Member Fixed-End Moments (Magnitudes)

Structural Analysis · FE Reference Handbook section

Structural Analysis

2 formulas

10 exam-style examples

~49 min

All Structural Analysis lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes)

A simply supported beam spans 26 ft under a uniform load of 4.50 kip/ft. Find the maximum shear and moment.

Given

$w = 4.50 kip/ft$
$L = 26 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 1 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 4.50(26)/2 = 58.50 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 4.50(26)²/8 = 380.3 kip·ft

Answer:

V_max = 58.50 kip; M_max = 380.3 kip·ft

Why the other options are there

1,521 kip·ft (cantilever formula used)
117.0 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 2

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (2)

A simply supported beam spans 32 ft under a uniform load of 4.75 kip/ft. Find the maximum shear and moment.

Given

$w = 4.75 kip/ft$
$L = 32 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 2 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (2)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 4.75(32)/2 = 76.00 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 4.75(32)²/8 = 608.0 kip·ft

Answer:

V_max = 76.00 kip; M_max = 608.0 kip·ft

Why the other options are there

2,432 kip·ft (cantilever formula used)
152.0 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 3

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (3)

A simply supported beam spans 22 ft under a uniform load of 4.25 kip/ft. Find the maximum shear and moment.

Given

$w = 4.25 kip/ft$
$L = 22 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 3 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (3)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 4.25(22)/2 = 46.75 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 4.25(22)²/8 = 257.1 kip·ft

Answer:

V_max = 46.75 kip; M_max = 257.1 kip·ft

Why the other options are there

1,029 kip·ft (cantilever formula used)
93.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 4

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (4)

A simply supported beam spans 36 ft under a uniform load of 5.00 kip/ft. Find the maximum shear and moment.

Given

$w = 5.00 kip/ft$
$L = 36 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 4 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (4)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 5.00(36)/2 = 90.00 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 5.00(36)²/8 = 810.0 kip·ft

Answer:

V_max = 90.00 kip; M_max = 810.0 kip·ft

Why the other options are there

3,240 kip·ft (cantilever formula used)
180.0 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 5

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (5)

A simply supported beam spans 18 ft under a uniform load of 4.25 kip/ft. Find the maximum shear and moment.

Given

$w = 4.25 kip/ft$
$L = 18 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 5 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (5)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 4.25(18)/2 = 38.25 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 4.25(18)²/8 = 172.1 kip·ft

Answer:

V_max = 38.25 kip; M_max = 172.1 kip·ft

Why the other options are there

688.5 kip·ft (cantilever formula used)
76.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 6

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (6)

A simply supported beam spans 29 ft under a uniform load of 1.50 kip/ft. Find the maximum shear and moment.

Given

$w = 1.50 kip/ft$
$L = 29 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 6 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (6)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 1.50(29)/2 = 21.75 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 1.50(29)²/8 = 157.7 kip·ft

Answer:

V_max = 21.75 kip; M_max = 157.7 kip·ft

Why the other options are there

630.8 kip·ft (cantilever formula used)
43.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 7

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (7)

A simply supported beam spans 30 ft under a uniform load of 2.75 kip/ft. Find the maximum shear and moment.

Given

$w = 2.75 kip/ft$
$L = 30 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 7 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (7)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 2.75(30)/2 = 41.25 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 2.75(30)²/8 = 309.4 kip·ft

Answer:

V_max = 41.25 kip; M_max = 309.4 kip·ft

Why the other options are there

1,238 kip·ft (cantilever formula used)
82.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 8

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (8)

A simply supported beam spans 43 ft under a uniform load of 2.50 kip/ft. Find the maximum shear and moment.

Given

$w = 2.50 kip/ft$
$L = 43 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 8 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (8)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 2.50(43)/2 = 53.75 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 2.50(43)²/8 = 577.8 kip·ft

Answer:

V_max = 53.75 kip; M_max = 577.8 kip·ft

Why the other options are there

2,311 kip·ft (cantilever formula used)
107.5 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 9

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (9)

A simply supported beam spans 19 ft under a uniform load of 2.50 kip/ft. Find the maximum shear and moment.

Given

$w = 2.50 kip/ft$
$L = 19 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 9 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (9)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 2.50(19)/2 = 23.75 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 2.50(19)²/8 = 112.8 kip·ft

Answer:

V_max = 23.75 kip; M_max = 112.8 kip·ft

Why the other options are there

451.3 kip·ft (cantilever formula used)
47.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 10

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (10)

A simply supported beam spans 36 ft under a uniform load of 5.00 kip/ft. Find the maximum shear and moment.

Given

$w = 5.00 kip/ft$
$L = 36 ft$

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure 10 — schematic for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (10)

Step-by-step solution

Reactions
$R = wL/2$
Substituting
$R = 5.00(36)/2 = 90.00 kip = V_max$
Midspan moment
$M = wL^{2}/8$
Substituting — M = 5.00(36)²/8 = 810.0 kip·ft

Answer:

V_max = 90.00 kip; M_max = 810.0 kip·ft

Why the other options are there

3,240 kip·ft (cantilever formula used)
180.0 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)