Axial Forces, Shearing Forces and Bending Moments

Assumptions (taken from the figure):

• Origin at A (hinge on the wall). Beam A–F–G–H–C–B: A to F = 1 m, F–G = 3 m, G–H = 3 m, H–C = 1 m, C–B = 1 m (B at x = 9 m).
• Loads (downward): triangle FG rising from 0 at G to 5 kN/m at F; triangle GH with 0 at G and 5 kN/m at H, i.e. total loads $\tfrac12\times3\times5=7.5$ kN each, acting at 1 m from F and 1 m from H.
• 5 kN at B acts at 60° to the beam, downward and away from the wall: $F_x = 5\cos60^\circ = 2.5$ kN (tension), $F_y = 5\sin60^\circ = 4.33$ kN (down).
• Bracket CD is rigid and vertical (D is 1 m above C, at x = 8 m); the tie rod DE joins D to E (0, 4 m), so its direction is $(-8, 3)/\sqrt{73}$.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

Support reactions

The tie force $T$ acts at D, 1 m above the beam: $T_x = \frac{8}{\sqrt{73}}T$ (to the left) and $T_y = \frac{3}{\sqrt{73}}T$ (upward). Moments about A: $8T_y + 1\cdot T_x = 7.5(2) + 7.5(6) + 4.33(9) = 98.97$, so $3.745\,T = 98.97$.

• $H_A$ = 22.24 kN →
• $V_A$ = 10.05 kN ↑
• $T$ = 26.43 kN (tension in the rod; components 24.74 kN ← and 9.28 kN ↑ on the beam)

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• Between loads V is constant; under a UDL it varies linearly and under a triangular load parabolically. M is one degree higher than V (linear, parabolic, cubic respectively); V jumps at point loads and reactions, M jumps at couples.
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum positive (sagging) M = 28.18 kN·m at x = 5.75 m
• Maximum negative (hogging) M = -4.31 kN·m at x = 8.00 m
• V = 0 at 5.75 m from A along AB: M = 28.18 (bottom) kN·m (local maximum)

Sketches (sagging +ve above the line)

SFD:

********
        ***
           *****                              ******
                **************                |
------------------------------*****-----------------
                                   ****       |
                                       *******|

BMD:

                      *******************
            **********                   *****|
      ******                                  |
  ****                                        |
**---------------------------------------------*****
                                              *

Principle of superposition: for a linear elastic structure (small deformations, Hooke's law valid), the total effect (reaction, shear force, bending moment, stress or deflection) produced by several loads acting together is equal to the algebraic sum of the effects produced by each load acting separately.

Explanation

• Each load is applied alone on the original, undeformed structure and its effect is found.
• The individual effects are then added, taking care of signs.
• It works because the equations of equilibrium and the stress–strain relation are linear: doubling a load doubles every effect, and the effect of one load does not change the effect of another.

Example

A simply supported beam of span $L = 6$ m carries a UDL $w = 2$ kN/m over the full span and a point load $P = 6$ kN at mid-span.

        P = 6 kN
 w = 2 kN/m  |
 ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
 A▲------- 3 m -------●------- 3 m -------▲B

Case 1 – UDL alone: $M_{mid,1} = \dfrac{wL^2}{8} = \dfrac{2\times6^2}{8} = 9$ kN·m.

Case 2 – point load alone: $M_{mid,2} = \dfrac{PL}{4} = \dfrac{6\times6}{4} = 9$ kN·m.

Superposing: $M_{mid} = 9 + 9 = 18$ kN·m. Similarly the reactions are $R_A = R_B = 6 + 3 = 9$ kN, which equals the sum of the reactions of the two cases ($6$ kN from the UDL and $3$ kN from the point load).

Hence a complicated loading is split into simple standard cases whose results are tabulated, and the answers are added.

For a beam carrying a distributed load of intensity $w$ (taken positive downward), the shear force $V$ and the bending moment $M$ at a section $x$ are related by

$$\frac{dV}{dx} = -w,\qquad \frac{dM}{dx} = V,\qquad \frac{d^2M}{dx^2} = -w$$

Derivation

Consider a small element of length $dx$ with load $w$ (acting down). Shear $V$ and moment $M$ act on the left face; $V + dV$ and $M + dM$ act on the right face.

        w (per unit length)
   ↓ ↓ ↓ ↓ ↓ ↓ ↓
  M ┌───────────┐ M + dM
 V ↑│           │↓ V + dV
    └───────────┘
        dx

Vertical equilibrium: $V - w\,dx - (V + dV) = 0 \Rightarrow \dfrac{dV}{dx} = -w$.

Moments about the right face (ignoring the second-order term $w\,dx^2/2$): $M + V\,dx - M - dM = 0 \Rightarrow \dfrac{dM}{dx} = V$.

Integrated forms between sections 1 and 2

$$V_2 - V_1 = -\int_{x_1}^{x_2} w\,dx\;(\text{= minus the area of the load diagram}),\qquad M_2 - M_1 = \int_{x_1}^{x_2} V\,dx\;(\text{= area of the shear force diagram})$$

Consequences

• The slope of the SFD at any point equals minus the load intensity there; the slope of the BMD equals the shear force.
• Bending moment is maximum or minimum where $V = 0$ (and changes sign).
• No load: SFD is horizontal and BMD is linear. UDL: SFD is linear and BMD is parabolic. Triangular load: SFD is parabolic and BMD is cubic.
• A point load causes a sudden jump in $V$ equal to the load; a couple causes a jump in $M$ equal to the couple, while the SFD is unaffected.

