    Each joint has three equations and six come from rigid-body motion.

  24. At an unloaded joint where two non-collinear members meet, the force in each member is

    1. zero
    2. equal in compression
    3. infinite
    4. equal in tension
    Answer

    A. zero

    Equilibrium in both directions forces both to be zero.

  25. In the method of sections for trusses, the section generally cuts through

    1. only one member
    2. at most three members with unknown forces
    3. all members of the truss
    4. exactly five members
    Answer

    B. at most three members with unknown forces

    Three equilibrium equations can solve for three unknowns.

  26. A three-hinged parabolic arch of span 20 m and rise 5 m carries a UDL 10 kN/m over the full span. The horizontal thrust is

    1. 400 kN
    2. 50 kN
    3. 100 kN
    4. 200 kN
    Answer

    C. 100 kN

    H = wL^2/8h = 10 x 400/40 = 100 kN.

  27. In moment distribution, the stiffness of a member with far end fixed is

    1. 2EI/L
    2. EI/L
    3. 3EI/L
    4. 4EI/L
    Answer

    D. 4EI/L

    For a hinged far end it is 3EI/L.

  28. The carry-over factor for a prismatic member with a fixed far end is

    1. 1
    2. 1/4
    3. 1/2
    4. zero
    Answer

    C. 1/2

    Half the applied moment is carried over to the fixed far end.

  29. At a joint, one member has stiffness 8 units and another 4 units. The distribution factor of the stiffer member is

    1. 1/3
    2. 2/3
    3. 1/2
    4. 3/4
    Answer

    B. 2/3

    8/(8 + 4) = 2/3.

  30. A propped cantilever of span 4 m carries a UDL of 16 kN/m. The reaction at the prop is

    1. 16 kN
    2. 32 kN
    3. 40 kN
    4. 24 kN
    Answer

    D. 24 kN

    R = 3wL/8 = 3 x 16 x 4/8 = 24 kN.

  31. The degree of static indeterminacy of a propped cantilever (vertical loads) is

    1. 0
    2. 2
    3. 1
    4. 3
    Answer

    C. 1

    Four reactions against three equations (planar).

  32. The absolute maximum bending moment in a simply supported beam due to a single moving point load occurs

    1. at midspan with the load at the centre
    2. at a support
    3. at the third point
    4. at the quarter span
    Answer

    A. at midspan with the load at the centre

    M = WL/4 with the load at midspan.

  33. Consider the statements. 1. A simply supported beam under downward loads only has no point of contraflexure. 2. An overhanging beam may have a point of contraflexure. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Overhang loads can reverse the moment sign.

  34. Consider the statements on trusses. 1. All members carry axial force only. 2. Loads are assumed to act at joints. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    These are the standard assumptions for pin-jointed trusses.

  35. Consider the statements. 1. A fixed beam is statically determinate. 2. A three-hinged arch is statically determinate. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    B. 2 only

    A fixed beam is indeterminate; the central hinge gives an extra equation for the arch.

  36. Consider the statements on moment distribution. 1. Distribution factors at a joint add up to 1. 2. The carry-over factor is 1/2 for a member with a hinged far end. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    With a hinged far end nothing is carried over to it.

  37. Consider the statements on influence lines. 1. They show the effect at a fixed section as a unit load moves. 2. They are drawn for moving loads. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    An ILD is the plot of a response against load position.

  38. Consider the statements on deflection. 1. Deflection is proportional to EI. 2. Doubling the depth of a rectangular beam reduces deflection to one eighth. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    B. 2 only

    Deflection is inversely proportional to EI.

  39. Consider the statements on the SFD and BMD. 1. Under a point load the SFD has a sudden jump. 2. A couple causes a jump in the BMD. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Both are standard features of the diagrams.

  40. Consider the statements. 1. The slope of the BMD equals the shear force. 2. The BMD is maximum where the SFD is at its steepest. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    The moment is maximum where the shear is zero.

  41. Match the beam with its maximum deflection. P) Cantilever with end load Q) SS beam with UDL R) SS beam with central load S) Cantilever with UDL

    1. P-WL^3/48EI, Q-wL^4/8EI, R-WL^3/3EI, S-5wL^4/384EI
    2. P-wL^4/8EI, Q-WL^3/48EI, R-5wL^4/384EI, S-WL^3/3EI
    3. P-5wL^4/384EI, Q-WL^3/3EI, R-wL^4/8EI, S-WL^3/48EI
    4. P-WL^3/3EI, Q-5wL^4/384EI, R-WL^3/48EI, S-wL^4/8EI
    Answer

    D. P-WL^3/3EI, Q-5wL^4/384EI, R-WL^3/48EI, S-wL^4/8EI

    Standard deflection results.

  42. A three-span continuous beam on four simple supports (vertical loads only) is statically indeterminate to the degree of

    1. 1
    2. 3
    3. 4
    4. 2
    Answer

    D. 2

    Reactions = 4, equations = 2 for a beam with no axial force, so the degree is 4 - 2 = 2.

  43. The maximum shear force in a cantilever of length L carrying a UDL w is

    1. wL
    2. wL^2/2
    3. wL/2
    4. wL/4
    Answer

    A. wL

    The shear at the fixed end equals the total load wL.

  44. Consider the statements. 1. In a simply supported beam with a central load the shear force is zero at the centre only. 2. The BMD of a cantilever with an end load is parabolic. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    D. Neither 1 nor 2

    The shear force is constant W/2 on each half and changes sign at the load; the cantilever BMD is linear.

  45. Match the beam with its maximum bending moment. P) Cantilever with UDL Q) SS beam with UDL R) SS beam with central load S) Fixed beam with central load

    1. P-WL/8, Q-WL/4, R-wL^2/8, S-wL^2/2
    2. P-wL^2/2, Q-wL^2/8, R-WL/4, S-WL/8
    3. P-wL^2/8, Q-wL^2/2, R-WL/8, S-WL/4
    4. P-WL/4, Q-WL/8, R-wL^2/2, S-wL^2/8
    Answer

    B. P-wL^2/2, Q-wL^2/8, R-WL/4, S-WL/8

    Standard results for these beams.

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