Shear force (SF)

The shear force at a section of a beam is the algebraic sum of all the vertical forces (loads and reactions) acting on either side of the section. It is the force tending to slide one part of the beam relative to the other along the section, and it is the same, in magnitude, whether taken from the left or the right part. Unit: N or kN.

Sign convention: shear force is positive when the part to the left of the section is pushed upward relative to the right part (i.e. the resultant of the vertical forces on the left of the section acts upward); this produces a clockwise shearing effect on the element.

Bending moment (BM)

The bending moment at a section is the algebraic sum of the moments, about the centroid of that section, of all the forces (loads and reactions) acting on either side of the section. It is the moment that tends to bend the beam and is resisted by the internal stresses. Unit: N·m or kN·m.

Sign convention: bending moment is positive (sagging) when it bends the beam concave upward, causing compression in the top fibres and tension in the bottom fibres; it is negative (hogging) when the beam bends concave downward, with tension on top.

 Positive (sagging)         Negative (hogging)
   ⌣  tension at bottom       ⌢  tension at top

Example: for a simply supported beam of span $L$ with a central load $P$, the shear force is $+P/2$ on the left half and $-P/2$ on the right half, and the bending moment at mid-span is $+PL/4$.

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 5 m; horizontal triangular load from 10 kN/m at A to 2 kN/m at B (→). Beam B–C–D–E: BC = CD = 3 m, DE = 2 m; hinge at C.
• 5 kN/m downward over B to D; 10 kN downward at E (overhang). Right column DF = 5 m, F hinged at ground level.
• The overhang DE is replaced at D by 10 kN downward plus a hogging moment $10\times2 = 20$ kN·m.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B---------------C--------------D
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
A                              F

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 21.67 kN ←
• $V_A$ = 1.94 kN ↑
• $H_F$ = 8.33 kN ←
• $V_F$ = 38.06 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V parabolic, M cubic
• BC: V linear, M parabolic
• CD: V linear, M parabolic
• DF: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 61.67 kN·m at D (on CD), tension on the top
• V = 0 at 2.79 m from A along AB: M = 27.32 (right face) kN·m (local maximum)
• V = 0 at 0.39 m from B along BC: M = 17.04 (bottom) kN·m (local maximum)
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).
• Overhang DE: V = 10 kN (constant), M = 0 at E varying linearly to 20 kN·m (hogging, tension on top) at D.

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 5 m; horizontal triangular load 10 kN/m at A to 0 at B (→). Beam B–C–D–E: BC = CD = 3 m, DE = 2 m; hinge at C.
• 5 kN/m downward over B to D; 10 kN downward at E (overhang, equivalent to 10 kN plus 20 kN·m hogging at D). Right column DF = 5 m, F hinged; 10 kN horizontal (←) at 2 m above F.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B---------------C--------------D
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
|                              |
A                              F

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 15.33 kN ←
• $V_A$ = 9.72 kN ↑
• $H_F$ = 0.33 kN →
• $V_F$ = 30.28 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V parabolic, M cubic
• BC: V linear, M parabolic
• CD: V linear, M parabolic
• DF: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 38.33 kN·m at D (on CD), tension on the top
• V = 0 at 1.89 m from A along AB: M = 13.37 (right face) kN·m (local maximum)
• Point of contraflexure (M = 0) at 4.30 m from A along AB
• V = 0 at 1.94 m from B along BC: M = 2.79 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 0.89 m from B along BC
• Point of contraflexure (M = 0) at 1.90 m from D along DF
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Degree of freedom

The degree of freedom (DOF) of a system is the number of independent displacement components (translations and rotations) by which the position of the system can change. For a plane structure a free body has 3 DOF: two translations ($x$ and $y$) and one rotation about the axis perpendicular to the plane. A support removes the DOF that it restrains; the number of reactions of a support equals the number of DOF it removes.

Can a statically determinate or indeterminate structure be geometrically unstable?

Yes. Static determinacy only compares the number of reactions $r$ with the equations of equilibrium (3 for a plane structure). Geometric stability depends on how the supports are arranged. Even if the count is right, the structure is unstable if the reactions cannot resist every possible load.

Example 1: determinate but unstable. A beam on three rollers (all vertical reactions), $r = 3 =$ equations, so it looks determinate. But there is no horizontal restraint; $\Sigma F_x = 0$ cannot be satisfied for a horizontal load, so the beam can slide. Another case: a beam with three reactions that are parallel or meet at one point (concurrent) cannot resist a moment about that point.

 ○──────────────○──────────────○   all reactions vertical:
 ▲              ▲              ▲   horizontal load → beam moves

Example 2: indeterminate but unstable. A beam supported on four vertical rollers has $r = 4 > 3$, so it is indeterminate by one degree (vertical reactions only), yet it is horizontally unstable.

Hence $r \ge 3$ is a necessary condition for stability, not a sufficient one. The supports must also be arranged so that the reactions are neither all parallel nor all concurrent, and no part of the structure forms a mechanism.

Assumptions (taken from the figure):

• A(0,0) hinged; column AB = 4 m; beam B–C–D = 2 m + 2 m; internal hinge at D (x = 4 m, y = 4 m).
• Inclined member DF slopes down at 60° to the horizontal to a hinged support F at the level of A, so F = (6.31 m, 0).
• 20 kN/m acts horizontally (→) over the full 4 m of AB; 100 kN downward at C; 30 kN acts perpendicular to DF at its mid-length, pointing down and to the left.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B------------C------------Do
|                          \\
|                           \
|                            \\
|                             \\
|                               \
|                               \\\
|                                 \
|                                  \\
|                                   \
|                                    \\
|                                     \\
|                                       \
A                                       \F

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 36.08 kN ←
• $V_A$ = 53.92 kN ↑
• $H_F$ = 17.94 kN ←
• $V_F$ = 61.08 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V linear, M parabolic
• BC: V constant (steps at point loads), M linear
• CD: V constant (steps at point loads), M linear
• DF: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 92.15 kN·m at C (on BC), tension on the bottom
• V = 0 at 1.80 m from A along AB: M = 32.54 (right face) kN·m (local maximum)
• Point of contraflexure (M = 0) at 3.61 m from A along AB
• Point of contraflexure (M = 0) at 0.29 m from B along BC
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• A(0,0) and D(7,0) both hinged at the same level; AB = 8 m; B(0,8) to C(7,10) inclined (rises 2 m in 7 m, length 7.28 m); CD = 10 m.
• With two hinged supports and no internal hinge the frame is indeterminate, so a hinge is assumed at the middle of BC (a three-hinged frame).
• 15 kN/m acts downward per metre of horizontal projection of BC (total 105 kN); the 75 kN horizontal force acts to the left on CD, 3 m above D.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

               ---C
          -----   |
   -------        |
B--               |
|                 |
|                 |
|                 |
|                 |
|                 |
|                 |
|                 |
|                 |
|                 |
A                 D

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 22.71 kN →
• $V_A$ = 84.64 kN ↑
• $H_D$ = 52.29 kN →
• $V_D$ = 20.36 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V linear, M parabolic
• CD: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 181.67 kN·m at B (on AB), tension on the left face
• V = 0 at 5.42 m from B along BC: M = 21.94 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 3.64 m from B along BC
• Point of contraflexure (M = 0) at 7.20 m from B along BC
• Point of contraflexure (M = 0) at 0.09 m from C along CD
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• A(0,0) hinged; AB = BC = 3 m; beam C–D–E–F with CD = DE = 3 m, EF = 2 m; internal hinge at D; column from E (6,6) down to hinged support G, taken 5 m long, so G = (6,1).
• 15 kN at B acts at 30° to the horizontal, to the right and downward; triangular load on CD rises from 0 at C to 5 kN/m at D; 25 kN·m couple at E is clockwise; 12 kN downward at F (overhang EF).
• The overhang EF is replaced at E by the 12 kN force plus a moment of $12\times2 = 24$ kN·m (hogging); moment in EF is $-12x$.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

C------------D------------E
|                         |
|                         |
|                         |
|                         |
|                         |
B                         |
|                         |
|                         |
|                         |
|                         |
|                         G
|
A

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 12.54 kN ←
• $V_A$ = 2.09 kN ↓
• $H_G$ = 0.45 kN ←
• $V_G$ = 29.09 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V constant (steps at point loads), M linear
• CD: V parabolic, M cubic
• DE: V constant (steps at point loads), M linear
• EG: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 51.26 kN·m at E (on DE), tension on the top
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).
• Overhang EF: V = −12 kN (downward, constant), N = 0, M = 0 at F rising linearly to 24 kN·m (hogging, tension on top) at E. At joint E the 51.26 kN·m of DE is balanced by 24 (overhang) + 25 (couple) + 2.26 (column EG) kN·m.

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 6 m; beam B–C = 4 m; column C–G = 6 m with G hinged; overhang C–E = 2 m carrying 15 kN downward at E.
• Load on AB: horizontal, varying from 10 kN/m at A to 5 kN/m at B (acting →); 5 kN/m downward on BC; 20 kN·m couple at C (clockwise); 10 kN at 30° to the horizontal acts at mid-height of CG, to the right and downward.
• Two hinged supports with no internal hinge are indeterminate, so a hinge is assumed at the mid-span of BC.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B----------------C
|                |
|                |
|                |
|                |
|                |
|                |
|                |
|                |
|                |
|                |
|                |
|                |
A                G

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 39.67 kN ←
• $V_A$ = 39.00 kN ↓
• $H_G$ = 14.00 kN ←
• $V_G$ = 79.00 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V parabolic, M cubic
• BC: V linear, M parabolic
• CG: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 107.99 kN·m at C (on BC), tension on the top
• V = 0 at 5.01 m from A along AB: M = 90.69 (right face) kN·m (local maximum)
• Point of contraflexure (M = 0) at 2.00 m from B along BC
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 8 m; 50 kN·m couple at B (clockwise); beam B–C = 10 m with 30 kN/m downward, internal hinge at C (10, 8).
• Inclined member CD: horizontal projection 3 m, vertical height 5 m, so D = (13, 3) is the hinged support (length 5.83 m).
• The 15 kN force is taken perpendicular to CD at mid-length, pointing down and to the left.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B-------------------------------Co
|                                \\
|                                 \\
|                                   \
|                                   \\\
|                                     \
|                                      \\
|                                       \\
|                                         D
|
|
|
|
A

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 68.75 kN →
• $V_A$ = 200.00 kN ↑
• $H_D$ = 55.89 kN ←
• $V_D$ = 107.72 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V linear, M parabolic
• CD: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 549.98 kN·m at B (on AB), tension on the left face
• V = 0 at 6.67 m from B along BC: M = 166.67 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 3.33 m from B along BC
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Principle of superposition: for a linear elastic structure under small deformation, the response (reaction, shear force, bending moment, deflection) caused by several loads acting together equals the algebraic sum of the responses caused by the loads acting one at a time.

Stepwise procedure for the bending moment of a beam

1. Check the conditions: the material is linearly elastic and deflections are small, so superposition is valid.
2. Break the loading into simple cases: split the given loading into separate cases, each containing a single load or a standard load (point load, UDL, triangular load, couple, support settlement) on the same beam with the same supports.
3. Find the reactions of each case using $\Sigma F_y = 0$ and $\Sigma M = 0$ (and the hinge condition, if any).
4. Write or draw the bending moment of each case as a function of $x$ (or as a diagram), using one sign convention for all cases (sagging +ve, hogging −ve).
5. Add the moments at each section (or add the ordinates of the diagrams): $M(x) = M_1(x) + M_2(x) + \dots$
6. Draw the resultant BMD, find where $V = dM/dx = 0$ for the maximum moment, and check that $M = 0$ at the free ends and simple supports.

Example: for a simply supported beam with a UDL $w$ and a central point load $P$, the mid-span moment is $\dfrac{wL^2}{8} + \dfrac{PL}{4}$, the sum of the two separate BMDs (parabolic and triangular).

Assumptions (taken from the figure):

• A(0,0) hinged; AB inclined, horizontal projection 2 m and rise 2 m, so B = (2, 2); BC = CD = 2 m horizontal.
• 20 kN/m acts vertically downward on the full inclined length AB (2.83 m, total 56.57 kN); 100 kN downward at C.
• D is taken as a roller (vertical reaction only), which makes the frame determinate.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

              /B--------------C--------------D
            ///
          ///
        ///
       //
     //
   //
 //
A

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

• $H_A$ = 0.00 kN ←
• $V_A$ = 80.47 kN ↑
• $V_D$ = 76.09 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V linear, M parabolic
• BC: V constant (steps at point loads), M linear
• CD: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 152.19 kN·m at C (on BC), tension on the bottom

Point of contraflexure

A point of contraflexure (inflexion point) is the point on a beam where the bending moment changes sign (passes through zero) so that the beam changes its curvature from sagging to hogging or vice versa. The bending moment there is zero (but a zero moment does not always mean contraflexure; the moment must change sign).

Relation between loading rate, shear force and bending moment

Take an element of length $dx$ of a beam on which the load intensity is $w$ per unit length, acting downward. On the left face act $V$ and $M$; on the right face $V + dV$ and $M + dM$.

          w
   ↓ ↓ ↓ ↓ ↓ ↓
  M ┌─────────┐ M + dM
  V ↑│         │↓ V + dV
    └─────────┘
        dx

Vertical equilibrium

$$V - w\,dx - (V + dV) = 0 \;\Rightarrow\; \frac{dV}{dx} = -w$$

Moment equilibrium about the right face (the term $w\,dx^2/2$ is negligible):

$$M + V\,dx - w\,dx\cdot\frac{dx}{2} - (M + dM) = 0 \;\Rightarrow\; \frac{dM}{dx} = V$$

Combining:

$$\frac{d^2M}{dx^2} = \frac{dV}{dx} = -w$$

Meaning: the rate of change of shear force equals the negative of the load intensity; the rate of change of bending moment equals the shear force. Moment is maximum where $V = 0$. A point of contraflexure occurs where $M = 0$ and $M$ changes sign, i.e. where the curvature (related to $d^2M/dx^2 = -w$) changes direction. For a beam with no load at that point ($w = 0$), it is the point where the straight-line BMD crosses the axis.

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 6 m; 50 kN horizontal (→) at B; beam BC = 8 m with 15 kN/m downward; right column from C(8,6): CD = 4 m, DE = 2 m (E at ground level).
• 20 kN horizontal force at D acts to the left; E is taken as a roller (vertical reaction only).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B----------------------------------C
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  D
|                                  |
|                                  |
|                                  |
A                                  E

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

• $H_A$ = 30.00 kN ←
• $V_A$ = 27.50 kN ↑
• $V_E$ = 92.50 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V linear, M parabolic
• CD: V constant (steps at point loads), M linear
• DE: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 205.21 kN·m at 1.83 m from B on BC, tension on the bottom
• V = 0 at 1.83 m from B along BC: M = 205.21 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 7.06 m from B along BC

Properties of an internal hinge

An internal hinge is a pin connection inside a structure (for example in a beam or frame) that joins two parts.

1. It transmits axial force and shear force from one part to the other, but cannot transmit bending moment, so the bending moment at the hinge is zero.
2. It allows free relative rotation of the two connected parts, so the slope (rotation) is discontinuous at the hinge.
3. It provides one extra equation of condition ($\Sigma M_{hinge} = 0$ for the forces on either side), which is why a hinge reduces the degree of static indeterminacy by one (it releases one internal moment) and can make a structure with extra reactions determinate (for example a three-hinged frame).
4. It must be located and counted carefully: too many hinges turn the structure into a mechanism (unstable).
5. A hinge is a point of zero bending moment, but not necessarily a point of contraflexure, because the moment need not change sign there.

Point of contraflexure

A point of contraflexure is a point on the span of a beam where the bending moment changes sign (from sagging to hogging or the reverse), passing through zero. The curvature of the elastic curve changes there. It is found by putting $M(x) = 0$. Such points are important in design, because the reinforcement or section may be curtailed or shifted, and the moment on either side has opposite sign, so tension moves from the bottom to the top face.

Assumptions (taken from the figure):

• Beam of 11 m: A (x = 0) free end; roller at x = 2 m; internal hinge at x = 5 m; fixed support at B (x = 11 m).
• Triangular load on 0–5 m, symmetrical, peak 20 kN/m at x = 2.5 m (total 50 kN); 80 kN downward at x = 6 m; 50 kN at 30° to the beam at x = 8 m (down and to the right: 43.30 kN axial tension, 25 kN down); 15 kN/m downward on 8–11 m (45 kN).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

Support reactions

Hinge condition: $M = 0$ at x = 5 m. Taking moments of the left part about the hinge gives the roller reaction; the fixed support then follows from $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

• $R_{roller}$ = 41.67 kN ↑
• $H_B$ = 43.30 kN ←
• $V_B$ = 158.33 kN ↑
• $M_B$ = 592.50 kN·m (clockwise)

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• Between loads V is constant; under a UDL it varies linearly and under a triangular load parabolically. M is one degree higher than V (linear, parabolic, cubic respectively); V jumps at point loads and reactions, M jumps at couples.
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum positive (sagging) M = 8.02 kN·m at x = 3.56 m
• Maximum negative (hogging) M = -592.50 kN·m at x = 11.00 m
• V = 0 at 3.56 m from A along AB: M = 8.02 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 2.50 m from A along AB
• Point of contraflexure (M = 0) at 5.00 m from A along AB
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Sketches (sagging +ve above the line)

SFD:

          *
**********-*****************------------------------
                            |
                            **********
                                      *******
                                             *******

BMD:

********************************--------------------
                                *******
                                       *****
                                            *****
                                                 ***

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 3 m, BC = 2 m; 50 kN horizontal (→) at B. Top member C–D–E: CD = 4 m (hinge at D), DE = 4 m.
• 100 kN downward at 2 m from C; 25 kN/m downward over DE (total 100 kN).
• Inclined member EF goes from E (8, 5) down to hinged support F (6, 0) at ground level, 2 m inside E.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

C--------------------D--------------------E
|                                        /
|                                       //
|                                      /
|                                     //
B                                     /
|                                   //
|                                   /
|                                  //
|                                 /
|                                //
|                                /
|                              //
A                              F

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 26.67 kN ←
• $V_A$ = 41.67 kN ↑
• $H_F$ = 23.33 kN ←
• $V_F$ = 158.33 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V constant (steps at point loads), M linear
• CD: V constant (steps at point loads), M linear
• DE: V linear, M parabolic
• EF: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 433.33 kN·m at E, tension on the right face
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 5 m; 50 kN horizontal (→) at mid-height (2.5 m). Top member B–C–D: BC = 4 m, CD = 9 m, hinge at C, D hinged support at beam level.
• 30 kN at B acts at 30° to the horizontal, to the right and downward; 20 kN/m downward on BC; triangular load on CD from 0 at C to 15 kN/m at D (total 67.5 kN).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B-------------C------------------------------D
|
|
|
|
|
|
|
|
A

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 25.00 kN →
• $V_A$ = 117.50 kN ↑
• $H_D$ = 100.98 kN ←
• $V_D$ = 45.00 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V linear, M parabolic
• CD: V parabolic, M cubic
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 250.00 kN·m at B (on AB), tension on the left face
• V = 0 at 5.20 m from C along CD: M = 77.94 (bottom) kN·m (local maximum)
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• Inclined leg A(0,0) to B(2,10) (length 10.2 m), A hinged; 20 kN horizontal (→) at mid-length (5 m height).
• Beam B–C = 10 m, triangular load 0 at B to 30 kN/m at C (total 150 kN downward). Column CD = 10 m, D fixed.
• A hinged and D fixed with no internal hinge is indeterminate (5 reactions), so hinges are assumed at B and C to make the frame determinate.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

     B-------------------------Co
     |                         |
    |                          |
    |                          |
   ||                          |
   |                           |
  ||                           |
  |                            |
  |                            |
  |                            |
 |                             |
 |                             |
||                             |
A                              D

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 0.00 kN →
• $V_A$ = 50.00 kN ↑
• $H_D$ = 20.00 kN ←
• $V_D$ = 100.00 kN ↑
• $M_D$ = 200.00 kN·m (anticlockwise)

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V parabolic, M cubic
• CD: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 200.00 kN·m at D (on CD), tension on the left face
• V = 0 at 5.77 m from B along BC: M = 192.45 (bottom) kN·m (local maximum)
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• A(0,0) hinged; inclined member A–B–C at 45°: B = (2,2), C = (5,5); 50 kN downward and 60 kN·m clockwise couple at B.
• Beam C–D = 6 m with 5 kN/m downward; column D–E = 4 m with E on a roller (vertical reaction only).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

                   /C------------------------D
                  /                          |
               ///                           |
              /                              |
           ///                               |
          /                                  |
       /B/                                   |
      /                                      |
   ///                                       E
  /
A/

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

• $H_A$ = 0.00 kN →
• $V_A$ = 43.64 kN ↑
• $V_E$ = 36.36 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V constant (steps at point loads), M linear
• CD: V linear, M parabolic
• DE: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 147.27 kN·m at B, tension on the right face

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 6 m; 80 kN horizontal (→) and 30 kN·m clockwise couple at B.
• Beam B–C–D–E: BC = 2 m (50 kN downward at C), CD = 1 m, internal hinge at D, DE = 9 m with triangular load from 0 at D to 15 kN/m at E (67.5 kN).
• Right column EF = 6 m, F hinged, with 10 kN/m horizontal (←) over its full height.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B-------C--D---------------------------------E
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
|                                            |
A                                            F

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 12.19 kN →
• $V_A$ = 31.04 kN ↑
• $H_F$ = 32.19 kN ←
• $V_F$ = 86.46 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V constant (steps at point loads), M linear
• CD: V constant (steps at point loads), M linear
• DE: V parabolic, M cubic
• EF: V linear, M parabolic
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 373.13 kN·m at E, tension on the right face
• Point of contraflexure (M = 0) at 1.39 m from B along BC
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

(i) Simply supported beam (span 4 m + 6 m = 10 m, 10 kN·m couple at 4 m from the left support)

Assume the couple is clockwise (if it is anticlockwise, reverse all signs).

Reactions: $\Sigma F_y = 0 \Rightarrow R_A + R_B = 0$. Moments about A: $R_B(10) - 10 = 0$ (the clockwise couple is balanced by an anticlockwise moment of $R_B$), so

$$R_B = +1\ \text{kN}\ (\uparrow),\qquad R_A = -1\ \text{kN}\ (\text{i.e. }1\ \text{kN}\ \downarrow)$$

Bending moment (sagging +ve):

• Left of the couple ($0 \le x < 4$): $M = R_Ax = -x$, so $M_A = 0$ and $M = -4$ kN·m just left of the couple (hogging).
• The clockwise couple causes a sudden rise of 10 kN·m: just right of the couple $M = -4 + 10 = +6$ kN·m.
• Right of the couple: $M = R_B(10 - x)$, falling linearly to 0 at B ($M = 1\times6 = 6$ at $x = 4$ ✓).

 BMD (kN·m, sagging above the line)
       A ........ +6 |-----.
  0 ----\         .   |      '---- B
         '-----.-4    |            0
   x=0         x=4 (couple)

The shear force is constant, $V = -1$ kN throughout (the couple does not affect it). The BMD is two straight lines with a jump of 10 kN·m at the couple.

(ii) Frame (column 5 m, horizontal member 4 m, 20 kN at the free end)

Assume the column is fixed at the base and the 20 kN load acts downward at the free end of the horizontal member.

• Horizontal member: $M = -20x$ (hogging, tension on top), from 0 at the free end to $-20\times4 = -80$ kN·m at the corner. The shear force is constant, $V = 20$ kN, and $N = 0$.
• Column: the load acts at a horizontal distance of 4 m from the column axis, so it transmits an axial force $N = 20$ kN (compression) and a constant moment $M = 20\times4 = 80$ kN·m along the full height, with tension on the outer face (the same side as the top of the beam). The shear force in the column is zero.
• Reactions at the base: $V = 20$ kN (↑), $H = 0$, $M = 80$ kN·m.

  free end ●───────┐  BMD: beam   0 → -80 (linear)
  20 kN            │       column -80 (constant)
                   │
                   │
                 ▒▒▒ fixed base

Answer: (i) $M = -4$ kN·m (left of couple) and $+6$ kN·m (right of couple), with $R_A = 1$ kN ↓, $R_B = 1$ kN ↑; (ii) beam moment linear from 0 to 80 kN·m (hogging) and a constant 80 kN·m in the column.

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 8 m; 30 kN horizontal (→) at B. Beam B–C–D: BC = CD = 6 m; hinge at C.
• Triangular load on BC from 0 at B to 30 kN/m at C; 20 kN/m on CD (both downward).
• Inclined member DE: horizontal 5 m, vertical 12 m downward, so E = (17, −4) is the hinged support (length 13 m).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B------------C------------D
|                          \
|                          \\
|                            \
|                            \\
|                             \
|                              \\
|                               \
|                               \\
A                                 \
                                  \\
                                   \
                                    \\
                                     E

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 47.25 kN →
• $V_A$ = 93.00 kN ↑
• $H_E$ = 77.25 kN ←
• $V_E$ = 117.00 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V parabolic, M cubic
• CD: V linear, M parabolic
• DE: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 378.00 kN·m at B (on AB), tension on the left face
• V = 0 at 0.15 m from C along CD: M = 0.23 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 0.30 m from C along CD
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• Units: tonne (t), t/m, t·m. A(0,0) hinged; AB = 4 m with 5 t/m horizontal (→) over the full height (20 t).
• Beam B–C–D = 3 m + 3 m with internal hinge at C; triangular load increasing uniformly from 0 at B to 1 t/m at D (total 3 t).
• Right column DE = 3 m, E hinged at (6, 1); 3 t/m horizontal (←) over the top 1 m of DE (3 t).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B-------------------C------------------D
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      |
|                                      E
|
|
A

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 12.57 t ←
• $V_A$ = 3.18 t ↓
• $H_E$ = 4.43 t ←
• $V_E$ = 6.18 t ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V linear, M parabolic
• BC: V parabolic, M cubic
• CD: V parabolic, M cubic
• DE: V linear, M parabolic
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 15.80 t·m at 2.51 m from A on AB, tension on the right face
• V = 0 at 2.51 m from A along AB: M = 15.80 (right face) t·m (local maximum)
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Principle of superposition: the total effect (reaction, force, moment, stress, strain, deflection) of a system of loads on a structure is the algebraic sum of the effects of the loads applied one at a time. It is valid only when the response is linear.

Example

A cantilever of length $L$ carries a point load $P$ at the free end and a UDL $w$ over the whole length. Free-end deflection:

$$\delta = \delta_P + \delta_w = \frac{PL^3}{3EI} + \frac{wL^4}{8EI}$$

and the fixed-end moment is $M = PL + \dfrac{wL^2}{2}$, the sum of the two separate cases.

Limitations

1. The material must obey Hooke's law (stress proportional to strain); it fails once the material yields or for non-linear materials.
2. Deformations must be small, so the geometry (lever arms of the loads) is the same before and after loading. If deflection changes the geometry appreciably (large deflection), the effects are no longer additive.
3. Deflection must not change the action of the loads: for a beam-column (axial load together with lateral load) the axial load produces extra bending $P\delta$, so the effects of axial and lateral loads cannot simply be added.
4. The supports and the structure must not change during loading (no slip, no support yielding, no contact that starts or stops as load increases).
5. It applies to quantities that are linear in load, e.g. deflection and bending moment of a linear structure, but not to strain energy, which depends on the square of the load.

Assumptions (taken from the figure):

• A(0,0) hinged at the base; AB = 4 m, internal hinge at its top B; horizontal triangular load from 20 kN/m at A to 0 at B (→).
• Beam B–C–D–E: BC = 1 m, CD = 1 m, DE = 2 m; 30 kN downward at C; 10 kN/m downward over DE (20 kN); 40 kN at 45° (down and to the right) at E.
• Inclined member EF descends 4 m over 1 m horizontally to the hinged support F = (5, 0).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B-----C------D------------E
|                         |
|                          |
|                          |
|                           |
|                           |
|                            |
|                            |
|                            ||
|                             |
|                             ||
|                              |
|                              ||
A                               F

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 26.67 kN ←
• $V_A$ = 1.64 kN ↓
• $H_F$ = 41.62 kN ←
• $V_F$ = 79.92 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V parabolic, M cubic
• BC: V constant (steps at point loads), M linear
• CD: V constant (steps at point loads), M linear
• DE: V linear, M parabolic
• EF: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 86.55 kN·m at E, tension on the right face
• V = 0 at 1.69 m from A along AB: M = 20.53 (right face) kN·m (local maximum)
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• A(0,0) hinged on the wall; B at x = 4 m, C at x = 6 m (BC = 2 m).
• Wire from the top of the wall (0, 6) to B: direction $(-4, 6)/\sqrt{52}$ from B (tension only).
• 2 kN/m downward on AB; triangular load on BC from 4 kN/m at B to 0 at C (4 kN).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

Support reactions

Moments about A (the horizontal wire component passes through the beam axis): $T_y\times4 = 8\times2 + 4\times4.667 = 34.667$, so $T_y = 8.667$ kN and $T = T_y\sqrt{52}/6$.

• $H_A$ = 5.78 kN →
• $V_A$ = 3.33 kN ↑
• $T$ = 10.42 kN (tension in the wire; components 5.78 kN ← and 8.67 kN ↑ on the beam)

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• Between loads V is constant; under a UDL it varies linearly and under a triangular load parabolically. M is one degree higher than V (linear, parabolic, cubic respectively); V jumps at point loads and reactions, M jumps at couples.
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum positive (sagging) M = 2.78 kN·m at x = 1.67 m
• Maximum negative (hogging) M = -2.67 kN·m at x = 0.00 m
• V = 0 at 1.67 m from A along AB: M = 2.78 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 3.33 m from A along AB

Sketches (sagging +ve above the line)

SFD:

**                                 **
  *****                            | ***
       *****                       |    *****
------------*****----------------------------*******
                 *****             |
                      *****        |
                           *****   |
                                ***|

BMD:

          **********
      ****          ***
   ***                 ***
 **                       **
*---------------------------**-------------*********
                              *        ****
                               **    **
                                 * **
                                  *

Assumed truss: a rectangular truss with two panels; each panel has a top chord, a bottom chord, vertical members at the ends and a pair of crossing diagonals (cross-bracing). So there are 3 joints on top and 3 at the bottom. The left support is a pin (2 reactions) and the right support is a hinge (2 reactions).

Count

• Joints $j = 6$
• Members: top chords 2 + bottom chords 2 + verticals 3 + diagonals 4 = $m = 11$
• Reactions $r = 2 + 2 = 4$

Static determinacy

For a plane truss, the condition for a determinate truss is $m + r = 2j$.

$$m + r = 11 + 4 = 15,\qquad 2j = 12$$

Since $m + r > 2j$, the truss is statically indeterminate, with degree of indeterminacy

$$D_s = (m + r) - 2j = 15 - 12 = 3$$

Split as external and internal: external degree $= r - 3 = 1$ (the pin plus hinge provide 4 reactions against 3 equilibrium equations), and internal degree $= m - (2j - 3) = 11 - 9 = 2$ (one extra diagonal in each panel).

Stability

• Each panel is a braced rectangle (made of triangles by the diagonals), so the truss is internally stable (rigid), with no mechanism.
• The two supports are both pinned, so the reactions are not all parallel or concurrent and the truss cannot move as a rigid body (no translation or rotation).

Hence the truss is stable but statically indeterminate to the third degree. (If only one diagonal were used in each panel, $m = 9$, $m + r = 13 > 12$ and the truss would be indeterminate to the first degree, externally only.)

Assumptions (taken from the figure):

• Left column AB = 2 m, with the internal hinge at its top B (the top-left corner); A(0,6) hinged; the 60 kN horizontal force acts at the support A and is resisted directly by it.
• Beam B–C = 6 m with 10 kN/m downward; right column C–D = 8 m, D(6,0) hinged; 2 kN vertical (down) at 5 m below C on CD.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B-------------------C
|                   |
|                   |
A                   |
                    |
                    |
                    |
                    |
                    |
                    |
                    |
                    |
                    |
                    D

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 60.00 kN ←
• $V_A$ = 30.00 kN ↑
• $H_D$ = 0.00 kN ←
• $V_D$ = 32.00 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V linear, M parabolic
• CD: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 45.00 kN·m at 3.00 m from B on BC, tension on the bottom
• V = 0 at 3.00 m from B along BC: M = 45.00 (bottom) kN·m (local maximum)
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Supports connect a structure to the ground and give the reactions needed for equilibrium. For a plane structure the common types are:

1. Roller support

Allows horizontal movement and rotation but prevents movement perpendicular to the supporting surface.

• Reaction: one force, perpendicular to the surface (normally vertical).

   ────────
     ○
  ▔▔▔▔▔▔▔▔   R_y

2. Hinged (pin) support

Prevents translation in any direction, but allows rotation.

• Reactions: two components, $H$ (horizontal) and $V$ (vertical), or one resultant of unknown magnitude and direction.

    ────────
      ▲   H, V
   ▔▔▔▔▔▔▔

3. Fixed (built-in) support

Prevents translation and rotation.

• Reactions: three: horizontal force $H$, vertical force $V$ and a moment $M$.

  ▒│─────────
  ▒│  H, V, M

4. Link or tie support (cable, wire or short strut)

A member pinned at both ends restrains movement along its own axis.

• Reaction: one force along the axis of the link.

5. Other: spring support and guided (sliding) support

• A guided or sliding support (roller that cannot rotate) gives one force perpendicular to the guide and a moment (two reactions).
• A spring support gives a force proportional to the displacement of the spring.

Assumptions (taken from the figure):

• A(0,0) hinged; AB = 6 m (4 m + 2 m); horizontal triangular load 10 kN/m at A to 0 at 4 m above A (→).
• Beam B–C–D: 8 m with 5 kN/m downward. Right column DE = 6 m, E hinged; 10 kN horizontal (←) at 4 m above E.
• With two hinged supports a hinge is assumed at the crown (mid-span of BD) to make the frame determinate.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B----------------C-----------------D
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
|                                  |
A                                  E

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate.

• $H_A$ = 7.78 kN ←
• $V_A$ = 21.67 kN ↑
• $H_E$ = 2.22 kN ←
• $V_E$ = 18.33 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V parabolic, M cubic
• BC: V linear, M parabolic
• CD: V linear, M parabolic
• DE: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 46.67 kN·m at B (on AB), tension on the left face
• V = 0 at 0.87 m from A along AB: M = 3.26 (right face) kN·m (local maximum)
• Point of contraflexure (M = 0) at 1.84 m from A along AB
• V = 0 at 0.33 m from C along CD: M = 0.28 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 0.67 m from C along CD
• Bending moment is zero at the internal hinge (value read from both sides: 0.00).

Assumptions (taken from the figure):

• A(0,0) hinged at the foot of a 2 m column; 2 kN horizontal (→) at 1 m above A. Beam from the column top: 4 m span to the roller B, then a 1.5 m overhang with 5 kN downward at the free end.
• Load on the 4 m span: triangular, symmetrical, peak 6 kN/m at 2 m from the column (total 12 kN).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

B--------------------------------C-----------D
|
|
|
|
|
|
|
A

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

• $H_A$ = 2.00 kN ←
• $V_A$ = 3.62 kN ↑
• $R_B$ = 13.38 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• AB: V constant (steps at point loads), M linear
• BC: V parabolic, M cubic
• CD: V constant (steps at point loads), M linear
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum bending moment = 7.50 kN·m at C (on BC), tension on the top
• V = 0 at 1.55 m from B along BC: M = 5.76 (bottom) kN·m (local maximum)
• Point of contraflexure (M = 0) at 3.05 m from B along BC

Assumptions (taken from the figure):

• Beam x = 0 to 7 m: hinge A at x = 0, roller at x = 4 m. A 1 m post at x = 1.5 m carries 2 kN horizontal (→) at its top, equivalent to 2 kN axial force on the beam plus a clockwise couple of 2 kN·m.
• 5 kN downward at x = 2.5 m; triangular load 0 at x = 4 m to 1 kN/m at x = 7 m (1.5 kN); clockwise couple 1 kN·m at the free end x = 7 m.

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

Support reactions

Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$.

• $H_A$ = 2.00 kN ←
• $V_A$ = 0.38 kN ↑
• $R_B$ = 6.12 kN ↑

Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$).

Axial force, shear force and bending moment at the key sections

Shape of the diagrams

• Between loads V is constant; under a UDL it varies linearly and under a triangular load parabolically. M is one degree higher than V (linear, parabolic, cubic respectively); V jumps at point loads and reactions, M jumps at couples.
• AFD: constant between loads on a member, changing only where a load component acts along the member.
• BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

• Maximum positive (sagging) M = 2.94 kN·m at x = 2.50 m
• Maximum negative (hogging) M = -4.00 kN·m at x = 4.00 m
• Point of contraflexure (M = 0) at 3.14 m from A along AB

Sketches (sagging +ve above the line)

SFD:

                              *****************
*******************----------------------------*****
                   |          |
                   |          |
                   |          |
                   ***********|

BMD:

           ********
           |       **
          *|         **
**********-------------*----------------------------
                        **                  ********
                          *           ******
                           **   ******
                             ***